TS Inter 2nd Year

Telangana TSBIE TS Inter 2nd Year Botany Study Material 9th Lesson Principles of Inheritance and Variation Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 9th Lesson Principles of Inheritance and Variation

Very Short Answer Type Questions

Question 1.
What is the cross between the Fa progeny and the homozygous recessive parent called? How is it useful?
Answer:

1. The cross between the Ft progeny -having dominant phenotype and the homozygous recessive parent is called Test cross.
2. The progenies of such a cross can be easily analysed and the genotype of the test organism can be determined.

Question 2.
Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome?
Answer:

1. Yes, Mendel’s law of independent assortment would not be true for the genes that are located on the same chromosome.
2. Linkage refers to the physical association of genes on a chromosome and are called linked genes.

Question 3.
Who proposed the Chromosome theory of Inheritance? [Mar. 2019, 17]
Answer:

1. Walter Sutton and Theodore Boveri.
2. They used chromosomal segregation during meiosis to explain Mendel’s laws.

Question 4.
Define true breeding. Mention its significance.
Answer:

1. A true breeding line is one that has undergone continuous self pollination.
2. It shows the stable trait inheritance and expression for several generations.

Question 5.
Explain the terms phenotype and genotype. [May ’17, Mar. ’14]
Answer:

1. The physical or external appearance of a character (trait) is called phenotype.
2. The genetic makeup of an individual is called genotype.

Question 6.
What is point mutation? Give an example. [Mar. ’18, May ’14]
Answer:

1. The mutation that occurs in a single base pair of DNA molecule is called gene mutation or point mutation.
2. A classical example for point mutation is sickle cell anemia.

Question 7.
A person has to perform crosses for the purpose of studying inheritance of a few traits / characters. What should be the criteria for selecting the organisms?
Answer:

1. Organism must have well defined characteristics, can be grown and crossed easily.
2. It possess bisexual flowers and can be self fertilized conveniently.
3. It must have short life cycle and produce large number of offsprings.

Question 8.
In order to obtain the F₁ generation, Mendel pollinated a pure-breeding tall plant with a pure breeding dwarf plant. But, to get the F₂ generation, he simply self-pollinated the tall F₁ plants. Why?
Answer:

1. Pure breedingtall and dwarf plants are homozygous and give the same on self pollination. So he bred pure tall with pure dwarf to produce a hybrid.
2. Mendel selfed the F₁ to understand the inheritance of tall and dwarf characters and to know fate of suppressed trait in F₂.

Question 9.
How are alleles of a particular gene differ from each other? Explain its significance.
Answer:

1. Alleles are slightly different forms of the same gene ie., they differ by a single base pair.
2. They are significant in understanding inheritance and behaviour of the genetic polymorphs.

Question 10.
In a monohybrid cross between red and white flowered plants, a geneticist got only red flowered plants. On self-pollinating these Ft plants, he got both red and white flow¬ered plants in 3 :1 ratio. Explain the basis of using RR and rr symbols to represent the genotype of plants of parental generation.
Answer:

1. Red flowered plant crossed with white flowered plant gave red flowered F₁ plant and hence red is dominant over white.
2. In general, first letter of the dominant allele (Red) is used as symbol (R) to denote the genotype and its respective recesssive allele (white) is denoted by small letter (r).

Question 11.
What is the genetic nature of wrinkled phenotype of pea seeds? [Mar. 2020]
Answer:

1. Wrinkled character of seed in pea is a recessive tract.
2. Hence genotype of wrinkled phenotype for pea seeds is ‘rr’

Short Answer Type Questions

Question 1.
In a Mendelian monohybrid cross, the F₂ generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.
Answer:
Incomplete dominance :
It is the phenomenon in which neither of the genes is completely dominant or completely recessive. As a result, the hybrid shows intermediate character. For example, the inheritance of flower colour in the dog flower. (Snapdragon or Antirrhinum sp). The cross between true breeding (homozygous) red flower (RR) and true breeding (homozygous) white flower plant (rr) F₁ (R r) was pink.

The phenotypic ratio deviates from Mendelian monohybrid ratio of 3 : 1 to 1 : 2 : 1 (Red flower – 1, Pink flowers – 2, White flower -1) since the heterozygous / hybrid shows a different phenotype genotype ratio remains the same as Mendelian ratio 1 : 2 : 1.

Question 2.
Mention the advantages of selecting pea plant for experiment by Mendel. [Mar. 2018, ’17 ’14]
Answer:

1. It is an annual plant that has well defined characteristics.
2. It can be grown and crossed easily.
3. It has bisexual flowers containing both female and male parts.
4. It can be self fertilized coveniently.
5. It has a short life cycle and produces large number of offsprings.

Question 3.
Differentiate between the following :
a) Dominant and Recessive [Mar. 2020]
b) Homozygous and Heterozygous [Mar. 2020]
c) Monohybrid and Dihybrid.
Answer:
a) The character which is expressed in F₁ generation is called dominant and that which is unexpressed is called recessive.

b) Parents carrying similar genes such as TT or tt are called homozygous and the parents with unlike genes like Tt are called heterozygous.

c) Cross involving two parents differing in only one character is called Monohybrid cross. For example, cross between tall (TT) and dwarf (tt) parents. Cross involving two parents differing in two characters is called a Dihybrid cross. For example, cross between pea plant having yellow round seeds (YYRR) with homozygous pea plant having green wrinkled seeds (yyrr).

Question 4.
Explain the Law of Dominance using a monohybrid cross. [May 2017]
Answer:
Mendel observed that in F₁ hybrids the character of tallness dominated or supress the dwarf character. The character which is expressed in F₁ generation is called dominant trait and that which remained unexpressed is called recessive trait.

Based on his observation on monohybrid crosses, Mendel proposed Law of Dominance.

Law of Dominance :

1. Characters are controlled by discrete units called factors.
2. Factors occur in pairs.
3. In a dissimilar pair of factors pertaining to a character one member of the pair dominates (dominant) the other (recessive).

The Law of Dominance is used to explain the expression of only one of the parental characters in a monohybrid cross in the F₁ and the expression of both in the F₂. It also explains the proportion of 3 : 1 obtained at the F₂.

The phenotypic ratio is 3 Tall: 1 Dwarf.
The genotypic ratio is 1 TT, 2Tt, 1 tt.

Question 5.
Define and design a test -cross.
Answer:

1. When the F₁ individuals are crossed with the recessive parent or organism similar in phenotype and genotype to the recessive parent, it is called test cross.
2. Test cross is used to test whether an individual is homozygous (pure) or heterozygous (hybrid).
3. A monohybrid test cross gives a phenotypic ratio of 1 : 1

Monohybrid Test cross

Question 6.
Using a punnet square, work out the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Answer:
Punnett square is a graphical representation to calculate the probability of the offsprings.

1. No recessive individuals are obtained in the progeny.
2. All the plants show phenotypically dominant character.
3. They show a genotypic ratio 1 : 1.
4. This is a test cross.

Question 7.
When a cross is made between tall plant with yellow seeds (It Yy) and a tall plant with green seeds (Tt yy). What proportion of phenotype in the offspring is expected to be
a) tall and green?
b) dwarf and green?
Answer:

a) Tall and green – 3/S
b) Dwarf and green – 1/8

Question 8.
Explain the following terms with examples. [May 2014]
a) Co-dominance
b) Incomplete dominance
Answer:
a) Co-dominance :
It is the phenomenon In which both the genes are equally dominant and so the character of both genes is well expressed in next generation. So in F1 generation regeneration resemble both parents.

Examples are different types of red blood cells that determine ABO blood grouping in human beings and seed coat pattern and size in lentil plants.

Lentil is a major grain legume (pulse) crop in N. America, A cross between pure-breed- ing spotted (having a few big irregular pathes) lentils and pure breeding dotted (having several small circular dots) lentils produce heterozygotes that are both spotted and dotted. Thus it shows the phenotypic features of both parents, which means that neither the spotted nor the dotted allele is dominant or recessive to the other.

b) Incomplete dominance :
It is the phenomenon in which neither of the genes is completely dominant or completely recessive. As a result the hybrid shows intermediate character. For example, the inheritance of flower colour in the dog flower. (Snapdragon or Antirrhinum sp). The cross between true breeding (homozygous) red flower (RR) and true breeding (homozygous) white flower plant (rr) F₁ (R r) was pink.

The phenotypic ratio deviates from Mendelian monohybrid ratio of 3 : 1 to 1 : 2 : 1 (Red flower -1, Pink flowers – 2, White flower -1) since the heterozygous / hybrid shows a different phenotype genotype ratio remains the same as Mendelian ratio 1 : 2 : 1.

Question 9.
Write a brief note on chromosomal mutations and gene mutations.
Answer:
Chromosomal Mutation :
Any change in structure and No. of chromosome is called chromosomal mutation.

1. Loss or gain of a segment of DNA, results in alternation in chromosomes. Because genes are located on chromosomes, alteration in chromosomes results in abnormalities.
2. Chromosomal alternations are common in cancer cells.

Gene Mutation :
Mutation that occurs due to change in a single base pair of DNA is called Gene Mutation or point mutation. E.g.: Sickle cell anaemia. Deletion and insertion of base pairs of DNA, Cancer from shift mutation.

Question 10.
How was it concluded that genes are located on chromosomes?
Answer:
Morgar hybridised yellow bodied and white eyed females to brown bodied, red eyed males and intercrossed their F₁ progeny. He observed that the two genes did not segregate independently with each other and the F₂ ratio deviated very significantly from the 9 : 3 : 3 : 1 ratio. He saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental combinations was much higher than the non – parental type. He attributed this due to the physical association or linkage of the two genes and coined the term linkage and the term recombination to describe the generation of non parental gene combinations. He also found that, even when genes were grouped on the same chromosome, some genes were tightly linked and some genes were loosely linked.

Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of distance between genes and mapped their position on the chromosome.

Question 11.
A plant with red flowers was crossed with one having yellow flowers. If F₁ showed all flowers in orange colour, explain the inheritance.
Answer:

The cross between red flower and yellow flower, showed all orange colour flowers in F₁ generation. This shows that neither of the genes is completely dominant or completely recessive. As a result, the hybrid shows intermediate character. This is incomplete dominance.

Question 12.
Define Law of Segregation and Law of Independent Assortment.
Answer:
Law of Segregation or law of purity of gametes :
Based on the results obtained from the monohybrid cross Mendel postulated the first law “the law of segregation or law of purity of gametes.” It states that the two alleles of a gene when present together in a heterozygous state, do not fuse or blend in any way, but remain distinct and segregate during meiosis or in the formation of gametes so that each meiotic product or gamete will carry only one of them.

Law of Independent Assortment :
Based on the results of the dihybrid crosses in pea plant, Mendel postulated his second law known as “the law of Independent Assortment.” Which states that “when the plants differ from each other in two or more pairs of contrasting characters or factors, then the inheritance of one pair of factors is independent to that of the other pair of factors.

Question 13.
In peas, tallness is dominant over dwarfness and violet colour of flowers is dominant over the white colour. When a tall plant bearing violet flowers was pollinated with a dwarf plant bearing white flowers, different phenotypic groups were obtained in the progeny in numbers mentioned against them.
Tall, violet = 138 Dwarf, violet = 136
Tall, white = 132 Dwarf, white = 128
Mention the genotypes of the two parents and of the four offspring types.
Answer:
When a tall plant bearing violet flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were in ratio nearly 1: 1:1: 1. This is Dihybrid test cross ratio. The genotype of two parents Tt Vv (Tall violet) and tt vv (Dwarf white) 

Question 14.
How do genes and chromosomes share similarity from the view point of genetical studies?
Answer:
Chromosomes are made up of DNA and proteins DNA is a special kind of molecules made up of nucleotides. There are four types of nucleotides. A DNA molecule consists of a long string of nucleotide pairs linked together. The sequence of nucleotide forms a code gene refer to a specific sequence of this DNA code that actually means characters all genes are DNA but not all the DNA are the genes.

Chromosomes are DNA molecule made up of a sequence of nucleotides. Genes are stretches of DNA that contain code to make protein. Thus both chromosome and genes are similar from the point of genetic studies.

Question 15.
With the help of an example differentiate between incomplete dominance and com-plete dominance.
Answer:
Incomplete dominance :
It is the phenomenon in which neither of the two alleles of a gene is completely dominant over the other.

1. The inheritance of flower colour in dog flower / snap dragon (Antirrhinum majus) and that in Mirabilis jalapa (4 o’ clock plant) are examples of this phenomenon.
2. When a cross is made between a red flowered plant with a white flowered plant of snap dragon the F₁ hybrid has pink flowers.
3. When the F₁ individual was self pollinated and an F₂ raised the F₂ contained individuals bearing red, pink and white flowers.

The phenotypic and genotypic ratio are same i.e., 1 Red : 2 Pink : 1 White.

Complete dominance is a phenomenon in which one allele of a gene expresses itself and suppresses the expression of the other allele of the same, when they are present to-gether in a hybrid.

1. When a cross was made between two individuals, one with tall stem (homozygous) and the other with dwarf stem, F₁ individual had tall stem.
2. When a F₁ individual is self pollinated the F₂ produced tall and dwarf individuals in the ratio of 3 : 1

The phenotypic ratio is 3 Tall : 1 Dwarf.
The genotypic ratio is 1 TT : 2 T t: 111.

Long Answer Type Questions

Question 1.
In a plant, tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents workout a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?
Answer:
In a plant tallness is dominant over dwarfness and red flower is dominant over the white cross between two parents with one parent, Tall with red flowers (TT RR) and other parent, Dwarf with white flowers (tt rr). Such a cross involving two parental plants differing in two character is called dihybrid cross.

When the F₁ hybrids were allowed to undergo self pollination, the F₂ generation progeny was showing four kinds of phenotypes in a definite pattern.

The phenotypic ratio is 9 : 3 : 3 : 1 and the genotypic ratio is 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.

The F₂ progeny showed not only the parental combination i.e., Tall with red flowers and Dwarf with white flowers but also the new combination i.e., Tall white and Dwarf red.

These new combination are produced due to genetic recombination.

The F₂ progeny are Tall purple 9 ; Tall white 3 ; Dwarf purple : 3; Dwarf white 1.
The dihybrid ratio is 9 : 3 : 3 : 1.
Yes, the values would deviate if the two genes interact with each other.

Intext Question Answers

Question 1.
What will be the phenotypic ratio in the offsprings obtained from the following crosses,
a) Aa × aa b) AA × aa c) Aa × Aa d) Aa × AA
Note : Gene ‘A’ is dominant over gene ‘a’.
Answer:
a) Aa × aa – P generation

The progeny shows 1 : 1 ratio.

b) AA × aa – P generation

The progeny shows Aa hybrid.

c) Aa x Aa – P generation

The progeny shows 1 : 2 : 1, 1 AA : 2 Aa : 1 aa genotype, 3 : 1 phenotype.
d) Aa × AA – generation

All the progeny shows dominant character.

Question 2.
In garden pea, the gene T for tall is dominant over its allele for dwarf. Give the geno-types of the parents in the following crosses.
a) tall × dwarf producing all tall plants.
b) tall × tall producing 3 tall and 1 dwarf plants.
c) tall × dwarf producing half tall and half dwarf number of plants.
Answer:
a) TT × tt → Tt all tall plants.
b) Tt × Tt → 3 tall and 1 dwarf plant.
c) Tt × tt → 1 tall (T t) and 1 dwarf (t t)

Question 3.
Mendel crossed pea plants producing round seeds with those producing wrinkled seeds. From a total of 7324 F₂ seeds, 5474 were round and 1850 were wrinkled. Using the symbols R and r for genes, predict the :
a) the parental (p) genotypes
b) the gametes
c) F₂ progeny
d) the cross between F hybrids
e) genotypes, phenotypes, genotypic frequency, phenotypic ratio of F₂ progeny.
Answer:
a) The parental genotypes are RR, rr (homozygous).
b) The gametes are (R) and (r).
c) F₁ progeny gene R r.
d) Cross between F₁ hybrids are R r × R r.
e) Genotypes are 3
Phenotypes are 2
Genotypic frequency 1 : 2 : 1
Phenotypic ratio is 3 : 1

Question 4.
The following data was obtained from an experiment on peas. The grey coloured seed is dominant over white coloured seed. Use the letter G for grey and g for white traits. Predict genotypes of the parents in each of the following crosses.

Answer:
a) Genotypes of the
Grey × white
Gg × gg

b) Grey × grey
Gg × Gg

c) White × white
gg × gg

d) Grey × grey
Gg × GG

Question 5.
In tomatoes, red fruit colour (R) is dominant to yellow (r). Suppose a tomato plant homozygous for red is crossed with one homozygous for yellow. Determine the appearance of the following.
a) the F₁ b) F₂, c) the offspring of a cross of the F₁ back to the red parent
d) the offspring of a cross of the F₁ back to the yellow parent.
Answer:
a) The appearance in F₁ are all red (Rr).

b) The appearance in F₂ are 3 : 1 (3 red; 1 yellow).

c) The appearance of the offspring of a cross of the F₂ back to the red parent are all red (homozygous red and heterozygous red). It is back cross.

d) The appearance of the offspirng of a cross of the F₁ back to the yellow parent are red and yellow in 1 : 1 ratio. It is test cross.

Question 6.
In pea, axillary position of flowers (T) is dominant over its terminal position (t). Coloured flowers (C) are dominant to white flowers (c). A true breeding plant with coloured flowers in axils is crossed to one with white terminal flowers. Give the phenotypes, genotypes and expected ratios of F₂, F₂ back cross and test cross progenies. What genotypic ratio is expected in the F₂ progeny?
Answer:
Axillary coloured flowers × Terminal white flowers – P

All the F₁ hybrids phenotypes are found to have Axillary coloured flowers.
All the F₁ hybrids genotypes are T t Cc.
The phenotype ratio of F₂ are 9 : 3 : 3 : 1.
The genotype ratio of F₂ are 1 : 2 : 2 :4 : 1 : 2 : 1 : 2 : 1.
Dihybrid test cross ratio is 1 : 1 : 1 : 1

In back cross when F₂ individuals are crossed with the parent having dominant traits no recessive individuals are obtained. All are Axillary coloured flowers.

Question 7.
In summer squash, a plant with white flowers and disc-shaped fruits is crossed to a plant with yellow flowers and sphere shaped fruits. The F₁ hybrids had white flowers and discshaped fruits. Which phenotypes are dominant? Give the genotypes of the parents and the hybrids. If these hybrids are selfed and 256 progeny are obtained. What would be the frequencies of the various phenotypes?
Answer:
The dominant phenotypes are white flowers with disc shaped fruits.
The dominant white flower (W) and recessive yellow flower (w).
The dominant disc shaped fruit (D) and recessive sphere shaped fruit (d).
The genotype of parents WWDD (white flowered disc shaped fruits) and wwdd (yellow flower and sphere shaped fruits).
The genotype of hybrid yellow flower and sphere shaped fruits is WwDd.
Phenotype frequency will be 9 : B : 3 : 1 ratio.

Question 8.
Give the ratios of the following:
a) Monohybrid test cross
b) Dihybrid test cross
c) F₂ phenotypic ratio of monohybrid cross
d) F₂ phenotypic ratio of dihybrid cross
e) F₂ Genotypic ratio of monohybrid cross and
f) F₂ genotypic ratio of dihybrid cross.
Answer:
a) Monohybrid test cross ratio 1 : 1
b) Dihybrid test cross ratio 1 : 1 : 1 : 1
c) F₂ phenotypic ratio of monohybrid cross 3 : 1
d) F₂ phenotypic ratio of dihybrid cross 9 : 3 : 3 : 1
e) F₂ genotypic ratio of monohybrid cross 1 : 2 : 1
f) F₂ genotypic ratio of dihybrid cross 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

Question 9.
A diploid organism is heterozygous for 4 loci. How many types of gametes can it produce?
Answer:
A diploid organism is heterozygous for 4 loci. It can produce sixteen types of gametes.

Question 10.
What is crossing over? In which stage of cell division crossing over occurs? What is its significance?
Answer:
The exchange of chromatids between non-sister chromatids leading to genetic recombination is known as crossing over. Crossing over occurs during pachytene sub stage of meiosis. Crossing over significance is

1. It is the cause of variation and genetic recombination.
2. It is essential for natural selection and evolution.

Question 11.
“Genes contain the information that is required to express a particular trait”. Explain.
Answer:
Mendel conducted experiments like monohyrid cross. He crossed tall and dwarf plants to study the inheritance of gene. He observed that only one of the parent traits was expressed in the F₁ generation while at the F₂ stage both the traits were expressed in the proportion 3 : 1. The constracting characters traits did not show any blending at.either F₁ or F₂ stage.

Based on these observations, Mendel proposed that something was being stably passed down, unchanged from parents to offspring through the gametes over successive generation. He called them as factors. Now a days we call them genes. Genes therefore are the units of inheritance. They contain the information that is required to express a particular trait in an organism.

Question 12.
For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.
Answer:

1. Two individuals with the same genotype may become different in phenotype when they come into contact with different environmental conditions like temperature, light, humidity are referred to as environmental variation or modifications.
2. The phenotype of any organism is necessarily a result of the interaction of a genotype with an environment, both are absolutely necessary.
3. Inbred lines are obtained in which genotype is uniform. One can measure the effect of environmental factors, such as amount and kind of fertilizer and type of soil in crop plants by growing members of an inbred line under different environmental conditions. They are different phenotypically.
4. For example, sweet potatoes grown in soil rich in potassium are round and fleshy but when this element is scare they became long and spindling.
5. Other example is the effect of temperature a hair colour in the Himalayan rabbit. The white hair on a small area of the back of this rabbit was pulled out and the animal was put in a cold room. The hair that grew in under cold condition was black. Hair is this region which grows under warm condition is white.

Then it proves that for the expression of traits gene provide only the potentiality and the environment provides the opportunity.

Question 13.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in the F₁ generation for a dihybrid cross?
Answer:
Linkage is defined as the coexistence of two or more genes are situated on the same chromosome and lie close to each other, then they are inherited together and are said to be linked genes. For example, a cross between yellow body and white eyes and wild type parent in a Drosophila will produce wild type and yellow white prggenies. It is because yellow bodied and white eyed genes are linked. Therefore, they are inherited together in progenies.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 8th Lesson Viruses Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 8th Lesson Viruses

Very Short Answer Type Questions

Question 1.
Mention the living and non-living characters of viruses.
Answer:
Living characters:

1. They contain nucleic acid.
2. They maintain genetic continuity through multiplication and undergo mutations.
3. The live as obligate intracellular parasites.

Non-living characters:

1. They do not exhibit most of the life processes like growth, irritability.
2. They are inert life less molecules.
3. They are acellular.
4. They do not have metabolic system.

Question 2.
What is the shape of T₄ phage? What is its genetic material?
Answer:

1. T₄ phage is tadpole shaped with a large head and a tail.
2. Genetic material is double stranded DNA.

Question 3.
What are virulent phages? Give an example.
Answer:

1. The viruses that attack the bacterium E.coli, causes lysis of the cells are called virulent phages.
2. Eg: T – even phages.

Question 4.
What is lysozyme and what is its function?
Answer:

1. The viral enzyme which dissolves the plasma membrane of the host cell (bacterial cell) is called lysozyme.
2. Lysozyme is synthesized within the cell and the bacterial cell wall breaks releasing the newly produced phage particles / virions.

Question 5.
Define ‘lysis’ and ‘burst size’ with reference to viruses and their effects on host cells.
Answer:
1. Lysis :
It is the final stage of lytic cycle, the host cell wall bursts during this phase and all the newly produced virions are released. This is known as lysis.

2. Burst size :
The number of newly synthesized phage particles released from a single host cell.

Question 6.
What is a prophage?
Answer:

1. Prophase : It is the phage DNA that is incorporated into bacterial DNA and remains latent during lysogenic cycle.
2. It is found in temperate phages and the prophase also undergoes replication.

Question 7.
What are temperate phages? Give one example.
Answer:

1. Temperate phage: A bacteriophage whose DNA is incorporated into the host DNA to form prophase during lysogenic cycle.
2. It does not cause immediate lysis and death of host, when they multiply Eg: Coliphase-λ.

Question 8.
Mention the differences between virulent phages and temperate phages.
Answer:
1. Virulent phages :
Bacteriophages that cause lysis of host at the end of replication (lytic cycle) Eg: T-even phages.

2. Temperate phages :
Bacteriophages, whose DNA is incorporated into host DNA to form prophases (lysogenic cycle) and do not cause immediate lysis of host. Eg: Coliphase-λ,.

Short Answer Type Questions

Question 1.
What is ICTV? How are viruses named?
Answer:
ICTV means International Committee on Taxonomy of Viruses. It regulates the norms of classification and nomenclature of viruses.

The ICTV scheme has only three hierarchial levels
– Family (including some sub – families)
– Genus
– Species

The family names end with the suffix ‘viridae’ while the genus names with ‘virus’ and the species names are common English expressions describing their nature.

Viruses are named after the disease they cause. Eg : Polio virus

Using the ICTV system, the virus that causes Acquired Immune Deficiency Syndrome (AIDS) in human beings is classified as :
Family : Retroviridae Genus : Lentivirus
Species : Human Immune Deficiency Virus (HIV).

Question 2.
Explain the chemical structure of viruses. [Mar. 2020]
Answer:

1. All viruses consists of two basic components 1) core 2) capsid.
2. Core is the nucleic acid that forms the genome. Capsid is the surrounding protein coat.
3. Capsid gives shape and protection. It is made up of protein subunits called capsomeres. The no. of capsomeres is characteristic for each type of virus.
4. Virus contains its genetic information in either double stranded (ds) DNA or single stranded (ss) DNA.
5. Generally, viruses that infect plants have single stranded RNA and viruses that infect animals have double stranded DNA.
6. Bacteriophages are usually ds DNA.
7. Viral nucleic acid molecules are either circular or linear.
8. Most viruses have a single nucleic acid molecule, but a few have more than one (Eg : HIV which has two identical molecules of RNA).

Question 3.
Write briefly about the symmetry of viruses.
Answer:
Helical virus are helical symmetry. They resemble long rods. Eg: RSbies virus and Tobacco mosaic virus.

Polyhedral symmetry :
Many plants and animals has polyhedral symmetry (many sides). Eg : Herpes simplex and polio viruses.

Binal symmetry :
Bacteriophage have both polyhedral symmetry in the head and helical symmetry in the tail sheath.
Spherical virus are enveloped virus. Eg : Influenza virus.

Question 4.
Explain the structure of TMV. [Mar. ’18 ’14; May ’17]
Answer:

1. TMV was the first virus to be crystallized by Stanley.
2. Fraenkel described the structure of TMV.
3. Tobacco Mosaic Virus is a rod shaped virus. It is about 300 nm long and 18 nm in diameter, with a molecular weight of 39 x 106 Daltons.
4. The capsid is made up of 2,130 subunits called capsomeres.
5. Capsomeres are arranged in a helical manner around a central core of 4 nm.
6. Each protein subunit is made up of a single polypeptide chain with 158 amino acids.
7. Single stranded RNA is present inside the capsid and is also spirally coiled.
8. RNA of TMV consists of 6,500 nucleotides.

Question 5.
Explain the structure of T – even bacteriophages. [Mar. 17, May’14]
Answer:

1. The body of T-even bacteriophage can be distinguished into head and tail regions joined by a collar.
2. The tail region includes a tail sheath, a base plate, pins and tail fibres which help the virus attach to host cell.
3. The tail sheath aids in injecting viral DNA into the host cell.

Question 6.
Explain the lytic cycle with reference to certain viruses. [Mar. 2019]
Answer:
T – even phages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle. It involves 5 step process. They are 1. attachment 2. penetration 3. biosynthesis 4. maturation and 5. release.

1. Attachment:

1. Contact of the virion to the surface of host bacterium is called attachment or adsorption.
2. The phages use tail fibres for attachment to the complementary receptor sites on the bacterial cell wall.

2. Penetration :

1. The injection of phage nucleic acid into the host cell is called penetration.
2. The phage DNA is injected into the bacterium through the tail core like a hypodermal syringe.
3. The capsid remains outside the bacterial cell and is referred to as ghost.

3. Biosynthesis :

1. Once the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes and capsid proteins are synthesized, using the cellular machinery of the host cell.
2. Host cell do not contain any complete infective viruses.

4. Maturation :

1. In this process bacteriophage DNA and capsids are assembled into complete virions.
2. This period of time between the infection by a virus and the appearance of the mature virus within the cell is called the eclipse period.

5. Release :

1. The final stage of viral multiplication is the lysis phase of the host cell and the release of virions from the host cell.
2. The plasma membrane of the host cell gets dissolved or lysed due to viral enzyme called lysozyme.
3. The bacterial cell wall breaks releasing the newly produced phage particles /virions.

Question 7.
Explain how temperate phages play a role in transduction.
Answer:

1. In lysogenic cycle, some bacteriophages such as X (Lambda) phages do not cause lysis
   during multiplication.
2. Instead, the phage DNA upon penetration into the E.coli gets integrated in to the circular bacterial DNA, becomes part of it and remains latent (inactive). Such phages are called temperate phages. This inserted phage DNA is now called prophage.
3. Every time the bacterial genetic material replicates, the prophage also gets replication. The prophage remains latent in the progeny cells.
4. However, rarely prophage gets disintegrated when they are exposed to UV light or some chemicals and enter into lytic cycle.
5. Thus temperate phase plays a role of transduction by the transfer of genetic material from one bacterium to another through bacteriophage.

Question 8.
Mention the differences between lytic and lysogenic cycles.
Answer:

Long Answer Type Questions

Question 1.
Write about the discovery and structural organization of viruses.
Answer:

1. Viruses have been causing diseases in humans, animals and plants from the ancient time.
2. ‘Germ theory of disease’ was put forth by Louis Pasteur but the agent responsible for the diseases are not known.
3. In 1892 for the first time, the Russian pathologist Dmitri Iwanowski, while studying tobacco mosaic disease, filtered the “sap of diseased tobacco leaf” through filter which was designed to retain bacteria. However the infectious agent passed through the pores of the filter. After injecting the filtered sap into the healthy plant, he found the development of symptoms of mosaic disease in it. Unable to see any microorganism in a sap, he reported that the filterable agent was responsible for the disease.
4. Martinus Beijerinck repeated Iwanowski’s experiments and concluded that the disease causing agent was a contagious living fluid (contagium vivum fluidum).
5. W.M. Stanley (1935) purified the sap and announced that the virus causing mosaic disease in tobacco could be crystallized. It was named as Tobacco Mosaic Virus (TMV).
6. Fraenkel Conrat (1956) confirmed that the genetic material of the TMV is RNA.

Question 2.
Describe the process of multiplication of viruses.
Answer:
The process of multiplication of viruses is done by two alternative mechanisms.
a) Lytic cycle b) Lysogenic cycle

a) Lytic cycle :
T – even phages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle. It involves 5 step process. They are 1. attachment 2. penetration 3. biosynthesis 4. maturation and 5. release.

1. Attachment:

1. Contact of the virion to the surface of host bacterium is called attachment or adsorption.
2. The phages use tail fibres for attachment to the complementary receptor sites on the bacterial cell wall.

2. Penetration :

1. The injection of phage nucleic acid into the host cell is called penetration.
2. The phage DNA is injected into the bacterium through the tail core like a hypodermal syringe.
3. The capsid remains outside the bacterial cell and is referred to as ghost.

3. Biosynthesis :

1. Once the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes and capsid proteins are synthesized, using the cellular machinery of the host cell.
2. Host cell do not contain any complete infective viruses.

4. Maturation :

1. In this process bacteriophage DNA and capsids are assembled into complete virions.
2. This period of time between the infection by a virus and the appearance of the mature virus within the cell is called the eclipse period.

5. Release :

1. The final stage of viral multiplication is the lysis phase of the host cell and the release of virions from the host cell.
2. The plasma membrane of the host cell gets dissolved or lysed due to the viral enzyme called lysozyme.
3. The bacterial cell wall breaks releasing the newly produced phage particles / virions.

b) Lysogenic cycle :

1. In lysogenic cycle, some bacteriophages such as X (Lambda) phages do not cause lysis during multiplication.
2. Instead the phage DNA upon penetration into the E.coli gets integrated in to the circular bacterial DNA, becomes part of it and remains latent (inactive). Such phages are called temperate phages. This inserted phage DNA is now called prophage.
3. Every time the bacterial genetic material replicates, the prophase also gets replication. The prophage remains latent in the progeny cells.
4. However, rarely prophage gets disintegrated when they are exposed to UV light or some chemicals and enter into lytic cycle.
5. Thus temperate phase plays a role of transduction by the transfer of genetic material from one bacterium to another through bacteriophage.

Intext Question Answers

Question 1.
When discussing the multiplication of viruses, Virologists prefer to call the process as replication, rather than reproduction. Why?
Answer:

1. Viral reproduction requires a living cell to takes place. Virus do not go through mitosis
   or cytogenesis nor do they have any mitotic machinery to produce new virus.
2. Virus cannot reproduce without a host cell or infect whereas a reproductive cell are show independent replication.
3. The cell infected by virus die immediately (lytic cycle) or after sometime (lysogenic cycle). Hence virologist prefer to call the multiplication of viruses as replication rather than reproduction.

Question 2.
In dealing with public health, the approach to deal with bacterial diseases is treatment. Can you guess the nature of the general public health approach to viral diseases ? What example do you cite to support your answer?
Answer:
The most effective medical approaches to viral diseases are vaccinations to provide immunity to infection and antiviral therapy to overcome drug resistance. Antibiotics have no effect on viruses. Ex : AIDS, Viral hepatitis and many more.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 7th Lesson Bacteria Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 7th Lesson Bacteria

Very Short Answer Type Questions

Question 1.
Write briefly on the occurrence of micro-organisms.
Answer:

1. Microorganisms are ubiquitous. They are found in soil, water, air, and inside living beings.
2. They occur in a variety of foods and can withstand extreme cold, heat, and drought conditions.

Question 2.
Define Microbiology.
Answer:

1. Microbiology is a branch of biological science that deals with the study of microorganisms like protozoa, microscopic algae, fungi (yeasts and molds) bacteria and viruses.
2. It is concerned with structure, function, classification, ways to control and using the activities of microorganisms.

Question 3.
Name the bacteria which is a common inhabitant of human intestine. How is it used in biotechnology? [May 2014]
Answer:

1. Escherichia coli (E.coli) is a common inhabitant of human intestine.
2. It is used in Recombinant DNA technology for the production of insulin harmone.

Question 4.
What are pleomorphic bacteria? Give an example. [May ’17, Mar. ’14]
Answer:

1. The bacteria that keep on changing their shape depending upon the type of environment and nutrients available are called pleomorphic bacteria.
2. Eg : Acetobacter.

Question 5.
What is sex pilus? What is its function?
Answer:

1. The process of conjugation requires a special conjugation apparatus called the conjugation tube or pilus or sex pilus.
2. Its function is to transfer F plasmid from F⁺ donor to F^– recipient.

Question 6.
What is a genophore? [Mar. 2019]
Answer:

1. Bacterial chromosome is also called as genophore.
2. It is the main genetic material of bacteria.

Question 7.
What is a plasmid? What is its significance?
Answer:

1. Plasmids : Small circular, double stranded DNA molecules present is addition to the bacterial chromosome (genophore) in Bacteria.
2. Plasmids contain genes, for resistance to drugs, production of toxins and enzymes. These are used as tools (vectors) in modern genetic engineering technique.

Question 8.
What is conjugation? Who discovered it and in which organism? [Mar. 2017]
Answer:

1. Conjugation is a process, in which two live bacteria come together and the donor cell directly transfers DNA to the recipient cell.
2. This process was first observed by Lederberg and Tatum (1946) in Escherichia coli.

Question 9.
What is transformation? Who discovered it and in which organism? [Mar. 2020]
Answer:

1. Transformation is uptake of naked DNA fragments from the surrounding environment and the expression of that genetic information in the recipient cell.
2. Frederick Griffith (1928) discovered it in Streptococcus pneumoniae.

Question 10.
What is transduction? Who discovered it and in which organism? [March 2010]
Answer:

1. The transfer of genetic material from one bacterium to anotherthrough bacteriophage is known as transduction,
2. It was discovered by Lederberg and Zinder (1951) in Salmonella typhimurium.

Short Answer Type Questions

Question 1.
Explain the importance of Microbiology.
Answer:
Microbiology is a branch of biological science that deals with the study of micro-organisms. Micro-organisms are useful in several ways for the welfare of human society.
1. Soil fertility :
Micro-organisms decompose dead plants and animals thereby enrich the soil with nutrients which are utilized by plants. Microbes also play vital role in recycling of elements like C, N, O, S, and P.

2. Antibiotics :
Alexander Fleming isolated an antibiotic Penicillin from a fungus, Penicillium notatum. Waksman isolated an antibiotic Streptomycin from a bacterium, Streptomyces griseus.

3. Industrial products :
Industrial products like enzymes, amino acids, vitamins, organic acids and alcohols are commercially produced using microbes.

4. Dairy products :
Lactobacillus, commonly known as Lactic Acid Bacteria (LAB) grows in milk and convert it to curd, which also improves its natural quality by increasing vitamin B12, food stuff, like cheese, yogurt are the byproducts of microbial growth.

5. Mining :
Metals like Uranium can be extracted with 50% reduced cost by using microbes.

6. Tools in Genetic Engineering :
Microbes are used as tools in altering the genetic make up of organisms.

7. Biocontrol agents :
Micro-organisms like Bacillus thuringiensis are used to control plant diseases and pests.,

8. Production of Biogas :
Biogas, a mixture of gases (predominantly methane) produced by the microbial activity. It may be used as fuel.

9. Exomicrobiology :
Microbes are used for the exploration of life in the outer space.

10. Sewage disposal :
Bacteria and fungi are also used in sewage disposal.

Question 2.
How are bacteria classified on the basis of morphology?
Answer:
Based on their shape bacteria are classified into following types.
1. Cocci :
Spherical bacteria are called cocci. They are 6 types.
a) Monococcus : A single spherical bacterium.
b) Diplococcus : A pair of spherical bacterium.
c) Tetracocci : A group of four spherical bacteria.
d) Streptococcus : A linear chain of spherical bacteria arranged in a single row.
e) Sarcinae : Cocci arranged in cubes of eight.
f) Staphylococci : A group of cocci bacteria forming irregular shapes producing bunches.

2. Bacillus :
Elongated rod shaped bacteria are called bacillus. They are 3 types.
a) Monobacillus : A single elongated rod shaped bacterium.
b) Diplobacillus : Paired cells of bacilli.
c) Streptobacillus : Chains of bacilli appearing like straws.

3. Spiral forms :
Vibrioid : Cells having less than one complete twist.
Spirillum : Cells that have more than one complete twist – a distinct helical shape. Spirochete : Cells are slender, long and cork-screw shaped.

4. Pleomorphic :
These bacteria change their shape depending upon the type of environment and nutrients available. This phenomenon is called pleomorphism.
Eg : Acetobacter.

Question 3.
How are bacteria classified on the basis of number and distribution of flagella?
Answer:
Based on the number and distribution of flagella bacteria are classified into 4 types.

1. Monotrichous : A single flagellum is present on one side of the cell.
2. Lophotrichous : A tuft of flagella are present on one side of the cell.
3. Amphitrichous : Tufts of flagella or a single flagellum on either end of the cell.
4. Peritrichous : Many flagella are distributed all over the cell surface.

Question 4.
What are the nutritional groups of bacteria based on their source of energy and carbon?
Answer:
Four major nutritional groups of bacteria based on the source of energy and carbon are described below.

Question 5.
Write briefly about chemoheterotrophs and their significance.
Answer:
Chemoheterotrophs derive both carbon and energy from organic compounds. Processing these organic molecules by respiration or fermentation releases energy in the form of ATP.

These bacteria are divided into saprophytes and parasites basing on how they obtain their organic compounds.

Saprophytes :
These are free living bacteria which grow on dead, organic matter are called Saprophytes. Eg : Bacillus

Parasites :
The bacteria which grow on or in living host and cause diseases are called parasites.
Eg : Xanthomonas, Salmonella.

Significance :
Most microbes of biomedical importance belong to these two categories. Bdellovibrio bacteriovorous grows as a parasite on some harmful bacteria and their abundance is supposed to be responsible for the microbial purity of Ganges waters.

Question 6.
Explain the conjugation in bacteria.
Answer:
Conjugation :

1. The transfer of genetic material (DNA) through direct cell to cell contact is known as conjugation.
2. It was first reported by Lederberg and Tatum in 1946 in Escherichia coli.
3. In E.coli, in addition to the bacterial chromosome or genophore which is the main genetic material, bacteria contain small circular, double stranded DNA molecules called plasmids or ‘F’ factor.
4. E.coli having F factor are called F⁺ cells or donor cells and the cells without F factor are called F^– cells or acceptor cells.
5. F⁺ cells have pilus or sex pilus. During conjugation the F⁺ and F^– strains come close together. Once contact is established, the pilus shortens to bring the two bacteria close together.
6. A conjugation.tube is established. The plasmid in F⁺ replicates and forms a copy of it, which moves to the acceptor cell (F). Thus donor bacterium generally retain a copy of genetic material that is being transferred.

Long Answer Type Questions

Question 1.
Explain different methods of sexual reproduction in Bacteria.
Answer:
Sexual reproduction :
True sexual reproduction is absent in bacteria. However the exchange of genetic material is reported through other methods.

Three types of genetic recombinations are reported in different species of bacteria. They are

1. Conjugation
2. Transformation and
3. Transduction.

1. Conjugation :

1. The transfer of genetic material (DNA) through direct cell to cell contact is known as conjugation.
2. It was first reported by Lederberg and Tatum in 1946 in Escherichia coli.
3. In E.coli, in addition to the bacterial chromosome or genophore which is the main genetic material, bacteria contain small circular, double stranded DNA molecules called plasmids or ‘F’ factor.
4. E.coli having F factor are called F⁺ cells or donor cells and the cells without F factor are called F^– cells or acceptor cells.
5. F⁺ cells have pilus or sex pilus. During conjugation the F⁺ and F strains come close together. Once contact is established, the pilus shortens to bring the two bacteria close together.
6. A conjugation tube is established, The plasmid in F⁺ replicates and forms a copy of it, which moves to the acceptor cell (F). Thus donor bacterium generally retain a copy of genetic material that is being transferred.

2. Transformation :
Transformation is uptake of naked DNA fragments from the surrounding environment and the expression of that genetic information in the
recipient cell. That is, the recipient cell has now acquired a characteristic that is previously lacked. This mode of bacterial genetic recombination was discovered by Frederick Griffith in streptococcus pneumoniae.

3. Transduction :
The transfer of genetic material from one bacterium to another through bacteriophage is known as transduction.

Question 2.
“Bacteria are friends and foes of man” – discuss.
Answer:
Bacteria are known to be the casual agents of plant, animal and human diseases. At the same time there are many bacteria which are directly or indirectly beneficial to man. Thus, these organisms can be considered both as ‘friends and foes of man’.

Beneficial activities:

1. Microbes are now used in extracting valuable metals like uranium from rocks. The process is known as Bio-mining. The use of microbes in mining reduces the cost of production by more than 50%.
2. DNA components from bacteria are used as Biosensors that can detect biologically active toxic pollutants.
3. Microbes also find application in medical diagnostics, food and fermentation operations.
4. The most important development in Biotechnology depends on the possibility of altering the genetic makeup of bacteria through genetic engineering.
5. Microbes in household products :
   a) A common example is the production of curd from milk micro-organisms such as Lactobacillus and others, commonly called lactic acid bacteria (LAB), grow in milk and convert into curd. It also improves its nutritional quality by increasing vitamin B12.
   b) Some of our food stuffs like cheese, yogurt are also actually the by- products of microbial growth.
6. Microbes are used as biocontrol agents. Biocontrol refers to the use of biological methods for controlling plant diseases and pests.
7. Many industrial products like enzymes, amino acids, vitamins, organic acid and alcohols are commercially produced by micro organisms.
8. Microorganism decompose dead plants and animals and enrich the soil nutrients which can be used by plants. They play an important role in recycling of elements.
9. Microbes cause diseases in plants and human beings. On the other hand, they help in creating disease free world by producing antibiotics and vaccinations. Penicillin was the antibiotic discovered by Alexander Fleming from a fungus, penicillin notatum. The antibiotic obtained from bacteria streptomyces griseus is known as streptomycin.
10. Biogas is a mixture of gases (containing predominantly methane) produced by the microbial activity and which may be used as fuel.

Harmful activities:
Some bacteria that cause human diseases are

Bacteria causes plant diseases :

Bacteria also cause animal diseases :

Intext Question Answers

Question 1.
Many people believe that bacteria do little more than cause human illness and infectious diseases. How does the information in this chapter help you correct that misconception?
Answer:
Most of the bacteria are beneficial activities than harmful activities.

Bacteria are known to be the casual agents of plant, animal and human diseases. At the same time there are many bacteria which are directly or indirectly beneficial to humans. Thus, these organisms can be considered both as ‘friends and foes of man’.”

Beneficial activities:

1. Microbes are now used in extracting valuable metals like uranium from rocks. The process is known as Bio-mining. The use of microbes in mining reduces the cost of production by more than 50%.
2. DNA components from bacteria are used as Biosensors that can detect biologically active toxic pollutants.
3. Microbes also find application in medical diagnostics, food and fermentation operations.
4. The most important development in Biotechnology depends on the possibility of altering the genetic makeup of bacteria through genetic engineering.
5. Microbes in household products :
   a) A common example is the production of curd from milk micro-organisms such as Lactobacillus and others, commonly called lactic acid bacteria (LAB), grow in milk and convert into curd. It also improves its nutritional quality by increasing vitamin Bir
   b) Some of our food stuffs like cheese, yogurt are also actually the by-products of microbial growth.
6. Microbes are used as biocontrol agents. Biocontrol refers to the use of biological methods for controlling plant diseases and pests.
7. Many industrial products like enzymes, amino acids, vitamins, organic acid and alcohols are commercially produced by microorganisms.
8. Microorganism decompose dead plants and animals and enrich the soil nutrients which can be used by plants. They play an important role in recycling of elements.
9. Microbes cause diseases in plants and human being. On the other hand, they help in creating disease free world by producing antibiotics and vaccinations. Penicillin was antibiotic obtained from bacteria streptomyces griseus is known as streptomycin.
10. Biogas is a mixture of gases (containing predominantly methane) produced by the microbial activity and which may be used as fuel.

Harmful activities:
Some bacteria that cause human diseases are

Bacteria causes plant diseases ;

Bacteria also cause animal diseases :

Question 2.
Humans produce about 50 grams of feces per day. Scientists estimated in broad terms about one -third of human feces is composed of bacteria. If one E.coli cell weighs 1 × 10^-12 g, how many bacteria are there in a day’s feces? How can this be possible?
Answer:
Feces per day = 50 gram
Bacteria present = \(\frac{1}{3}\) × 50 gm
Each bacteria weight = 1 × 10^-12 gm
Bacteria in a day’s feces = \(\frac{1}{3}\) × 50 × 1 × 10^-12 gm
i.e., 16.66 × 10^-12(approx)

Question 3.
An organism is described as a peritrichous bacillus. How might you translate this bacteriological language into a description of the organism?
Answer:
Bacillus indicates that bacteria are rod elongated in shape. Peritrichous indicates that bacteria is having many flagella distributed all over the cell surface flagella are locomotory in function.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development

Very Short Answer Type Questions

Question 1.
Define plasticity. Give an example.
Answer:

1. Plasticity is the ability of plants to follow different pathways in response to the environment or phases of life to form different kinds of structures.
2. Heterophylly is an example of plasticity.

Question 2.
What is the disease that formed the basis for the identification of gibberellins in plants? Name the causative fungus of the disease.
Answer:

1. “Bakane” (foolish seedling) disease of rice seedlings.
2. It is caused by a fungal pathogen Gibberella fujikuroi.

Question 3.
What is apical dominance? Name the growth hormone that causes it.
Answer:

1. Apical dominance : The growing apical bud inhibits the growth of lateral (axillary) buds.
2. It is caused by Auxins.

Question 4.
What is meant by bolting ? Which hormone causes bolting?
Answer:

1. Bolting : Elongation of internodes just prior of flowering in plants with rosette habit (Eg : Beat, Cabbage).
2. Gibberellins are responsible for bolting.

Question 5.
Define respiratory climactic. Name the PGR associated with it.
Answer:

1. The rise in the rate of respiration during the ripening of the fruits is known as respiratory climatic.
2. Ethylene is responsible for it.

Question 6.
What is ethephon? Write its role in agricultural practices.
Answer:

1. Ethephon in an aqueous solution. It is readily absorbed and transported within the plant and releases ethylene slowly.
2. Ethephon hastens fruit ripening in tomatoes and apples and accelerates abscission in flowers and fruits. It also promotes female flowers in cucumbers and thereby increasing the yield.

Question 7.
Which of the PGR (Plant Growth Regulator) is called stress hormone and why?
Answer:

1. Abscisic acid (ABA) is called as the stress hormone.
2. It stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Question 8.
What do you understand by vernalisation? Write its significance.
Answer:

1. Vernalisation is the method of inducing early flowering in plants by pre-chilling treatment of their seeds or young shoots.
2. Vernalization of winter varieties of wheat, barley and rye useful to early harvesting of crop.

Question 9.
Define the terms quiescence and dormancy.
Answer:

1. Quiescence is the condition of a seed which is unable to germinate only because favourable external conditions normally required for growth are not present.
2. Dormancy is the condition of a seed when it fails to germinate because of internal conditions, even though external conditions are suitable.

Short Answer Type Questions

Question 1.
Write a note on agricultural/horticultural applications of auxins. [Mar. ’17,’14; May ’14]
Answer:

1. Auxins help to initiate rooting in stem cuttings, an application widely used for plant propagation in horticulture.
2. Auxins promote flowering e.g. in pineapples.
3. Auxins help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
4. Removal of shoot tips usually results in the growth of lateral buds. This is widely applied in tea plantations.
5. Auxins also induce parthenocarpy. e.g. in tomatoes.
6. Auxins (2, 4D) are widely used as herbicides.
7. Auxins control xylem differentiation and help in growth.

Question 2.
Write the physiological responses of gibberellins in plants. [Mar. 2019]
Answer:

1. Gibberellins promote bolting (internode elongation just before flowering) in beet, cabbages and many plants with rosette habit.
2. The ability to cause an increase in the length of axis is used to increase the length of grapes stalks.
3. Gibberellins cause fruits like apple to elongate and improve their shape.
4. Gibberellins delay senescence. Thus, fruits can be left on the tree longer so as to extend the market period.
5. GA₃ is used to speed up the malting process in brewing industry.
6. Spraying Gibberellins on sugarcane crop increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre.
7. Spraying juvenile conifers with Gibberellins fastens the maturity period, thus leading to early seed production.

Question 3.
Write any four physiological effects of cytokinins in plants. [Mar. ’18, May ’17]
Answer:

1. Cytokinins have specific effects on cytokinesis (that means division of cytoplasm during cell division). It helps in occuring rapid cell division for example: root apices, developing shoot buds, young fruits etc.
2. Cytokinins help to produce new leaves, chloroplasts in leaves, lateral shoot growth and adventitious shoot formation.
3. Cytokinins help to overcome the apical dominance.
4. Cytokinins promote nutrient mobilisation which helps in the delay of leaf senescence.

Question 4.
What are the physiological processes that are regulated by ethylene in plants? [Mar. 2020]
Answer:

1. Ethylene accelerates the ripening of fruits. It is regarded as fruit ripening hormone.
2. Effects of ethylene on plants include horizontal growth of seedlings, swelling of the axis and apical hook formation in dicot seedlings.
3. Ethylene promotes senescence and abscission of plant organs, especially of leaves and flowers.
4. Ethylene breaks seed and bud dormancy. It initiates germination in peanut seeds and sprouting of potato tubers.
5. Ethylene promotes rapid internode/petiole elongation in deep water rice plants.
6. Ethylene helps leaves/upper parts of the shoot to remain above water.
7. Ethylene promotes root growth and root hair formation, thus helping plants to increase their absorption surface.
8. Ethylene is used to initiate flowering and for synchronising fruit – set in pineapples. It also induces flowering in mango.

Question 5.
Write short notes on seed dormancy.
Answer:

1. Dormancy is the condition of seed when it fails to germinate because of internal conditions, even though external conditions (e.g. temperature, moisture) are suitable.
2. The dormancy may be caused by hard seed coats that prevents uptake of oxygen or water (e.g. fabaceae).
3. Dormancy caused by hard seed coats can be broken by scarification.
4. Seeds of certain plants (e.g.: tomato) contain chemical compounds, which inhibit their germination.
5. Many seeds (e.g.: Polygonum) will not germinate untill they have been exposed to low temperatures in moist conditions in the presence of oxygen for weeks to months.
6. Motet seeds respond to high temperatures and several seeds respond best when daily temperatures alternate between high and low.

Question 6.
Which one of the plant growth regulators would you use if you are asked to
a) Induce rooting in a twig
b) QCiickly ripen a fruit
c) Delay leaf senescence
d) Induce growth in axillary buds
e) ‘Bolt’ a rosette plant
f) Induce immediate stomatal closure in leaves
g) Overcome apical dominance
h) Kill dicotyledonous weeds.
Answer:
a) Auxins like IBA, NAA
b) Ethylene
c) Cytokinin
d) Cytokinins
e) Gibberellins
f) Abscisic acid (ABA)
g) Cytokinins
h) Auxins (2-4 D)

Long Answer Type Questions

Question 1.
Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Answer:
1) Growth :
Growth is defined as a permanent or irreversible increase in dry weight, size, mass or volume of a cell, organ or organism.

2) Differentiation :
The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed as differentiation.

3) Dedifferentiation :
The living differentiated cells, that have lost the capacity to divide, can regain the capacity of division under certain conditions. This phenomenon is called dedifferentiation, e.g.: Formation of meristems – interfascicular cambium and cork cambium from parenchyma cells.

4) Redifferentiation :
In the process of redifferentiation meristems/tissues are able to divide and produce cells that once again lose the capacity to divide but mature to perform specific functions. That means they get redifferentiation.

5) Determinate growth :
Leaves are specialised organs characterised by defined developmental destiny and determinate growth.

6) Meristem :
It is located at root and shoot tips. The meristem refers to the cells that remain dividing.

7) Growth rate :
Increased growth per unit time is termed as growth rate.

Question 2.
Describe briefly
a) Arithmetic growth
b) Geometric growth
c) Sigmoid growth curve
d) Absolute and relative growth rates
Answer:
a) Arithmetic growth :
In arithmetic growth, following mitotic cell division, only one daughter cell continues to divide while the other differentiates and matures.

The simplest expression of arithmetic growth is exemplified by a root elongating at a constant rate. On plotting the length of the organ against time, a linear curve is obtained. Mathematically, it is expressed as

Constant linear growth, a plot of length L against time t
L₁ = L₀ + rt
L₁ = Length at time’t’
L₀ = Length at time ‘zero’
r = growth rate/elongation per unit time.

b) Geometric growth :
In several systems, the initial growth is slow (lag phase). Growth increases rapidly thereafter at an exponential rate (log or exponential phase) S – curve is observed

Diagrammatic representation of: (a) Arithmetic (b) Geometric growth and (c) Stages during embryo development showing geometric and arithmetic phases

c) Sigmoid curve growth :
A sigmoid curve is a characteristic of living organism growing in a natural environment. It is typical for all cells, tissues and organs of a plant.

An idealised sigmoid growth curve typical of cells in culture, and many higher plants and plant organs

It consists of 3 phases namely
(a) Lag phase
(b) log phase
(c) Stationary phase

d) Absolute and relative growth rate :
Measurement and comparison of the total growth per unit time is called the absolute growth rate.

The growth of the given system per unit time expressed on a common basis e.g. per unit initial parameter is called the relative growth rate.

Question 3.
List five natural plant growth regulators. Write a note on discovery, physiological functions an agricultural/horticultural applications of any one of them.
Answer:

Note : Write any one of the above.

Intext Question Answers

Question 1.
Fill in the blanks with appropriate word/words.
a) The phase in which growth is most rapid is …………… .
b) Apical dominance as expressed in dicotyledonous plants is due to the presence of more …………… in the apical bud than in the lateral ones.
c) In addition to auxin, a ………….. must be supplied to the culture medium to obtain a good callus in plant tissue culture.
d) …………… of vegetative plants are the sites of photoperiodic perception.
Answer:
a) Log phase
b) Auxins
c) Cytokinins
d) Shoot apex of plant

Question 2.
A primary root grows from 5 cm to 19 cm in a week. Calculate the growth rate and relative growth rate over the period.
Answer:
Growth rate is = \(\frac{19-5}{7}=\frac{14}{7}\) = 2 cm

Question 3.
Gibberellins promote the formation of ………….. flowers on genetically …………… plants in Cannabis whereas ethylene promotes formation of …………. flowers on genetically …………… plants.
Answer:
a) male
b) dwarf
c) female
d) dwarf

Question 4.
Classify the following plants into Long day plants (LDP), Short day plants (SDP) and Day neutral plants (DNP)
Xanthium, Spinach, Henbane (Hyoscyamus niger), Rice, Strawberry, Bryophyllum, Sunflower, Tomato, Maize.
Answer:
Xanthium – Short day plant
Spinach – Long day plant
Henbane – Short day plant
Rice – Short day plant
Strawberry – Short day plant
Bryophyllum – Short day plant
Sunflower – Day neutral plant
Tomato – Day neutral plant
Maize – Long day plant

Question 5.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers. Which plant growth regulator can be applied to achieve this?
Answer:
Auxins.

Question 6.
Where are the following hormones synthesized in plants?
a) IAA
b) Gibberellins
c) Cytokinins
Answer:
a) Shoot apex and to some extent root apex also contribute to the synthesis of auxins.

b) Gibberellins are synthesised from Acetyl co enzyme A. They are mostly synthesized in apical tissues. They also synthesised in developing seeds, fruits, young leaves of developing apical bud, and elongation shoots.

c) Natural cytokinin are synthesised in the regions where rapid cell division occurs for example root apices, developing shoot buds, young fruits, etc.

Question 7.
Light plays an important role in the life of all organisms. Name any three physiological processes in plants which are influenced by light.
Answer:

1. Photosynthesis
2. Photoperiodism
3. Growth

Question 8.
Growth is one of the characteristics of all living organisms. Do unicellular organisms also grow? If so, what are the parameters?
Answer:
Yes. If we plot the parameter of growth against time, we get a typical sigmoid or . S-curve.

Question 9.
Rice seedlings infected with fungus Gibberella fujikuroi are called foolish seedlings. What is the reason?
Answer:
The reason is the rice plants grew excessively tall and very little grain production. The plants were taller, thinner, and paler with little tillering as compared to healthy plants.

Question 10.
Why isn’t any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Answer:
Because growth depends upon cell division, nutrient supply, time, etc.

Question 11.
‘Both growth and differentiation in higher plants are open.’ Comment.
Answer:
Plants retain the capacity for unlimited growth throughout the life due to the presence of meristems at certain locations of their body. The cells of such meristems have the capacity to divide and self perpetite. The product, however, soon loses the capacity to divide and such cells make up plant body.

This form of growth where in new cells are always being added to the plant body by the activity of the meristem is called the open form of growth.

The cells divided from root apical and shoot apical meristems and cambium differentiate the nature to perform specific functions. It is called differentiation.

Differentiation in plants is open because cells/tissues arising out of the same meristems have different structures at maturity. The final structure at maturity of cell tissue is also determined by the location of the cell with in.

Question 12.
‘Both a short day plant and a long day plant can produce flowers simultaneously in a given place.’ Explain.
Answer:
A long day plant required the exposure to light for a period exceeding well defined critical duration while short day plant must be exposed to light for a period less than this critical duration before the flowering is initiated in them. It means that not only the duration of light but also the duration of dark period is of equal importance. Hence flowering depends not only a combination of light and dark exposure but also their relative durations.

That is why, both a short day and a long day plant produce flowering simultaneously in a given place when they are exposed to necessary inductive photoperiod required by them in artificial condition.

Question 13.
Would a defoliated plant respond to photoperiodic cycle? Why?
Answer:
No, a defoliated plant would not respond to photoperiodic cycle. Because the site of perception of light/dark duration is the leaves. In defoliated plants leaves are absent.

It has been hypothesised that there is a hormonal substances that is responsible for flowering. This hormonal substance migrates from leaves to shoot apices for inducing flowering only when the plants are exposed to the necessary inductive photoperiod.

Question 14.
What would be expected to happen if
(1) GA₃ is applied to rice seedlings.
(2) Dividing cells stop differentiating.
(3) A rotten fruit gets mixed with unripe fruits
(4) You forget to add cytokinin to the culture medium
Answer:
(1) Elongation of coleoptile occurs quickly.
(2) Undifferentiated mass of cells are formed.
(3) All the fruits will be mature.
(4) Shoot growth is inhibited.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants

Very Short Answer Type Questions

Question 1.
Energy is released during the oxidation of compounds in respiration. How is this energy stored and released as and when it is needed?
Answer:

1. Energy contained in respiratory substrates is released in a series of slow step wise reactions controlled by enzymes and stored as ATP.
2. ATP is broken down whenever and wherever energy needs to be utilised.

Question 2.
Explain the term ‘Energy currency’. Which substance acts as energy currency in plants and animals?
Answer:

1. ATP is broken down whenever and wherever energy needs to be utilised is cells of living organisms.
2. ATP acts as energy currency in plants and animals.

Question 3.
Different substrates get oxidised during respiration. How does respiratory quotient (RQ) indicate which type of substrate i.e., carbohydrate, fat or protein is getting oxidised?
RQ = A/B
What do A and B stand for?
What type of substrates have RQ of 1, <1, >1?
Answer:
1. RQ is an index for type of substrate being used in respiration RQ = A/B, in which A stands for volume of CO₂ evolved and B stands for volume of O₂ consumed. Thus RQ = volume of CO₂ evolved/Volume of O₂ consumed.

2. RQ is 1, when carbohydrates are used as substrate. It is less than 1, when fats or proteins are used as substrate.

Question 4.
What is the specific role of F₀ – F₁ particles in respiration?
Answer:

1. F₀ is the integral membrane protein complex that forms the channel through which protons cross enter into matrix by crossing the inner membrane.
2. F₁ head piece is the peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate.

Question 5.
When does anaerobic respiration occur in man and yeast?
Answer:

1. Anaerobic respiration occurs during exercise in the muscles of man when oxygen is inadequate for cellular respiration.
2. In yeast, anaerobic respiration occurs during fermentation.

Question 6.
Distinguish between obligate anaerobes and facultative anaerobes.
Answer:
1) Obligate anaerobe :
An organism that cannot grow in the presence of oxygen. It neither requires O₂ to grow nor does it tolerate it.

2) Facultative anaerobe :
An organism can grow in either the presence or absence of oxygen. It does not require oxygen to grow but it does tolerate its presence.

Question 7.
Explain the economic importance of fermentation.
Answer:

1. Fermentation is useful in making bread.
2. It is useful for industrial production of organic acids and alcohol.

Question 8.
What is the common pathway for aerobic and anaerobic respirations? Where does it take place?
Answer:

1. Glycolysis (EMP pathway) is the common pathway for both aerobic and anaerobic respirations.
2. It occurs in cytoplasm of the cell and takes place is all living organisms.

Question 9.
Why are mitochondria termed as the power houses of the cell?
Answer:

1. Mitochondria are found in eukaryotic cells and are seat of Krebs cycle and oxidative phosphorylation.
2. They produce energy currency of the cell (ATP).

Question 10.
What is the reason for describing ATP synthesis in FQ – Fx particles of mitochondria as oxidative phosphorylation?
Answer:

1. Oxygen acts as the final Hydrogen acceptor is ETS of aerobic respiration.
2. Production of proton gradient and synthesis of ATP by ATP synthase in respiration are oxygen driven, hence it is called oxidative phosphorylation.

Question 11.
Which substance is known as the connecting link between glycolysis and Krebs cycle? How many carbons does it have?
Answer:

1. Acetyl CoA is the connecting link between glycolysis and Krebs cycle.
2. It consists of 2 carbon atoms.

Question 12.
What cellular organic substances are never used as respiratory substrates?
Answer:
Pure proteins or fats are never used as respiratory substrates.

Question 13.
Why is the RQ of fats less than that of carbohydrates?
Answer:
1. It requires more oxygen for complete oxidation, since C : O ratio in fats in very high.

Question 14.
What is meant by ‘Amphibolic pathway’?
Answer:

1. The pathway involved both in catabolism and anabolism is termed as Amphibolic pathway.
2. Eg. Respiratory pathway.

Question 15.
Name the mobile electron carriers of the respiratory electron transport chain in the inner mitochondrial membrane.
Answer:

1. Cytochrome C that transfer electrons between complex III and IV of ETS.
2. U biquinone that acts as electron carries in between complex II and III of ETS.

Question 16.
What is the final acceptor of electrons in aerobic respiration? From which complex does it receive electrons?
Answer:

1. Oxygen is the final acceptor of electrons in aerobic respiration.
2. It receives electrons from complex IV (cytochrome ‘c’ oxidase).

Short Answer Type Questions

Question 1.
What is meant by the statement ‘Aerobic respiration is more efficient’?
Answer:

1. Aerobic respiration is the process that leads to a complete oxidation of organic substance in the presence of oxygen.
2. It releases CO₂, water and a large amount of energy present in the substrate.
3. 686 k.cal of energy is released where as in anaerobic respiration only 56 k.Cal. is released.
4. In aerobic respiration 36 ATP molecules are formed. Where as in anaerobic respiration only 2 ATP molecules are formed.

Question 2.
Pyruvic acid is the end product of glycolysis. What are the three metabolic fates of pyruvic acid under aerobic and anaerobic conditions?
Answer:
The fate of pyruvate depends upon the availability of O₂ and the type of organism. The three metabolic fates of pyruvic acid are

1. lactic acid fermentation
2. alcoholic fermentation under anaerobic condition and
3. aerobic respiration under aerobic condition.

Question 3.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic condition like human beings and angiosperms?
Answer:

1. Under the conditions of oxygen scarcity anaerobic respiration takes place in organism that live in aerobic conditions.
2. For example because Muscle tissue under intense use muscle demands too much energy (ATP) and consume much more Oxygen to produce that energy. This high consumption leads to oxygen scarcity and the muscle cells begin to make lactic acid by anaerobic respiration to fulfill their energy needs.
3. The first cells on this planet lived in an atmosphere that lacked O₂. Even today all living organisms retain the enzymatic machinery to partially oxidise glucose without the help of oxygen during glycolysis.

Question 4.
Oxygen is an essential requirement for aerobic respiration but it enters the respiratory process at the end. Discuss.
Answer:

1. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process.
2. The presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system.
3. Oxygen acts as the final hydrogen acceptor.
4. In respiration, it is the energy of oxidation-reduction utilised for Phosphorylation. It is for this reason that the process is called oxidative phosphorylation.

Question 5.
Respiration is an energy releasing and enzymatically controlled catabolic process which involves a stepwise oxidative breakdown of organic substances? Inside living cells.
In this statement about respiration, explain the meaning of i) step wise oxidative breakdown ii) organic substance (used as substrates).
Answer:
i) a) Respiration involves a stepwise oxidative breakdown of organic substances.
b) In the process of aerobic respiration it is divided into 4 phases. They are Glycolysis, TCA cycle, Electron Transport System and Oxidative Phosphorylation.
c) It is generally assumed that the process of respiration and production of ATP in each phase takes place in a stepwise manner.
d) The product of one pathway forms the substrate of the other pathway.
e) Various molecules produced during respiration are involved in other biochemical processes.
f) The respiratory substrate enters and withdraws from pathway.
g) ATP gets utilised whenever required and enzymatic rates are generally controlled.
h) Stepwise breakdown of organic substances makes the system more efficient in extracting and storing energy.

ii) Usually carbohydrates, glucose are oxidised to release energy but proteins, fats and even organic acids can be used as respiratory substrates in some plants under certain conditions.

Question 6.
Comment on the statement – Respiration is an energy producing process but ATP is used in some steps of the process.
Answer:

1. Though respiration is an energy producing process ATP is utilised in two steps during glycolysis.
2. First in the conversion of glucose into glucose 6 phosphate.
   Glucose + ATP → Glucose 6 phosphate + ADP.
3. Second in the conversion of fructose 6 phosphate into Fructose 1-6 bisphosphate.
   Fructose 6 phosphate + ATP → fructose 1, 6 bisphosphate + ADP.

Question 7.
Explain briefly the process of glycolysis.
Answer:
Glycolysis occurs in the cytoplasm of all living organisms. In this process, glucose undergoes partial oxidation to form two molecules of pyruvic acid. The scheme of glycolysis was given by Gustav Embden, Meyerhof and J. Parnas and is often referred to as the EMP pathway.

In glycolysis, a chain of ten reactions, under the control of different enzymes takes place to produce pyruvate from glucose. They are

1) Phosphorylation :
Glucose is phosphorylated to give rise to glucose 6-phosphate one ATP is utilised.

2) Isomerisation :
Glucose 6 phosphate converts into isomer fructose 6 phosphate in the presence of phosphohexose isomerase.

3) Phosphorylation :
The enzyme phosphofructokinase uses another ATP molecule to transfer a phosphate group to fructose 1, 6 biphosphate.

4) Cleavage :
The enzyme aldolase splits fructose 1,6 bisphosphate into two sugars that are isomers to each other. They are Glyceraldehyde 3 phosphate & Dihydroxy acetone phosphate (DHAP).

5) Isomerisation :
Enzyme Triose phosphate isomerase rapidly inter-converts the molecules of DHAP and G3P.

6) Oxidation :
Glyceraldehyde-3-phosphate undergoes oxidation and phosphorylation in the presence of Glyceraldehyde dehydrogenase resulting in the dehydrogenase formation of 1, 3 bisphosphoglyceric acid and NADH.

7) Dephosphorylation :
Enzyme phosphoglycerokinase catalyses the dephosphorylation of 1, 3 bisphosphoglyceric add resulting 3 PGA. ATP is formed.

8) Intramolecular shift (Isomerisation) :
Enzyme phosphoglyceromutase transfers phosphate group from 3-carbon position to 2-carbon position resulting 2 PGA.

9) Dehydration :
The enzyme enolase catalyses the removal of one water molecule from 2 PGA resulting in PEP.

10) Dephosphorylation :
Phosphoenol pyruvic acid undergoes dephosphorylation in the presence of enzyme pyruvic kinase and results in the formation of pyruvic acid ATP is formed.

Question 8.
Why is the respiratory pathway referred to as an amphibolic pathway? Explain.
Answer:
a) Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyi CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway acts as both breakdown and the synthesis of fatty acids.

b) Similarly during the breakdown and the synthesis of protein too, respiratory intermediates forms the link.

c) The breaking down process within the living organism constitute catabolism, which synthesis is anabolism.

d) Because the respiratory pathway is involved in both anabolism and catabolism, it would be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

Question 9.
We commonly call ATP the energy currency of the cell. Can you think of some other energy carriers present in the cell? Name any two.
Answer:

1. NADH and FADH₂
2. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP.
3. Oxidation of one molecule of FADH₂ give rise to 2 molecules of ATP.
4. This takes place in electron transport system for mitochondria.

Question 10.
ATP produced during glycolysis is a result of substrate level phosphorylation. Explain.
Answer:
Substrate level phosphorylation is a type of metabolic reaction that results in the formation of ATP by the direct transfer and donation of phosphate group to ADP. This occurs twice in Glycolysis.

1. 1, 3 bis phospho glyceric acid form 3 phosphoglyceric acid. The enzyme phosphoglycero kinase, catalyses the phosphorylation. ADP molecule accepts the phosphate released in this reaction and gets converted into ATP.
2. Phosphoenol pyruvic acid undergo dephosphorylation in the presence of the enzyme pyruvic kinase and results in the formation of pyruvic acid. ADP is converted to ATP.

Question 11.
Do you know of any step in Krebs cycle where there is a substrate level phosphorylation? Explain.
Answer:
In Krebs cycle, Succinyl coenzyme A splits into succinic acid and co-enzyme A by the catalytic activity of ‘Succinic acid thiokinase’. The energy released in this reaction is utilized to form ATP from ADP and inorganic phosphate (Pi). Because ATP formation is linked directly to conversion of substrate, this reaction is an example of substrate level phosphorylation.

Question 12.
When a substrate is being metabolised. Why doesn’t all the energy that is produced get released in one step ? Instead it is released in multiple steps. What is the advantage of step wise release of energy?
Answer:

1. The stepwise release of chemical bond energy enables the cell to utilize that energy in ATP synthesis.
2. It minimises the wastage of energy.
3. It keeps the temperature of the cell low to prevent its burning.
4. The activities of enzymes for different steps of respiration can be controlled.
5. The intermediates of respiratory pathway can be used for the synthesis of other biomolecules.
   Thus stepwise release of energy makes the system more efficient in extracting and storing energy.

Question 13.
Respiration requires O₂. How did the first cells on earth manage to survive in an atmosphere that lacked oxygen?
Answer:
The first cells on earth manage to survive in an atmosphere that lack oxygen are anaerobic bacteria. The facultative anaerobic organism is an organism usually bacterium, that makes ATP by aerobic respiration if oxygen is present but is also capable of switching to fermentation under anaerobic conditions. Thus facultative anaerobes is an organism that can grow in either the presence or absence of oxygen. It does not require O₂ to grow but it does tolerate its presence. An obligate anaerobe is an organism that cannot grow in the presence of organism. It neither require O2 to grow nor does it tolerate it.

In any case, all living organisms remain the enzymatic machinery to partially oxidise glucose without the help of oxygen.

Question 14.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Explain.
Answer:

1. The energy yield in terms of ATP during aerobic respiration is 38 ATP molecules.
2. Glucose molecule gets completely oxidised in the presence of Oxygen.
   C₆H₁₂O₆ + 6O₂ → 6 CO₂ + 6 H₂O + energy
3. Aerobic respiration involves Glycolysis, Krebs cycle and electron transport. 2 ATP molecules are utilised during Glycolysis. Hence net gain in aerobic respiration is 36 ATP molecules.
4. The energy yield in terms of ATP during an aerobic respiration is only 2 ATP molecules. It involves Glycolysis and Fermentation Glucose molecule gets incomplete oxidation in the absence of oxygen.
   C₆H₁₂O₆ → CO₂ + Ethyl alcohol + Energy

Question 15.
RUBP carboxylase, PEPase, pyruvate dehydrogenase ATPase, cytochrome oxidase, Hexokinase, Lactic dehydrogenase.
Select the enzymes from the list above which are involved in
a) Photosynthesis
b) Respiration
c) Both photosynthesis and respiration.
Answer:
a) Enzymes in Photosynthesis : RUBP caboxylase, PEPase.

b) Enzymes in Respiration : Lactic dehydrogenase, Pyruvic dehydrogenase, Hexokinase, Cytochrome oxidase.

c) Enzymes in both photosynthesis and respiration : ATPase.

Question 16.
How does a tree trunk exchange gases with the environment although it lacks stomata?
Answer:
In stems, the living cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support. Thus, most cells of a plant have at least a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems and roots, which provide an interconnected network of air spaces.

Question 17.
Write about two energy yielding reactions of glycolysis.
Answer:
1) Phosphoglycerokinase enzyme catalyses the 1, 3 – bisphosphoglyceric acid resulting in the formation of 3-phosphoglyceric acid. ADP molecule accepts the phosphate released in this reaction and get converted into ATP. This process of ATP is known as substrate level phosphorylation.
phosphoglycerokinase

2) In the final step of glycolysis, phosphoenol pyruvic acid undergoes dephosphorylation in the presence of enzyme ‘pyruvic kinase’ and results in the formation of pyruvic acid. ADP is converted into ATP. This is also substrate level phosphorylation.

Question 18.
Name the site(s) of pyruvate synthesis. Also write the chemical reaction wherein pyruvic acid dehydrogenase acts as a catalyst.
Answer:
Sites of pyruvate are Cytosol
Pyruvate which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after entering into mitochondrial matrix, undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvate dehydrogenase require the participation of several coenzymes, including NAD+ and coenzyme A.

Question 19.
Mention the important series of events of aerobic respiration that occur in the matrix of the mitochondrion as well as the one that takes place in the inner membrane of the mitochondrion.
Answer:

1. The important events that occur in the matrix of mitochondrion is TCA cycle or Krebs cycle or citric acid cycle.
2. During citric acid cycle complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms leavings three molecules of CO₂, Eight molecules of NADH + H⁺ and two molecules of ATP are formed.
3. The important events that takes place in the inner membrane of the mitochondrion is Electron Transport System (ETS) and Oxidative Phosphorylation.
4. During electron transport system the passing one of the electrons removed as part of the hydrogen atoms to molecules O₂ with simultaneous synthesis of ATP takes place.

Question 20.
The respiratory pathway is believed to be a catabolic pathway. However, the nature of TCA cycle is amphibolic. Explain.
Answer

1. Respiration has been considered as catabolic process and the respiratory pathway as a catobolic pathway.
2. Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Thus the respiratory pathway comes into the picture for both during the breakdown and synthesis of fatty acids.
3. Similarly during the breakdown and synthesis of proteins too, respiratory intermediates form the link.
4. The breakdown process within the living organism is called catabolism. The synthesis of his process is called anabolism.
5. Thus the respiratory pathway includes both anabolism and catabolism. Hence it is better to consider respiration as an amphibolic pathway rather than catabolic one.

Question 21.
The net gain of ATP for the complete aerobic oxidation of glucose is 36. Explain.
Answer:
ATP produced during complete aerobic oxidation of one molecule of glucose is as follows.

I) Glycolysis:
i) ATP produced by substrate level phosphorylation

ii) From NADH generated in glycolysis

II) Oxidative decarboxylation of Pyruvic acid

III) Krebs’s Cycle

Long Answer Type Questions

Question 1.
In the following flow chart, replace the symbols a, b, c and d with appropriate terms. Briefly explain the process and give any two applications of it.

Answer:

This process is Fermentation.
In fermentation, the pyruvic acid the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions, where pyruvic acid is converted to C02 and ethanol. In some bacteria lactic acid is formed from pyruvic acid.

Applications :

1. Fermentation is used for preparing alcohol by using yeast cell.
2. In animal cells, like muscles during exercise when oxygen is inadequate for cellular respiration, pyruvic acid.is reduced to lactic acid by lactate dehydrogenase.

Question 2.
Explain Mitchell’s chemiosmosis in relation to oxidative phosphorylation.
Answer:
The synthesis of ATP in respiration is associated with the consumption of oxygen, hence it is referred as oxidative phosphorylation. The mechanism of mitochondrial ATP synthesis can be explained by Mitchell’s chemiosmosis. The transfer of electrons from NADH or FADH to oxygen through electron transport system results in proton transfer from matrix to inner membrane space of mitochondria. Due to this, proton concentration gradient is established across the inner mitochondrial membrane (more number of H+ on inner membrane space side and less on the matrix side).

The inner membrane Of mitochondria is virtually impermeable to protons and thus prevents the return of proton into the matrix.

However the ATP synthase (complex V) consists of the major components F₁ and F₀ .

F₀ is the integral membrane protein complex that forms the channel through which protons cross the inner membrane.

F₁ head piece is a peripheral membrane protein complex and contains the site for the synthesis of ATP from ADP and inorganic phosphate.

When H⁺ move down the gradient, energy is released some amount of energy helps in combining ADP and iP leading to the form of ATP. The energy of 3H⁺ moving down the potential gradient is sufficient to form one ATP molecule.

Diagramatic presentation of ATP synthesis in mitochondria

Question 3.
Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Answer:
The energy stored in NADH + H⁺ and FADH₂ are oxidised through electron transport system. The electrons are passed on to O₂ resulting in the formation of H₂O.

The metabolic pathway through which an electron passes from one carrier to another is called the Electron Transport System (ETS). It is present in the inner mitochondrial membrane.

1. Electrons from NADH produced in the mitochondrial matrix during the citric acid cycle are oxidised by an NADH dehydrogenase (complex I) and the electrons are transferred to ubiquinone located within the inner membrane.
2. Ubiquinone also receives reducing equivalents via FADH₂ (complex II) that is generated during oxidation of succinate in the citric acid cycle.
3. The reduced ubiquinone is then oxidised with the transfer of electrons to cytochrome c via cytochrome bcx complex (Complex III).
4. Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for the transfer of electrons between complex III and IV.
5. Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3 and two copper centres.

When the electrons pass from one carrier to another via complex I to IV in the electron transport chain, they are coupled to ATP synthase (complex V) for the production of ATP from ADP and inorganic phosphate.

The number of ATP molecules synthesised depends upon the nature of electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of FADH₂ produces 2 molecules of ATP. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. The presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system.

Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilised for the production of proton gradient required for phosphorylation.

In respiration it is the energy of oxidation – reduction utilised for the same process. It is for this reason that the process is called oxidative phosphorylation.

Question 4.
Enumerate the assumptions that we undertake in making the respiratory balance sheet. Are these assumptions valid for a living system ? Compare fermentation and aerobic respiration in this context.
Answer:
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions. These assumptions are :

1. There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
3. None of the intermediates in the pathway are utilised to synthesise any other compound.
4. Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

These kinds of assumptions are not really valid in a living system because (1) all pathway work simultaneously and do not take place one after another. (2) Substrates enter the pathways and are withdrawn from it as and when necessary. (3) ATP is utilised as and when needed. (4) Enzymatic rates are controlled by multiple means.

Comparison of fermentation and aerobic respiration :

1. Fermentation accounts for only a partial breakdown of glucose whereas in aerobic respiration glucose completely degraded to CO₂ and H₂O.
2. In fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid whereas many more molecules of ATP are generated under aerobic conditions.
3. NADH is oxidised to NAD⁺ rather slowly in fermentation; however the reaction is very vigorous in the case of aerobic respiration.

Question 5.
Give an account of glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration. [Mar. ’20, 18, May ’17]
Answer:
Glycolysis occurs in the cytoplasm. The end product of glycolysis is pyruvic acid.

Glycolysis is the process in which glucose undergoes partial oxidation to form two molecules of pyruvic acid. It is as follows.
1) In the first of glycolysis, a phosphate group is added to glucose molecule in the formation of glucose-6-phosphate. ATP is utilised. This reaction is catalysed by hexokinase enzyme.

2) Phosphohexokinase enzyme catalyses the conversion of glucose-6-phosphate to its isomer fructose-6-phosphate.

3) Fructose-6-phosphate undergoes phosphorylation resulting Fructose-1, 6 bisphosphate. In this step ATP is utilised. This reaction is done by enzyme phosphofructokinase.

4) Fructose 1-6 bisphosphate with the help of enzyme aldolase splits into 2 molecules. They are glyceraldehyde 3 phosphate and dihydroxyacetone phosphate (DHAP).

5) Among the two trioses, only Glyceraldehyde 3 phosphate directly participates in the subsequent reactions.

On the other hand, dihydroxyacetone phosphate gets converted into glyceraldehyde- 3-phosphate in the presence of triose phosphate isomerase and can participate in the next reaction of glycolysis.

6) Glyceraidehyde-3-phosphate undergoes oxidation as well as phosphorylation in the presence of glyceraldehyde 3 phosphate dehychoganase resulting in the formation of 1-3-bisphospho glyceric acid and NADH.

7) Phosphoglycerokinase enzyme catalyses the dephosphorylation of 1,3 bisphosphoglyceric acid resulting in the formation of 3 phosphoglyceric acid. ADP molecule accepts the phosphate released in this reaction and gets converted into high energy compound ATP. This process of formation is known as substrate level phosphorylation.

8) Phosphoglyceromutase enzyme catalyses the transfer of phosphate group from 3-carbon position of 3 PGA to 2-carbon position leading to the formation of 2PGA.

9) The enzyme enolase catalyses the removal of one water molecule from 2-phospho- glyceric acid resulting in the formation of phosphoenol pyruvic acid.

10) Phosphoenol pyruvic acid undergoes dephosphorylation in the presence of the enzyme pyruvic kinase and results in the formation of pyruvic acid. ADP is converted to ATP. This is substrate level of

Fate of Pyruvic acid :
The ultimate fate of pyruvic acid which is an end product of glycolysis depends upon the availability of oxygen. In the presence of oxygen it is completely oxidized into C02 and H20. If oxygen is not available, the pyruvic acid is converted to ethyl alcohol or other organic substances by fermentation during anaerobic respiration.

Question 6.
Explain the reactions of Kreb’s cycle.
Answer:
The reaction sequence of Kreb’s cycle is as below.
1. Condensation :
Acetyl co-enzyme A condenses with oxaloacetic acid and results in the formation of citric acid and co-enzyme A. This condensation reaction is catalysed by the enzyme ‘citric synthetase.’

2. Dehydration :
The enzyme ‘aconitase’ catalyses the removal of one molecule of water from citric acid leading to the formation of cis-aconitic acid.

3. Hydration :
Addition of one molecule of water is done in the presence of aconitase. Cis-aconitic acid forms Isocitric acid.

4. Oxidation -1 :
Isocitric acid undergoes dehydrogenation (oxidation) in the presence of ‘isocitric dehydrogenase’ enzyme, leading to the formation of oxalosuccinic acid. In this reaction NAD is reduced to NADH + H⁺.

5. Decarboxylation :
Oxalosuccinic acid releases one molecule of CO₂ in the presence of oxalosuccinic decarboxylase enzyme and forms a – Ketoglutaric acid.

6. Oxidation – II :
α-ketoglutaric acid undergoes oxidation (dehydrogenase) decarboxylation and condensation with one molecule of CoA leading to the formation of succinyl coenzyme A.

7. Cleavage :
Succinyl coenzyme A splits into succinic acid and coenzyme A by the enzyme activity of succinic acid thiokinase. Energy released in this reaction is utilised to form ATP from ADP and inorganic phosphate.

8. Oxidation III :
Succinic acid then undergoes oxidation and forms Fumeric acid. Instead of NAD+, FAD serves as hydrogen acceptor in this reaction. Therefore FAD is reduced to FADH₂. The enzyme which catalyses this reaction is known as succinic dehydrogenase.

9. Hydration :
The enzyme Fumerase mediates the addition of one water molecule to fumaric acid and leads to the formation of malic acid.

10. Oxidation IV :
In the presence of Malic dehydrogenase, Malic acid releases two hydrogen atoms and gives rise to oxaloacetic acid. In this step NAD+ acts as hydrogen acceptor and is converted into NADH + H⁺.

Intext Question Answers

Question 1.
Differentiate between
a) Respiration and Combustion
b) Glycolysis and Krebs cycle
c) Aerobic respiration and Fermentation
Answer:

a) Differences between Respiration and combustion :

b) Differences between Glycolysis and Kreb’s cycle :

c) Differences between Aerobic respiration and Fermentation.

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
The compounds that are oxidised during respiration are known as respiratory substrate. The most common respiratory substrate is glucose. Usually carbohydrates are oxidised to release energy but proteins, fats and even organic acids can be used as respiratory substrate.

Question 3.
Give the schematic representation of glycolysis.
Answer:

Question 4.
What are the main steps in aerobic respiration ? Where does it take place?
Answer:
The main steps in aerobic respiration are glycolysis, Krebs cycle, electron transport system and oxidative phosphorylation.

Glycolysis occurs in the cytosol of the cell, Krebs cycle occurs in the matrix of mitochondrion. Electron transport system and oxidative phosphorylation takes palce in the inner membrane of mitochondria.

Question 5.
Give the schematic representation of an overall view of Krebs cycle.
Answer:

Question 6.
Explain ETS.
Answer:
The metabolic pathway through which an electron passes from one carrier to another is called the electron transport system. It is present in the inner mitochondrial membrane.

1. Electrons from NADH produced in the mitochondrial matrix during the citric acid cycle are oxidised by an NADH dehydrogenase (Complex – I)
2. The electrons are then transferred to ubiquinone located within the inner membrane. Ubiquinone also receives reducing equivalents via FADH₂ (complex II) that is generated during oxidation of succinate in the citric acid cycle.
3. The reduced ubiquinone (ubiquinol) is then oxidised with the transfer of electrons to cytochrome c via cytochrome bc₁ complex (complex III).
4. Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for the transfer of electrons between complex III and IV.
5. Complex IV refers to cytochrome c oxidase complex containing cytochrome a and a₃ and two copper centres.

When the electrons passes from one carrier to another via complex I to IV in the electron transport chain, they are coupled to ATP synthase (Complex V) for the production of ATP from ADP and inorganic phosphate.

The number of ATP molecules synthesised depends on the nature of the electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while the molecule of FADH₂ produces 2 molecules of ATP.

Aerobic respiration takes place in the presence of oxygen. Oxygen drives the whole process by removing hydrogen from the system. Oxygen acts as a final hydrogen acceptor. Synthesis of ATP from ADP and ip coupled to electron transport from substrate to molecular oxygen is called oxidative phosphorylation.

Question 7.
Distinguish between the following :
a) Aerobic respiration and Anaerobic respiration.
b) Glycolysis and Fermentation.
c) Glycolysis and Citric acid cycle.
Answer:
a) Differences between Aerobic respiration and Anaerobic respiration :

b) Differences between Glycolysis and Fermentation.

c) Differences between Glycolysis and C trie acid cycle:

Question 8.
What are the assumptions made during the calculation of net gain of ATP?
Answer:
The assumptions are

1. There is a sequential, orderly pathway functioning with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
3. None of the intermediates in the pathway are utilised to synthesise any other compound.
4. Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

Question 9.
Discuss “The respiratory pathway is an amphibolic pathway.”
Answer:
The respiratory pathway is involved in both anabolism and catabolism. So it is considered as amphibolic pathway rather than as a catabolic one.
e.g. : Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway shows break down and synthesis of fatty acids.
e.g.: Similarly during the breakdown and the synthesis of proteins too, respiratory intermediate form the link. Thus the breaking down process with in the living organisms constitute catabolism while synthesis is anabolism.

Question 10.
Define RQ. What is its value for fats?
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called the Respiratory Quotient (RQ) or respiratory ratio.

Question 11.
What is oxidative phosphorylation?
Answer:
Synthesis of ATP from ADP and iP coupled to electron transport from substrate to molecular oxygen is called oxidative phosphorylation.

Question 12.
What is the significance of stepwise release of energy in respiration?
Answer:

Question 13.
Find the correct ascending sequence of the following, on the basis of energy released in respiratory oxidation.
a) 1 gm of fat
b) 1 gm of protein
c) 1 gm of glucose
d) 0.5 gm of protein + 0.5 gm of glucose
Answer:
Ascending order
a) 1 gm of fat
b) 1 gm of protein
c) 0.5 gm of protein + 0.5 gm of glucose
d) 1 gm of glucose

Question 14.
Name the products, respectively, in aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast.
Answer:
The product of glycolysis in aerobic skeleton muscle is — The product of anaerobic fermentation in yeast is C02 and ethanol.

Question 15.
If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese, sandwich, which might have more energy. Why?
Answer:
Energy is released quickly in case of glucose or fruit juice, it is a quick process.

Question 16.
In a way green plants and cyanobacteria have synthesised ail the food on earth. Comment.
Answer:
Green plants and cyanobacteria prepare food materials by photosynthesis. They are producers. Consumers depend on them directly or indirectly. Thus green plants and cyanobacteria prepare food materials but not all the food on earth.

Question 17.
It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Answer:
Red muscle fibres are striated muscle. They contain large number of mitochondria which release energy. They work very speed and active for longer period but with little force. As they work, they are well developed according to Lamark’s theory. Smooth muscle fibres works more longer periods than red muscle fibres.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 4th Lesson Photosynthesis in Higher Plants Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 4th Lesson Photosynthesis in Higher Plants

Very Short Answer Type Questions

Question 1.
Name the processes which take place in the grana and stroma regions of chloroplasts.
Answer:
1. Grana :
Light reactions – trapping the light energy by membrane systems and synthesis of ATP and NADPH.

2. Stroma :
Dark reactions – carbon fixation leading to synthesis of sugars.

Question 2.
Can chloroplasts be passed on to progeny? How?
Answer:

1. Yes, nearly equal number of chloroplasts among 2 daughter cells formed from a mother cell during mitosis.
2. Maternal inheritance (cytoplasm of egg) in a sexually reproducing plant.

Question 3.
Where does the photolysis of H₂O occur? What is its significance? [Mar. ’20, 17; May ’14]
Answer:

1. Membrane of grana thylakoid and is associated with PSII.
2. Oxygen liberated by splitting of H₂Ois the main source of atmospheric O₂, electrons released during photolysis replace those removed for PSI.

Question 4.
Where is the enzyme NADP reductase located? What is released if the proton gradient breaksdown?
Answer:

1. The enzyme NADP reductase is located on the stroma side of the grana thylakoid membrane.
2. Energy is released if the proton gradient breaks down and is stored in ATP.

Question 5.
Which tissue transports photosynthates ? What experiments prove this?
Answer:

1. Phloem
2. Ringing experiment

Question 6.
How many molecules of ATP and NADPH are needed to fix a molecule of CO₂ in C₃ plants? Where does this process occur?
Answer:

1. In C₃ plants, 3 molecules of ATP and 2 molecules of NADPH are required to fix one CO₂ molecule.
2. This process occurs by Calvin cycle in the stroma of chloroplast.

Question 7.
Explain the terms:
a) Hatch-Slack pathway
b) Calvin cycle
c) PEP carboxylase
d) Bundle sheath cells
Answer:
a) Hatch-Slack pathway :
It is also called as C₄ cycle because the first stable compound in this cycle is a four carbon compound OAA (Oxalo Acetic Acid).

b) Calvin cycle :
It is also called as C₃ pathway, because the first stable compound is this cycle is a three carbon compound PGA (phosphoglyceric acid)

c) PEP carboxylase :
The enzyme responsible for primary fixation of CO₂ in C₄ plants. It mediates the formation of C₄ acids by reaction of CO₂ with phosphoenolpyruvate.

d) Bundle sheath cells :
The large cells with thick walls impervious to gaseous exchange and no intercelliilo spaces present around the vascular bundles of the C₄ pathway plants. These cells are characterised by a large number of agranular chloroplasts.

Question 8.
What is the role of NADP reductase in the development of proton gradient?
Answer:

1. The NADP reductase enzyme is located on the stroma side of the membrane.
2. It mediates the reduction of NADP⁺ to NADH + H⁺, creates protos gradient across the granathylakoid membrane.

Question 9.
Mention the components of ATPase enzyme. What is their location? Which part of the enzyme shows conformational change?
Answer:

1. The ATPase enzyme consists of two parts : a) F₀ (stalk) is embedded in thylakoid membrane and b) F₁ (head) that protrudes into the stroma.
2. F₁ particle of the ATPase undergoes conformational changes to synthesize ATP molecules.

Question 10.
What products drive Calvin Cycle? What process regenerates them?
Answer:

1. ATP and NADPH
2. Light reaction – phostophosphorylation.

Question 11.
What is the basis for designating. C₃ and C₄ pathway of photosynthesis?
Answer:
1. C₃ Pathway :
In plants with this pathway, the first product of CO₂ fixation is a 3 Carbon containing acid (3 – phosphoglyceric acid / 3 – PGA).

2. C₄ Pathway :
In plants with this pathway, the first product of CO₂ fixation is a 4 Carbon containing acid (oxaloacetic acid / OAA).

Question 12.
Distinguish between action spectrum and absorption spectrum.
Answer:

1. Action spectrum is the graph showing rate of photosynthesis as a function of different wavelengths of light.
2. Absorption spectrum is the graph shows the absorption of different wavelengths of light by photosynthetic pigments.

Question 13.
Of the basic raw materials of photosynthesis, what is reduced? What is oxidised?
Answer:

1. CO₂ and H₂O are the two raw materials of photosynthesis.
2. CO₂ is reduced and H₂O is oxidized during photosynthesis.

Question 14.
Define the law of limiting factors proposed by Blackman. [Mar. 2019, May 2017]
Answer:
If a process (like photosynthesis) is conditioned as to its rapidity by a number of separate factors, the rate of the process is limited by the factor that is present in a relative minimal value.

Question 15.
What is Joseph Priestley’s contribution to the study of photosynthesis?
Answer:

1. Plants restore to the air whatever breathing animals and burning candles remove, i.e., O₂ is evolved during photosynthesis.
2. His experiments revealed the essential role of air in the growth of green plants.

Question 16.
Comment on the contribution of Van Niel to the understanding of photosynthesis.
Answer:

1. Experiments by Van Niel, on purple and green bacteria demonstrated that photosynthesis is essentially a light – dependent reaction.
2. Niel inferred that the O₂ evolved by the green plant comes from H₂O, not from CO₂.
   

Question 17.
With reference to photosystem, bring out the meaning of the terms a) antennae b) reaction centre.
Answer:
a) Antennae :
All the pigments of a photosynten that form a light harvesting system. Those pigments absorb different wavelengths of light,

b) Reaction centre :
A special chlorophyll a in a photosystem forms. It converts light energy into chemical energy. In PS I, the reaction centre is P700 that has an absorption peak at 700 nm. In PS II, the reaction centre is P680 that has an absorption maximum at 680nm.

Question 18.
Why is photosynthetic electron transport from H₂O to NADP⁺, named as Z-scheme?
Answer:

1. The electrons released by photolysis are transported to NADP⁺ via PS II, electron carriers, PS I and electron carriers in non cyclic manner.
2. In this, when all the carriers are placed in a sequence on a redox potential scale it appears like ‘Z’ and hence called as z – scheme.

Question 19.
What is the primary acceptor of CO₂ in C₃ plants? What is the first stable compound formed in a Calvin cycle?
Answer:

1. The primary acceptor of CO₂ in C₃ plants is RuBP (Rubilose bi phosphate).
2. The first stable compound formed in a Calvin cycle is PGA (phosphoglyceric acid).

Question 20.
What is the primary acceptor of CO₂ in C₄ plants? What is the first compound formed as a result of primary carboxylation in the C₄ pathway? [Mar. 2018]
Answer:

1. The primary acceptor of CO₂ in C₄ plants is PEP (phosphoenol pyruvate).
2. The first compound formed due to primary carboxylation in the C₄ pathway is OAA (oxaloacetic acid).

Short Answer Type Questions

Question 1.
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic CO₂ requirements?
Answer:

1. Succulent plants have only one kind of photosynthetic cell in which CO₂ is fixed during night and used to make glucose during day.
2. In succulent plants there is an alternative pathway of CO₂ fixatibn called Crassulacean Acid Metabolism (CAM). E.g.: Cacti Crassulaceae is one family of succulent plants.
3. CAM pathway is similar to C₄ pathway in that CO₂ is trapped by highly efficient PEP carboxylase during night time.
4. During day time, the malic acid undergoes oxidative carboxylation to form Pyruvic acid & CO₂. CO₂ is used for photosynthesis.

Question 2.
Chlorophyll ‘a’ is the primary pigment for light reaction. What are accessory pigments? What is their role in photosynthesis?
Answer:

1. Chlorophyll ‘a’ is the primary pigment for light reaction other thylakoid pigments like chlorophyll b, xanthophylls and carotenoids are accessory pigments.
2. Accessory pigments also absorb light and transferthe energy to chlorophyll a.
3. They utilise wider range of wavelength of incoming light for photosynthesis.
4. They also protect chlorophyll a from photo oxidation.

Question 3.
Does ‘dark reaction’ of photosynthesis require light? Explain.
Answer:

1. No. Dark reaction does not depend on the presence of light but is dependent on the products of the light reaction i.e., ATP and NADPH besides H₂O and CO₂.
2. ATP and NADPH are used to drive the process leading to the synthesis of food, sugars. This is called biosynthetic phase of photosynthesis.
3. By convention, it is called dark reaction. However this should not be taken to mean that they occur in darkness.
4. This could be verified. It is simple. Immediately after light becomes unavailable, the biosynthetic process continues for sometime and then stops. If then, light is made available, the synthesis starts again.

Question 4.
How are photosynthesis and respiration related to each other?
Answer:

1. Photosynthesis is an anabolic process. Simple substances like carbondioxide and water combine and yield complex carbohydrates. E.g.: Glucose
   Respiration is an catabolic process. Complex carbohydrates (food materials) are broken into simpler substances like CO₂ and H₂O.
2. Photosynthesis is a reductive process where as respiration is a oxidative process.
3. Photosynthesis is an endogonic process where as respiration is endergonic process.
4. In photosynthesis O₂ is a by product. In Respiration O₂ is utilised.
5. Thus photosynthesis is opposite to respiration Photosynthesis equation is
   
   Respiration equation is
   C₆H₁₂0, + 6O₂ → 6CO₂ + 6H₂O + 686 k.Cal

Question 5.
What conditions enable “RuBisCO” to function as oxygenase? Explain the ensuing process.
Answer:
RuBisCO is characterised by the fact that its active site can bind tb both CO₂ and O₂. It is the relative concentration of O₂ and CO₂ that determines which of the two will bind to the enzyme. If O₂ concentration is more, RuBisCO function as Oxygenase, it binds with O₂ and instead of forming 2 molecules of PGA, it forms one molecule of phosphoglycerate and phosphoglycolate. This pathway is called photorespiration.

In photorespiration pathway, sugars or ATP are not formed, moreover, there is a release of CO₂ with the utilisation of ATP. Therefore photorespiration is a wasteful process.

Question 6.
Why does the rate of photosynthesis decrease at higher temperatures?
Answer:

1. The effect of temperature is linked with the optimum range of enzymatic activity.
2. The dark reactions, being enzymatic, they are temperature controlled. Though the light reactions are also temperature sensitive they are affected to a much lesser extent.
3. Tropical plants (C₄ plants) respond to higher temperatures than temperate plants (C₃ plants).
4. As enzymes gets denatured at higher temperatures the rate of photosynthesis decreases.

Question 7.
Explain how, during light reaction of photosynthesis, ATP synthesis is a chemiosmotic phenomenon.
Answer:

1. ATP synthesis in the chloroplast is a chemiosmotic phenomenon.
2. Like in respiration, in photosynthesis also ATP synthesis is linked to development of a proton gradient across the membrane.
3. During splitting of water and also during oxidation of PQ. (plastoquinol) protons are released into the lumen of thylakoid.
4. Hence the concentration of protons is many times greater than that of stroma, thus leading to proton concentration gradient.

Because of the concentration difference the protons try to move into the stroma, but thylakoids are impermeable to protons (H⁺).

However an enzyme complex ATPase enzyme is present in the membrane. It allows th’e movement of H⁺ through it into the stroma.

The ATPase enzyme consists of two parts.

1. F₀ is embeded in the membrane and forms a transmembrane channel that carries out facilitated diffusion of protons across the membrane.
2. F₁ that comes out of the stroma.
3. The breakdown of the gradient provides enough energy to cause a conformational change in the F₁ particle of the ATPase, which makes the enzyme synthesis several molecules of energy – packed ATP.

Question 8.
Explain how Calvin worked out the complete biosynthetic pathway for the synthesis of sugar.
Answer:

1. The process leading to the synthesis of sugars is called biosynthetic phase of photosynthesis.
2. ATP and NADPH are used in biosynthetic phase CO₂ combines with H₂O toTorm (CH₂O) or sugar.
3. The use of radioactive ¹⁴C by him in algal photosynthesis led to the discovery that the first CO₂ fixation product was a 3 – carbon organic acid.
4. Calvin also contributed to working out the complete biosynthetic pathway. Hence it is called Calvin cycle.
5. Calvin cycle occurs in all photosynthetic plants.
6. Calvin cycle can be described in three stages.

a) Carboxylation :
During which CO₂ combines with ribulose 1, 5 bisphosphate and form 3 phosphoglyceric acid.

b) Reduction :
During which carbohydrate is formed at the expense of the photochemically made ATP and NADPH.

c) Regeneration :
During which the CO₂ acceptor ribulose-1, 5 bisphosphate is formed again so that the cycle continues.

Question 9.
Six turns of Calvin cycle are required to generate one molecule of glucose. Explain.
Answer:

1. With each turn of Calvin cycle, one CO₂ molecule enters the system.
2. With three turns, uptake of 3 molecules of CO₂ and one triose (G – 3P) is available.
3. Hence six turns, results in the uptake of six molecules of CO₂ accounts for two triose (G – 3P), ultimately forms a glucose molecule.
4. Thus the fixation of six molecules of CO₂ and 6 turns of the cycle are required for the removal (net gain) of two molecules of triose (= one glucose molecule) from the pathway.
5. For every CO₂ molecule entering the Calvin cycle 3 molecules of ATP and 2 of NADPH are required. To make one molecule of glucose 6 turns of the cycle are required. Then for 6CO₂, 18 ATP and 12 NADPH are required.

Question 10.
With the help of diagram, explain briefly the process of cyclic photo – phosphorylation. [May 2014]
Answer:

1. In cyclic photo phosphorylation only PS I is functional, the electron is circulated within the photosystem and the phosphorylation occurs due to the cyclic flow of electrons.
2. This takes place in stroma lamellae.
3. Electrons in the reaction centre of PS I gets excited when they receive red light of wavelength 700 nm.
4. In this process, the electrons from the reaction centre of P700 are conveyed to e- acceptor.
5. The excited electron is cycled back to PS I complex through electron transport system.
6. The cyclic flow results in the synthesis of ATP.
7. Cyclic photo phosphorylation also occurs when only light of wavelength beyond 680 nm is available.
8. In green plants, cyclic photo phosphorylation is an additional source of ATP required for chloroplast activities over and above that is required in the Calvin cycle.

Question 11.
In what type of plants do you come across ‘Kranz’ anatomy? To which conditions are those plants better adapted? How are these plants better adapted than the plants, which lack this anatomy?
Answer:
a) C₄ plants.
b) C₄ plants are better adapted to dry tropical regions, they tolerate high temperatures.
c) C₄ plants are special. They have a special type of leaf anatomy.
They tolerate higher temperatures.
They show a response to high light intensities.
They lack a process called photorespiration.
They have greater productivity of biomass.

Question 12.
Explain the structure of the chloroplast. Draw a neat labelled diagram.
Answer:
Chloroplasts are double membrane bound. The space limited by the inner membrane of the chloroplast is called stroma. In the stroma, flattened membrane sacs called Thylakoids are present. Thylakoids are arranged in stacks like a piles of coins called grana. In addition there are flat membranous tubules called the stroma lamellae connecting the thylakoids of the different grana. The space in the thylakoids is called lumen. The stroma contains enzymes for the synthesis of carbohydrates and protein Dark reaction takes place in stroma. It also contain circular DNA and ribosomes. Thylakoids contain photosynthetic pigment. In it light reaction occurs.

Diagrammatic representation of an electron micrograph of a section of chloroplast

Question 13.
Explain why 12 molecules of water are used as substrate, instead of 6 molecules of water, in the following equation.

Answer:
By the middle of nineteenth century the key features of plant photosynthesis were know, namely, that plants could use light energy to make carbohydrates from CO₂ and water.

In the equation when 6 molecules of water is used it denotes that O₂ is evolved from CO₂.

But Cornelius Van Neil of based on his studies (1897 -1985) purple and green bacteria demonstrated that photosynthesis is – light essentially a dependent reaction in which hydrogen from a suitable oxidisable compound reduces CO₂ to carbohydrates. This is expressed as

In green plants H₂O is the hydrogen donor and is oxidised to O₂. Some organisms do not release O₂ during photosynthesis. When H₂S is the hydrogen donor for purple and green sulphur bacteria, the ‘oxidation’ product is sulphur or sulphate depending on the arganism and not O₂. Hence Niel inferred that the O₂ evolved by the green plant comes from H₂O, not from CO₂. This was later proved by using isotopic techniques. The correct equation is

Question 14.
Compare and contrast the absorption spectrum of chlorophylls and carotenoids.
Answer:
Absorption spectrum is a graph showing the absorption of light by pigments at different wavelengths. Chlorophyll a shows the maximum absorption of light and also shows higher rate of photosynthesis in the blue and red regions. Hence we can conclude that chlorophyll a is the chief pigment associated with photosynthesis.

Carotenoid is an accessory pigment. It also absorbs light and transfer the energy to chlorophyll a.

Question 15.
Which group of plants exhibits two types of photosynthetic cells? What is the first product of carboxylation? What carboxylating enzyme is present in bundle sheath cells and mesophyll cells?
Answer:

1. C₄ plants exhibit two types of photosynthetic cells.
2. OAA (oxaloacetic acid) is the first product of carboxylation.
3. PEP carboxylase is present in mesophyll cells Ribulose biphosphate carboxylase- oxygenase (RuBisCO) is present in bundle sheath cells.

Question 16.
A cyclic process is occurring in a C₃ plant, which is light dependent and needs O₂. This process does not produce energy rather it consumes energy.
a) Can you name the given process?
b) Is it essential for survival?
c) What are the end products of this process?
d) Where does it occur?
Answer:
a) Yes, it is photorespiration.
b) Yes, though it is wasteful process, it is essential, as it protects from photo oxidative damage.
c) CO₂ is the end product.
d) It occurs in chloroplast, peroxysomes and mitochondria.

Question 17.
Suppose Euphorbia and Maize are grown in the tropical area.
a) Which one of them do you think will be able to survive under such conditions?
b) Which one of them is more efficient in terms of photosynthetic activity?
c) What differences do you think are there in the leaf?
Answer:
a) Euphorbia (C₄ plants grown in tropical area)
b) Maize CO₂ fixation is done both in Mesophyll cell and Bundle sheath cell.
c) Kranz anatomy in Maize leaf.

Long Answer Type Questions

Question 1.
The entire process of photosynthesis consists of a number of reactions. Where in the cell do each of these take place?
a) Synthesis of ATP and NADPH
b) Photolysis of water
c) Fixation of CO₂
d) Synthesis of sugar molecule
e) Synthesis of starch
Answer:
a) This occur in Grana thylakoid.
b) PS II – Oxygen Evolving Complex (OEC) is associated with the PS II, which itself is physically located on the inner side of the membrane of the thylakoid.
c) Stroma of chloroplast.
d) Cytoplasm
e) Cytoplasm

Question 2.
Which property of pigments is responsible for its ability to initiate the process of photosynthesis? Why is the rate of photosynthesis higher in red and blue regions of the spectrum of light?
Answer:

1. Ability to absorb light at specific wavelength initiates the process of photosynthesis.
2. Absorption spectrum of chlorophyll a show maximum absorption of light at blue and red regions.
3. Action spectrum shows the rate of photosynthesis it is maximum at blue and red region.
4. Absorption spectrum and action spectrum clearly indicate that chlorophyll a is the chief pigment associated with photosynthesis. The rate of photosynthesis is higher in red and blue regions of the spectrum due to it absorption of light wavelengths.

Question 3.
Under what conditions are C₄ plants superior to C₃?
Answer:

Because of these above differences C₄ plants are considered to be superior than C₃ plants.

4. a) How can we plot an action spectrum? What does the action spectrum indicate? Explain with an example.
b) How can we derive an absorption spectrum for any substance?
c) If chlorophyll ‘a’ is responsible for light reaction of photosynthesis, why do the action spectrum and absorption spectrum not overlap?
Answer:
a) Rate of photosynthesis is measured by O₂ released. By drawing a graph between the rate of photosynthesis on x-axis and the wavelength of light on y – axis we can plot an action spectrum.

It indicates the wavelengths at which maximum photosynthesis occurs in plants. Graph shows that chlorophyll a show maximum rate of photosynthesis in the blue and the red regions. Hence chlorophyll ‘a’ is considered as chief pigment associated with photosynthesis.

b) The absorption spectrum for any substance is the ability of the substance to absorb lights at different wavelength.

c) The action spectrum and absorption spectrum does not overlap because the rate of photosynthesis at different wave length is not uniform. Accessory pigment absorbs light at different wavelengths.

Question 5.
What are the important events and end products of light reactions?
Answer:
In light reaction, important events are light absorption, water splitting, oxygen release and the end products are ATP and NADPH.

Light absorption :
The molecules that absorb light are called pigments. The pigments are organised into two discrete photochemical light harvesting complexes (LHC) within the photosystem I (PS I) and photosystem II (PS II). Each photosystem has all the pigments forming a light harvesting system called antennae. The antennae serves to absorb radiant energy. They supply this energy finally to reaction centre. Reaction centre is chlorophyll ‘a’ molecule. It converts light energy into chemical energy. The reaction centre in PS I is P700. The reaction centre in PS II is P680.

Cyclic and Non-cyclic Photo phosphorylation :
The process of forming ATP by cells is named phosphorylation. As it is done in the presence of light, it is called photo phosphorylation. It is of two types.

1. Cyclic photo phosphorylation
2. Non-cyclic photo phosphorylation.

1) Cyclic photo – phosphorylation :

1. In cyclic photo phosphorylation only PS I is functional. The electron is circulated within the photosystem and the phosphorylation occurs due to cyclic flow of electrons.
2. It takes place in stroma lamellae.
3. The electrons in the reaction centre of PS I gets excited when they receive red light of wave length 700 nm.
4. In this process, the electrons from the reaction centre of PS I are conveyed to e“ acceptor.
5. The excited electron is cycled back to PS I through electron transport system.
6. The cyclic flow results in the synthesis of ATP.

2) Non-cyclic photo phosphorylation :

1. When the two photosystems work in a series, first PS I! and then PS I a process called non-cyclic photo phosphorylation occurs.
2. In PS II, the reaction centre chlorophyll a absorbs 680 nm wavelength of red light causing electrons to become excited and jump.
3. The electrons are picked up by an electron acceptor.
4. From there it passes through electron transport system consists of cytochrome and reaches PS I.
5. Simultaneously electrons in the reaction centre PS I also gets excited when they receive red light of 700 nm wavelength.
6. The electrons from the PS i are conveyed to e acceptor.
7. These electrons then are moved downhill again, to a molecule of energy rich NADP⁺.
8. The addition of these electrons reduces NADP⁺ to NADPH + H⁺.
   Non-cyclic photo phosphorylation is also called Z-scheme.

Splitting of water:

1. The electrons that are moved from PS II are supplied with electrons continuously by splitting of water.
2. The electrons removed from PS I are provided by PS II.
3. Water splitting complex (oxygen evolving complex) is responsible for photolysis of water.
   2H₂O → 4H⁺ + O₂ + 4e^–

Thus oxygen released is one of the net products of photosynthesis.
The end products of light reaction are ATP, NADPH and O₂.

Question 6.
Explain various aspects of Mitchell chemiosmotic hypothesis, with the help of diagrams.
Answer:
Mitchell proposed chemiosmotic hypothesis to explain ATP synthesis.
ATP synthesis is linked to development of proton gradient across the membrane.

a) Since splitting of the water molecule takes place on the inner side of the membrane, the protons or hydrogen ions that are produced by the splitting of water accumulate within the lumen of the thylakoids.

b) As electrons move through the transport chain, protons are transported across the membrane. This happens because the primary acceptor of electrons which are located towards the outer side of the membrane transfers its electron not to electron carrier, but to (H⁺) proton carrier (PQ). Hence it removes protons, from the stroma which transporting an electron. When this molecule passes its electron, to the electron carrier on the inner side of the membrane, the proton is released into the inner side or lumen of the membrane. Then proton gradient across the membrane increases due to quinones.

c) The NADP reductase enzyme is located on the stroma side of the membrane. Along with the electrons that come from the acceptor of electrons of PS I, protons are necessary for the reduction of NADP⁺ to N ADPH + H⁺. These protons are also removed from the stroma. Thus protons in the stroma decrease in number. Where as in lumen protons are accumulated. This creates a proton gradient across the thylakoid membrane and the pH in the lumen decreases.

The proton gradient is broken down due to the movement of protons across the membrane to the stroma through the transmembrane channel of ATPase. ATPase consists of two parts.

1. One is F₀. F₀ is embeded in the membrane that forms transmembrane channel it carries facilitated diffusion of protons across the membrane.
2. Other one is F₁ It protrudes on the outer surface of the thylakoid membrane.

Diffusion of protons across the membrane releases enough energy to activate ATPase enzyme that catalyses the formation of ATP.

Question 7.
Comment on the dual role of RuBisCO. What is the basis for its oxygenation activity? Why is this activity absent or negligible in C₄ plants?
Answer:
RuBisCO is the most abundant enzyme in the world. It is characterised by the fact that its active site can bind to both CO₂ and O₂. Thus it shows dual role. It is written a Ribulose bisphosphate carboxylase – oxygenase. Generally, RuBisCo has greater affinity for CO₂ than O₂. This binding is competitive. It is the relative concentration of O₂ and CO₂ that determines which of the two will bind to the enzyme.

In C₃ plants some O₂ does bind to RuBisCO and hence CO₂ fixation is reduced. Here RuBP instead of forming 2 molecules of PGA forms one molecule of phosphoglycerate and phosphoglycelate. This pathway is called photorespiration. In this photorespiration sugar or ATP or NADP are not formed. Moreover, CO₂ is released with the utilisation of ATP molecule. Hence it is wasteful process.

In C₄ plants photorespiration is negligible or absent because mechanism that increases the concentration of CO₂ at the enzyme site. C₄ acid from the mesophyll is broken down in the bundle sheath cells to release CO₂. This results in increasing the intracellular concentration of CO₂. This makes RuBisCO function as carboxylase, minimising oxygenase activity.

Intext Question Answers

Question 1.
By looking at a plant externally, can you tell whether a plant is C₃ or C₄? Why and how?
Answer:
In C₃ plants photorespiration occurs. In C₃ plants transpiration is more. They occur in all types climates. They have much lower temperature optimum. C₃ plants respond to increased CO₂ concentration. C₄ plants are found in dry tropical and subtropical regions. E.g. : Sugarcane, maize, sorghum, amaranthus etc. They tolerate higher temperatures. They show a response to high light intensities. They lack photorespiration. They have greater productivity of biomass. They transpire less.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Answer:
By looking at the vertical section of leaf we can tell whether a plant is C₃ or C₄. The C₄ plant leaves have Kranz anatomy. Large cells around the vascular bundle called bundle sheath cells are present. The bundle sheath cells have a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.
E.g.: Maize, sorghum etc.

In C₃ plants, chloroplast dimorphism is not present, cells participating will have only one type of chloroplast. In C₄ plants, chloroplast dimorphism is present. Cells participating will have two types of chloroplasts (agranal and granal).

Question 3.
Even though a very few cells in a C₄ plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?
Answer:
RuBisCO has an active site which can bind to both CO₂ and O2. RuBisCO has greater affinity for CO₂ than for O₂. It is the relative concentration of 02 and C02 that determine which of the two will bind to the enzyme. In C₄ cycle, RuBisCO is present in bundle sheath cells, C02 concentration is more in bundle sheath cells. Hence productivity is high. In C₃ cycle some O₂ does bind to RuBisCO, leading to photorespiration. So CO₂ fixation is decreased.

Question 4.
RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C₄ plants?
Answer:
RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. In C₄ plants RuBisCO is present only in bundle sheath cells. So oxygenation of RuBisCO is avoided. These plants have CO₂ concentrating mechanism. That is why RuBisCO carries more carboxylation in C₄ plants.

Question 5.
Suppose there were plants that had a high concentration of chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
Answer:
Plants having high concentration of chlorophyll b, but lacked chlorophyll a would not carry photosynthesis. Chlorophyll a shows maximum absorption in the blue and red regions where maximum photosynthesis occurs. Hence chlorophyll a is the chief pigment of photosynthesis. Plants have chlorophyll b and other accessory pigments to absorb light and transfer the energy to chlorophyll a and protect chlorophyll a from photo oxidation.

Question 6.
Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
Answer:
Colour of the leaf is not due to single pigment but due to four pigments. We can separate these pigments by chromatographic separation.

1. Chlorophyll a – bright or blue green
2. Chlorophyll b – yellow green
3. Xanthophylls – yellow
4. Carotenoids – yellow to yellow orange
   Chlorophyll a is the most stable pigment.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Answer:
The leaves of the plant potted in the sunlight will be more green as compared to the leaves of the plant potted in the shade. This is because the amount of chlorophyll in a leaf has a direct relationship with the rate of photosynthesis. In dark, the photosynthesis decreases. So it appears pale.

Question 8.
Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions :
a) At which point/s (A, B or C) in the curve is light a limiting factor?
b) What could be the limiting factor/s in region A?
c) What do C and D represent on the curve?

Answer:
a) At higher light intensities the rate of photosynthesis become limiting factor – A/B.

b) Light

c) Saturation of light (C).
The rate of photosynthesis is not increased by increasing intensity of light (D).

Question 9.
Give comparison between the following :
a) C₃ and C₄ pathways.
b) Cyclic and non-cyclic photophosphorylation,
c) Anatomy of leaf in C₃ and C₄ paints.
Answer:

a)

b)

c)

Question 10.
Cyanobacteria and some other photosynthetic bacteria do not have chioroplasts. How do they conduct photosynthesis?
Answer:
Due to presence of chlorophyll pigments bacteria xanthophyll, phycobillins. They conduct photosynthesis in cyanobacteria. In xanthophyll, photosynthetic bacteria have bacterial chlorophyll, chlorobial chlorophyll conducts photosynthesis.

Question 11.
Does moonlight support photosynthesis?
Answer:
Yes, at very low rate.

Question 12.
Why does photorespiration not occur in C₄ plants?
Answer:
In C₄ plants RuBisCO is present only in bundle sheath cells. So oxygenation of RuBisCO is avoided. So photorespiration does not occur.

Question 13.
Tomatoes, chillis and carrots are red in colour due to the presence of one pigment. Name the pigment. Is it a photosynthetic pigment?
Answer:

1. Carotenoids.
2. It absorbs light of different wavelength and also protect of photo-oxidation.

Question 14.
If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as supplement to maintain its growth or survival?
Answer:
No, it cannot carry photosynthesis.
Yes, artificial light.

Question 15.
Photosynthetic organisms occur at different depths in the ocean. Do they receive qualitatively and quantitatively the same light? How do they adapt to carry out photosynthesis under these conditions?
Answer:
No, they don’t receive same light.
Diffuse light.

Question 16.
In tropical rain forests, the canopy is thick and shorter plants growing below it, receive filtered light. How are they able to carry out photosynthesis?
Answer:
They get modified into epiphytes.

Question 17.
Why do you believe chloroplast and mitochondria to be semi-autonomous organelle.
Answer:
Due to the presence of circular double stranded DNA.

Question 18.
Is it correct to say that photosynthesis occurs only in the leaves of a plant ? Besides leaves, what are the other parts that may be capable of carrying out photosynthesis ? Justify.
Answer:
Young plants which are green in colour Herbs.
Herbs contain green colour stem.
Photosynthetic roots. E.g.: Taeniophyllum.

Question 19.
What can we conclude from the statement that the action and absorption spectrum of photosynthesis overlap? At which wavelength do they show peaks?
Answer:
Maximum absorption by chlorophyll a and the rate of photosynthesis is highest in the blue and the red regions. It is at its peak in blue and red wavelengths.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 3rd Lesson Enzymes Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 3rd Lesson Enzymes

Very Short Answer Type Questions

Question 1.
How are prosthetic groups different from co-factors?
Answer:
1. Prosthetic group :
An organic co-factor that is tightly bound to the apoenzyme.
Eg. Hemee in peroxidase enzyme.

2. Cofactor :
Non protein part of a holo enzyme. It may be a metal ion or orgain compound.
Eg. Zinc in carboxy peptidase.

Question 2.
What is meant by ‘feedback inhibition’?
Answer:

1. Feedback inhibition: The end product of a chain of enzyme catalysed reactions inhibits the enzyme of the first reaction as part of homeostatic control of metabolism.
2. Eg. Pyruvic acid in Glycolysis.

Question 3.
Why are ‘oxido reductases’, so named?
Answer:
Oxido reductases are the enzymes which catalyse oxido reduction between two substrates S and S.
E.g.: S reduced + S’ oxidised → S oxidised + S’

Question 4.
Distinguish between apoenzyme and cofactor. [March 2014]
Answer:

1. The protein part of a holoenzyme is called apoenzyme.
2. The non-protein part of a holoenzyme is called cofactor, it may be a metal ion or an organic compound.

Question 5.
What are competitive enzyme inhibitors? Mention one example. [May 2014]
Answer:

1. Competetive inhibitor: An inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme
2. E.g.: Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure.

Question 6.
What are non-competitive enzyme inhibitors? Mention one example.
Answer:

1. Non-competitive enzyme inhibitor has no structural similarity with the substrate and forms an enzyme inhibitor complex at a point other than its active site, So that the globular structure of the enzyme is changed and catalysis cannot take place.
2. E.g.: Metal ions of copper, mercury, silver, etc.

Question 7.
What do the four digits of an enzyme code indicate?
Answer:

1. The four digits of an enzyme code helps to identify individual enzyme.
2. For example : Glucose-6- phosphotransferase has the enzyme code (E.C.) 2.7.I.2. The first, second, third and fourth of the code indicate the major class, subclass, sub-sub class and serial number of the enzyme respectively.

Question 8.
Who proposed ‘Lock and Key hypothesis’ and Induced fit hypothesis?
Answer:

1. ‘Lock and Key’hypothesis was proposed by Emil frsher. (1884)
2. ‘Induced – Fit’ hypothesis was proposed by Daniel E. Koshland. (1973)

Short Answer Type Questions

Question 1.
Explain how pH affects enzyme activity with the help of a graphical representation.
Answer:

1. Generally enzymes function in a narrow range of pH value.
2. Each enzyme shows its highest activity at a particular pH called optimum pH.
3. Activity declines both below and above the optimum value.

Question 2.
Explain the importance of (ES) complex formation.
Answer:
The formation of (ES) complex is essential for catalysis.

1. First, the substrate binds to the active site of the enzyme, fitting into the active site.
2. The binding of the substrate induces the enzyme to alter its shape; fitting more tightly around the substrate.
3. The active site of the enzyme, now in close proximity to the substrate, breaks the chemical bonds of the substrate and the new enzyme product complex is formed.
4. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and runs through the catalytic cycle once again.

Question 3.
Write briefly about enzyme inhibitors. [Mar. 2019, ’18, ’17 ; May ’17]
Answer:
a) The chemicals that block the enzyme activity are called inhibitors. The process is called inhibition.
b) Inhibitors are three types. They are
i) Competitive inhibitors
ii) Non-competitive inhibitors
iii) Feedback inhibitors

i) Competitive inhibitors :
These are the substances which are structurally similar to substrate molecules and compete for the. active sites of an enzyme. E.g : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure.

ii) Non-competitive inhibitors :
These chemicals do not resemble the substrate in structure. They bind to an enzyme at locations other than the active sites and makes it inactive. So no new products are formed. E.g: Metal ions of copper, mercury, silver etc.

iii) Feedback inhibitors :
In many biochemical reactions, the accumulation of end products of reactions will inhibit the first step of reaction. It is a part of homeostatic control of metabolism.

Question 4.
Explain different types of cofactors.
Answer:
a) Enzymes having non-protein part along with protein part is called Holoenzyme. The non-protein part is called co-factor and protein part is called apoenzyme.
Co-factor + Apoenzyme = Holoenzyme
b) Co-factors are of three kinds. They are
i) Prosthetic groups
ii) Co-enzymes
iii) Metal ions

i) Prosthetic groups :
Prosthetic groups are the organic co-factors which are tightly bound to the apoenzyme. For example : In peroxidase and catalase, which catalyse the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is the active part of enzyme.

ii) Co-enzymes :
Coenzymes are the organic molecules which are loosely associated with the apoenzyme. These co-enzymes are derived from wajer soluble vitamins. Eg: Coenzyme NAD and NADP contain vitamin niacin.

iii) Metal ions :
Mostly enzymes require metal ions for their activity which form coordinate bonds with side chains at the active site and at the same time form one or more coordination bonds with the substrate. E.g. : Zinc is a cofactor for the proteolytic enzyme carboxypeptidase.

Long Answer Type Questions

Question 1.
Write an account of the classification of enzyme.
Answer:
Enzymes have been classified into different groups based on the type of reactions they catalyse. Enzymes are divided into 6 classes. Each class is again divided into sub-class and sub-subclasses. They are

1) Oxidoreductases / dehydrogenases :
Enzymes which catalyse oxidoreduction between two substrates S and S’.
E.g.: S.reduced + S’ oxidised → S oxidised + S’ reduced

2) Transferases :
Enzymes catalysing a transfer of a group G (other than hydrogen) between a pair of substrate S and S’.
E.g.: S — G + S’ → S + S’ — G

3) Hydrolases :
Enzymes catalysing hydrolysis of ester, ether, peptide, glycosidic, C – C, C – halide or P – N bonds.

4) Lyases :
Enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds.

5) Isomerases :
Includes all enzymes catalysing inter – conversion of optical, geometric or positional isomers.

6) Ligases :
Enzymes catalysing the linking together of 2 compounds. E.g.: enzymes which catalyse joining of C – O, C – S, C – N, P – O etc., bonds.

The above classification provides for a four digit code to identify individual enzymes. For E.g.: Glucose – 6 – Phosphotransferase has the enzyme code (E.C.) 2,7.1.2
The first digit of code indicates the major class.
The second digit of code indicates the subclass.
The third digit of code indicates the sub-subclass.
The fourth digit indicates the serial number of the enzyme in a particular sub-subclass.

Question 2.
Explain the mechanism of enzyme action. [Mar. 2020]
Answer:

1. During the enzyme action the enzyme (E) combines with its specific substrates (S) to form a enzyme – substrate complex (E – S) which is short – lived.
2. Energy that is required for a substrate to react inorder to get converted into end product is called “activation energy”.
3. This activation energy is available in different forms like heat, ATP etc.
4. The activation energy of the formation of this E – S complex is low, hence many molecules can react and participate in the reaction, leading to the formation of products (P).
5. The E – S complex dissociates into its products P and the unchanged enzyme E with an intermediate formation of the enzyme – product complex (EP).
6. The formation of the ES complex is essential for catalysis. E + S → (ES) → (EP) → E + P
7. Formation of (ES) complex has been explained with ‘Lock and Key’ hypothesis by Emil Fisher and later with “Induced Fit” hypothesis by Daniel E. Koshland (1973).
8. According to this theory every enzyme possess “Active sites”.
9. The substrate (S) gets attached to the active site of the enzyme (E) and forms an enzyme substrate (ES) complex.
10. The enzyme remains unchanged while the substrate is broken into products (P).

Intext Question Answers

Question 1.
Enumerate the properties of enzymes.
Answer:
Enzymes show following properties :
a) Catalytic property :
An enzyme is organic catalyst. It does not undergo any change during a reaction, catalysed by it. It only speeds up a rate of a reaction.

b) Specificity :
Enzymes are specific and act only on specific substances. For example, sucrase acts only on sucrose.

c) Active in minute quantity :
Enzymes are active in small quantities. The number of substrate molecules to be converted into products by one molecule of enzyme per minute is called turn over number.

d) Reversibility :
Most of the enzymes are reversible in their action. They can speed up a particular reaction either in forward or in backward direction.

e) Thermolability :
Enzymes are heat sensitive. At high temperature, enzymes are denatured and at low temperature, they are inactive because enzymes are Chemically proteins.

f) Sensitivity of pH :
The enzyme activity is by pH controlled by pH concentration. Most of the enzymes work at neutral pH.

g) Proteinaceous nature :
All enzymes are chemically proteins, having high molecular weights, ranging from 10,000 to several million deltron. Based on the composition enzymes are of two types. 1) Simple enzymes 2) Conjugated or Holoenzymes.

Question 2.
What is Michaelis constant?
Answer:
a) The Michaelis constant Km. is very important in determining enzyme substrate interaction.
b) The value of enzyme range widely and often dependent on environmental conditions such as pH, temperature and ionic strength.
c) The Michaelis constant is able to detect two factors.

One is the concentration of the substrate when the reaction velocity is half that of the maximal velocity, thus Michaelis constant measures the concentration of substrate required for a significant catalysis to take place.

Second is the Michaelis constant is able to detect the strength of the enzyme-substrate complex (ES).

Question 3.
Distinguish between feedback inhibition and allosteric inhibition.
Answer:
In many biochemical reactions, the accumulation of end products of reactions will inhibit the first step of reaction. This is called Feedback inhibition.

Allosteric inhibition :
Some enzymes, possess allosteric (alios = other; steros = site) sites. The enzymes which possess allosteric sites are known as allosteric enzymes. The binding of substance to allosteric site may stimulate or inhibit enzyme action. The substances that reduce the activity of an enzyme by binding at allosteric site are known as allosteric inhibitors.

Question 4.
What are isoenzymes?
Answer:
Isoenzymes are proteins with different structure which catalyze the same reaction.

Question 5.
What is turnover number? What is the fastest acting enzyme?
Answer:
The number of substrate molecules converted into products by one molecule of an enzyme in one minute time is called turn over number (TON). The fastest acting enzyme is carbonic- anhydrase.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 2nd Lesson Mineral Nutrition Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 2nd Lesson Mineral Nutrition

Very Short Answer Type Questions

Question 1.
Define hydroponics.
Answer:

1. The technique of growing plants in a specified nutrient solution is known as hydroponics.
2. Julius Von Sachs (1860) demonstrated this technique for the first time.

Question 2.
How do you categorize a particular essential element as a macro or micronutrient?
Answer:
1. Macronutients :
Elements needed in high quantities and are present is large amounts (in excess of lOmmole Kg^-1 of dry matter) in plant tissues.

2. Micronutrients :
Trace elements that are needed in very small amounts and present is less than lOmmole Kg^-1 of dry matter in plants.

Question 3.
Give two examples of essential elements that act as activators for enzymes.
Answer:
1. Mg 2⁺ is an activator of RuBp carboxylase oxygenase.
2. Zn 2⁺ is an activator of alcohol dehydrogenase.

Question 4.
Name the essential mineral elements that play an important role in photolysis of water.
Answer:
Calcium, Manganese and chlorine are the mineral elements that help is splitting of water molecules to liberate O₂ during photosynthesis.

Question 5.
Out of the 17 essential elements which elements are called non-mineral essential elements?
Answer:

1. Carbon, Hydrogen and Oxygen obtained from CO₂ and H₂O are called non – mineral essential elements.
2. These frame work elements are not absorbed from the soil as mineral nutrients. Name two amino acids in which sulphur is present.

Question 6.
Name two amino acids in which sulphur is present.
Answer:

1. Cysteine and methionine are sulphur containing amino acids.
2. They help in formation of disulphide bridges and stabilizing the protein structure.

Question 7.
When is an essential element said to be deficient?
Answer:

1. The concentration of an essential element below which plant growth is retarded is called critical concentration.
2. The element is said to be deficient when present below the critical concentration.

Question 8.
Name two elements whose symptoms of deficiency first appear in younger leaves.
Answer:

1. Sulphur and Iron.
2. Deficiency symptoms of immobile elements first appear in young leaves.

Question 9.
Explain the role of the pink colour pigment in the root nodule of legume plants. What is it called?
Answer:

1. The pink colour pigment in root nodule of legume plant is Leg-haemoglobin.
2. Enzyme nitrogenase is highly sensitive to the molecular oxygen and requires anaerobic condition. Leg haemoglobin which is an oxygen scavenger protects enzyme nitrogenase from oxygen.

Question 10.
Excess Mn in soils leads to deficiency of Ca, Mg and Fe. Justify.
Answer:

1. Manganese competes with iron and Mg for uptake and with magnesium for binding with enzymes. It also inhibits calcium translocation in the shoot apex.
2. Therefore, excess of manganese induce deficiencies of iron, magnesium and calcium.

Question 11.
What acts as a reservoir of essential elements for plants? By what process is it formed?
Answer:

1. Soil acts as a reservoir of essential elements for plants.
2. It is formed due to weathering and breakdown of rocks.

Question 12.
Which element is regarded as the 17^th essential element? Name a disease caused by its deficiency.
Answer:

1. Nickel is regarded as the 17^th essential element.
2. Its deficiency causes mouse ear in pecan.

Question 13.
Nitrogen fixation is shown by prokaryotes only. Why not by eukaryotes?
Answer:

1. Nitrogenase enzyme, which is capable of nitrogen reduction is present exclusively in prokaryotes.
2. In eukaryotes, nitrogenase enzyme is absent and so they cannot fix Nitrogen.

Question 14.
Give an example for each of the aerobic and anaerobic nitrogen fixing prokaryotes.
Answer:

1. Aerobic nitrogen fixing prokaryotes. Eg: Azotobacter, Beijernickia.
2. Anaerobic nitrogen fixing prokaryotes. Eg: Rhodospirillum.

Question 15.
Non-legume plants also form root nodules. Justify.
Answer:

1. The microbe Frankia, produces nitrogen fixing nodules on the roots of non-leguminous plants Eg: Alnus.
2. Frankia is free living in soil, but as a symbiont can fix atmospheric nitrogen.

Question 16.
Name the essential elements present in nitrogenase enzyme. What type of essential elements are they?
Answer:

1. Nitrogenase enzyme consists of 2 essential elements Molybdenum and Iron.
2. They are micronutrients.

Question 17.
Write the balanced equation of nitrogen fixation.
Answer:
N₂ + 8H⁺ + 8e^– + 16 ATP → 2NH₃ + H₂ + 16 ADP + 16 Pi

Question 18.
How many ATPs of energy is required to fix one molecule of atmospheric nitrogen by biological mode? What is the source of that energy?
Answer:

1. 16 ATP molecules, are required to fix one N₂ molecule and to produce 2 NH₃ molecules.
2. Source of that energy is obtained from the respiration of the host cells.

Question 19.
Why are amides transported through xylem?
Answer:

1. Amides contain more nitrogen than amino acids and are highly toxic to living cells.
2. Hence, they are transported to other parts of the plant through non living xylem vessels. Name any two essential elements and the deficiency diseases caused by them.

Question 20.
Name any two essential elements and the deficiency diseases caused by them.
Answer:

1. Zinc deficiency – mottled leaf in citrus.
2. Molybdenum deficiency – whip tail in cauliflower.

Short Answer Type Questions

Question 1.
“All elements that are present in a plant need not be essential for its survival.” Justify.
Answer:
Most of the elements present in the soil enter plants through roots. All the elements that are present in a plant need not be essential for its survival. The criteria for essentiality of elements are given below :

1. The element must be absolutely necessary for supporting normal growth and reproduction. In the absence of the element the plant do not complete their life cycle or set the seeds.
2. The requirement of the element must be specific and not replaceable by another element. In other words, deficiency of any one element cannot be met by supplying some other element.
3. The element must be directly involved in the metabolism of the plant.

Question 2.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
Answer:
Deficiency symptoms in plants are chlorosis, necrosis, inhibition of cell division, delay flowering.

Chlorosis :
Chlorosis is the loss of chlorophyll leading to yellowing in leaves. This symptom is caused by the deficiency of elements N, K, Mg, S, Fe, Mn, Zn and Mo.

Necrosis :
It is the death of tissue, particularly leaf tissue. It is due to the deficiency of Ca, Mg, Cu, K.

Inhibition of cell division :
Cell division stops. It is due to the deficiency of lack or low level of N, K, S, Mo.

Delay flowering :
Flowering is delayed due to low concentration of N, S, Mo.

Mottled leaf in citrus :
It is due to deficiency of Zn,

Question 3.
Explain the steps involved in the formation of root nodule. [Mar. 2019, 18, 17, ’14]
Answer:
Steps involved in the formation of root nodule :

1. Rhizobia attracted by the sugars, amino acids etc., released by the host legume, multiply and colonise the surroundings of roots and get attached to epidermal and root hair cells.
2. The root hairs curl and the bacteria invade the root hair.
3. An infection thread is produced, carrying the bacteria into the cortex of the root.
4. Bacteria initiate nodule formation in the cortex of the root.
5. Then the bacteria released from the thread into the cortical cells of the host stimulate the host cells to divide. Thus leads to the differentiation of specialised nitrogen fixing cells.
6. The nodule thus formed establishes a direct vascular connection with the host for the exchange of nutrients.

Development of root nodules in soyabean :
(a) Rhizobium bacteria contact a susceptible root hair, divide near it. (b) Successful infection of the root hair causes it to curl, (c) Infected thread carries the bacteria to the inner cortex. The bacteria get modified into rod-shaped bacteroids and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation, (d) A mature nodule is complete with vascular tissues continuous with those of the root.

Question 4.
Some angiospermic plants have adapted to absorb molecular nitrogen from atmosphere. Explain, citing two examples.
Answer:

1. Plants cannot use atmospheric nitrogen directly. But some of the plants in association with N₂ – fixing bacteria, especially roots of legumes, can fix atmospheric nitrogen.
2. Leguminous plants (Eg : alfalfa, sweet clover, sweet pea etc.,) are associated with Rhizobium bacteria in nodular roots. Non-leguminous plants (Eg: Alnus) are associated with Frankia in nodula roots.
3. Both Rhizobium and Frankia are free – living in soil but as symbionts can fix atmospheric nitrogen.
4. Molecular nitrogen (N₂) is available abundant in air. Only prokaryotic species are capable of nitrogen fixation.
   Reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation.
5. The enzyme nitrogenase is capable of nitrogen reduction. It is present only in prokaryotes. Such microbes are called N₂ – fixers.
   

Question 5.
Write in brief how plants synthesize amino acids.
Answer:
NH₄⁺ is used to synthesize amino acids in plants. There are two main ways in which this can take place :
1) Reductive amination
2) Transamination.

1) Reductive amination :
In these processes, ammonia reacts with α-ketoglutaric acid and forms glutamic acid as indicated in the equation given below

2) Transamination :
It involves the transfer of an amino group from an amino acid to the keto group of a keto acid. Glutamic acid is the main amino acid from which the transfer of NH₂, the amino group, takes place and other amino acids are formed through transamination. The enzyme transaminase catalyses all such reactions. For example

Question 6.
What will happen if a healthy plant is supplied with excess essential elements? Explain.
Answer:

1. Excess essential element may inhibit the uptake of another element.
2. For example, the prominent symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins.
3. Manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes.
4. Manganese also inhibits calcium translocation in the shoot apex.
5. Therefore, excess of manganese induce deficiencies of iron, magnesium and calcium.

Question 7.
Explain in brief how plants absorb essential elements.
Answer:
1. The process of absorption of elements can be demarcated into two main phases. In the first phase, there is an initial rapid up take of ions into the free space or outer space of cells – the apoplast. It is a passive process. In the second phase of uptake, the ions are taken in slowly into the ‘inner space’ – the symplast of the cells.

2. The movement of ions into the apoplast along the concentration gradient is passive process. The movement of ions to and from the symplast against the concentration gradient requires the expenditure of metabolic energy. It is an active process.

Question 8.
Nitrogen is fixed into the soil not only by biological processes. Elaborate. [May 2014]
Answer:
Nitrogen is fixed into the soil not only by biological processes, it is also done by abiological processes.

a) In nature, due to thunders and lightening dinitrogen is converted into nitrogen oxides.
N₂ + O₂ → 2NO (nitric oxide)
2NO + O₂ → 2NO₂ (nitrogen dioxide)
2NO₂ + H₂O → HNO₂ / HNO₃ (nitrous/nitric acid)
HNO₃ + Ca/K salts → Ca/K nitrates
When nitrous/nitric acid reach the soil reacts with the alkali radical to form nitrates.

b) Industrial combustions, forest fires, automobile exhausts and power-generating stations are also the sources of atmospheric nitrogen oxides.

Long Answer Type Questions

Question 1.
Explain the nitrogen cycle, giving relevant examples. [Mar. 2020]
Answer:
The cyclic movement of nitrogen from the atmosphere to soil and from soil back into the atmosphere through plants, animals and micro-organisms is termed as nitrogen cycle. Nitrogen cycle involves five steps :

1. Nitrogen fixation
2. Nitrogen assimilation
3. Ammonification
4. Nitrification
5. Denitrification

1) Nitrogen fixation :
The process of conversion of molecular nitrogen (N₂) to ammonia or nitrogen oxides, nitrites and nitrates is termed as nitrogen – fixation. It occurs both by biological and physical method.

Biological method :
Conversion of molecular nitrogen into ammonia by prokaryotes is called biological methed.

Eg : Free – living nitrogen – fixing aerobic microbes – Azotobacter – Beijernickia.
Free living nitrogen – fixing anaerobic microbes – Rhodospirillum
Cyanobacteria (blue green algae) – Nostoc & Anabaena
Symbiotic bacteria – Rhizobium (roots of leguminous plant)
Symbiotic bacteria – Frankia (roots of non – leguminous plant)

Physical or abiological method :
In nature lightning and ultraviolet radiation provide enough energy to convert nitrogen to nitrogen oxides (NO, NO₂, NO₃). Industrial combustions, forest fires, automobile exhausts and power – generating stations are also sources of atmospheric nitrogen oxides.
N₂ + O₂ → 2NO
2NO + O₂ → 2NO₂
2NO₂ + H₂O → HNO₂ + HNO₃
HNO₃ + Ca/K salts → Ca or K nitrates

2) Nitrogen assimilation :

1. The process of absorbing nitrates, ammonia to produce organic nitrogen constitutes is called nitrogen assimilation.
2. Nitrates and ammonia formed in 1st step are absorbed by plants and converted into organic nitrogen constitute like proteins, enzymes, nucleic acid etc.
3. When plants are eaten by animats, this organic nitrogen is passed into animal body.

3) Ammonification :

1. Decomposition of organic nitrogen of dead plants and animals into ammonia is called ammonification.
2. Bacteria responsible for this are called ammonifying bacteria.

4) Nitrification :
The conversion of ammonia into nitrites and nitrates by bacteria is called nitrification. Such bacteria are called nitrifying bacteria (chemo auto trophs). It occurs in two steps.

1. Ammonia is first oxidised to nitrite by the bacteria Nitrosomonas and Nitrococcus.
   2NH₃ + 3O₂ → 2NO₂^– + 2H⁺ + 2H₂O
2. The nitrite is further oxidised to nitrate with the help of the bacterium Nitrobacter.
   2NO₂^– + O₂ → 2NO₃^–

The nitrate thus formed is absorbed by plants and is transported to the leaves. In leaves, it is reduced to form ammonia that finally forms the amine group of amino acids.

5) Denitrification :
Conversion of nitrates from soil into molecular nitrogen is called denitrification. Denitrification is done by bacteria like Pseudomonas and Thiobacillus.

Question 2.
Trace the events starting from the coming in contact of Rhizobiurri with a leguminous root till nodule formation. Add a note on the importance of leg haemoglobin.
Answer:
Various stages of nodule formation :

1. Roots of legume plant secrete sugars, amino acids etc.
2. Attracted by this, Rhizobium bacteria move to the root. It multiply and colonise the surroundings of roots and get attached to epidermal and root hair cells.
3. The root – hairs curl and the bacteria invade the root – hair.
4. An infection thread is produced, carrying the bacteria into the cortex of the root.
5. Bacteria initiate nodule formation in the cortex of the root.
6. Then the bacteria released from the thread into the cortical cells of the host stimulate the host cells to divide. Thus leads to the differentiation of specialised nitrogen fixing cells.
7. The nodule thus formed establishes a direct vascular connection with the host for exchange of nutrients.

Development of root nodules in soyabean :
(a) Rhizobium bacteria contact a susceptible root hair, divide near it. (b) Successful infection of the root hair causes it to curl, (c) Infected thread carries the bacteria to the inner cortex. The bacteria get modified into rod-shaped bacteroids and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation, (d) A mature nodule is complete with vascular tissues continuous with those of the root.

The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leg-haemoglobin. The enzyme nitrogenase is a Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia, the first stable product of nitrogen fixation.
N₂ + 8H⁺ + 8e^– + 16 ATP → 2NH₃ + H₂ + 16 ADP + 16 Pi

The enzyme nitrogenase is highly sensitive to the molecular oxygen, it requires anaerobic conditions. To protect these enzymes, the nodule contains an oxygen scavenger called leg -haemoglobin. These microbes live as aerobes under free living conditions but during nitrogen – fixing events, they adapt to anaerobic conditions, thus protecting the nitrogenase enzyme.

Intext Question Answers

Question 1.
Who should be credited for initiation of hydroponics?
Answer:
Julius von Sachs

Question 2.
Are all the essential elements required by plants mineral elements? Explain.
Answer:
Yes. Whenever the supply of an essential element becomes limited plant growth is retarded.

Question 3.
Which essential element is needed to activate the enzymes required for CO₂ fixation?
Answer:
Magnesium and Manganese.

Question 4.
Name a cation and an anion that maintain osmotic balance in cells.
Answer:
Potassium K⁺ and Chlorine Cl^–

Question 5.
Which element is required for the formation of mitotic spindle?
Answer:
Calcium

Question 6.
What is the role of sulphur in plant life?
Answer:
Sulphur is obtained in the form of sulphate SO₄^2- ions. Sulphur is present in two amino acids – cysteine and methionine and is the main constituent of several coenzymes, vitamins (thiamine, biotin, coenzyme A) and Ferredoxin. Sulphur forms disulphide bridges which help in stabilizing the protein structure.

Question 7.
Which microelement is required in more quantity than the other micronutrients ?
Answer:
Iron

Question 8.
Which element is necessary for the synthesis of the chief photosynthetic pigment without being its structural component?
Answer:
Iron and Mg

Question 9.
Which micronutrient necessary for photolysis of water is absorbed by plants in anionic form?
Answer:
Cl

Question 10.
Which enzyme is activated by the 17^th essential element?
Answer:
Nickel is the 17^th essential element which acts as an activator for Urease.

Question 11.
When is an element considered to be toxic?
Answer:
Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered to toxic.

Question 12.
Which element when supplied in excess leads to appearance of brown spots surrounded by chlorotic veins?
Answer:
Magnesium

Question 13.
Name an anaerobic, free living, photo-heterotrophic nitrogen fixing bacterium.
Answer:
Rhodospirillum

Question 14.
Which microorganism produces nitrogen-fixing nodules in Alnus?
Answer:
Frankia

Question 15.
When the cross section of root nodules of ground – nut plants are observed under microscope, they appear pinkish. Why?
Answer:
Due to presence of leg haemoglobin or leguminous haemoglobin.

Question 16.
Apart from the cortical cells, which other cells are stimulated to divide by the bacteroids inside the root nodules?
Answer:
Pericycle cells.

Question 17.
What is the ratio of electrons and protons required for the fixation of atmospheric mplecular nitrogen through biological mode?
Answer:
8 protons and 8 electrons i.e., 1:1 ratio.

Question 18.
What acts as oxygen scavenger in the legume-root nodule combination?
Answer:
Leg-haemoglobin

Question 19.
In what way does asparagine differ from aspartic acid?
Answer:
Asparagine is an amide found in plants as a structural part of protein. It is formed from aspartic acid by additional of another amino group.

Question 20.
Through which tissue the amino acids are transported inside the plant body?
Answer:
Amino acids are transported to other parts of the plant body via xylem vessels.

Question 21.
Plants like the Pitcher and Venus fly trap have special nutritional adaptations. Name the essential element and its source for which they show such adaptations.
Answer:

1. N₂ (Nitrogen)
2. Nitrogen is absorbed from the insect body.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 1st Lesson Transport in Plants Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 1st Lesson Transport in Plants

Very Short Answer Type Questions

Question 1.
What are porins? What role do they play in diffusion? [March 2018]
Answer:

1. Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria.
2. Molecules upto the size of small proteins to pass through the channels formed by porins due to facilitated diffusion.

Question 2.
Define water potential. What is the value of water potential of pure water?
Answer:

1. Water potential is a relative term, which refers to the chemical potential of pure water to that of chemical potential of water in a solution.
2. Water potential of pure water not under pressure is zero (0) at standard temperature.

Question 3.
Differentiate osmosis from diffusion. [March 2017]
Answer:
1. Osmosis :
Movement of water (solvent) from a region of its higher concentration to a region of lower concentration through a selectively permeable membrane, due to pressure and concentration gradient.

2. Diffusion :
Movement of gases and liquids from a region of thejr higher concentration to the region of lower concentration due to concentration gradient.

Question 4.
Compare transpiration and evaporation.
Answer:

1. Transpiration is a physiological process, in which water is lost as vapour from the plant surface.
2. Evaporation is a physical process, in which water is lost from any free surface in the form of vapour.

Question 5.
What are apoplast and symplast? [March 2019]
Answer:

1. Apoplast is the continuous system (network) of cell walls and intercellular spaces in a plant body. This allows movement of water molecules without crossing the cell membranes,
2. Symplast is the system of interconnected protoplasts by means of plasmodesmata. This allows movement of water from one cell to another through their cytoplasm.

Question 6.
How does guttation differ from transpiration?
Answer:
1. Guttations :
The loss of water from the leaves of grasses and many herbaceous plants In liquid form.

2. Transpirations :
The loss of water from the plant surface in the form of vapour.

Question 7.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:

1. Water potential increases, if a pressure greater than atmospheric pressure is applied to pure water or a solution.
2. It is equivalent to pumping water from one place to another.

Question 8.
Explain what will happen to a plant cell if it is kept in a solution having higher water potential?
Answer:

1. Endosmosis occurs when a plant cell is placed in a solution having higher water potential (hypotonic solution)
2. Hence, the cytoplasm build up a pressure against the cell wall, which is called turgor pressure.

Question 9.
What are the physical factors responsible for the ascent Of sap through xylem in plants?
Answer:
1. Cohesion :
Mutual attraction between water molecules.

2. Adhesion :
Attraction of water molecules to polar surfaces (such as the surface of tracheary elements)

3. Transpiration pull :
Driving force for upward movement of water.

Question 10.
Explain why xylem transport is unidirectional while that in phloem is bidirectional.
Answer:

1. Xylem transport is unidirectional due to one way flow of water in transpiration.
2. Food in phloem sap can be transported bidirectionally so long as there is a source of sugar and a sink which is able to use, store or remove the sugar.

Question 11.
With reference to transportation within plant cells, what are source and sink?
Answer:

1. Source is understood to be that part of the plant which prepare food i.e., the leaf.
2. Sink is the part that needs or store of food i.e., root, stem or fruits.

Question 12.
Does transpiration occur at night? Give an example.
Answer:

1. Yes, transpiration occurs at night is succulent plants (E.g.: Cacti, Bryophyllum)
2. Guard cells is those plants become turgid and open at night due to accumulation of organic acids.

Question 13.
Compare the pH of guard cells during the opening and closing of stomata.
Answer:

1. The pH of guard cells increases during the opening of stomata due to influx of K⁺ ions and efflux of H⁺ ions.
2. The pH of guard cells decreases during closing of stomata due to effux of K⁺ and influx of H⁺ ions.

Question 14.
In the wake of transpirational loss, why do the C₄ plants are more efficient than C₃ plants?
Answer:

1. C₄ plant loses only half as much water as a C₃ plant for the same amount of CO₂ fixed.
2. C₄ plants are more efficient than C₃ plants in minimizing water loss.

Question 15.
What is meant by transport saturation? How does it influence facilitated diffusion?
Answer:

1. Transport saturation : A protein transporters are being used (saturation) and so transport rate reaches maximum.
2. During transport saturation all the active sites of carrier proteins are occupied with transporting molecules.

Question 16.
Pressure potential in plant systems can be negative. Elaborate.
Answer:

1. Pressure potential in xylem is negative and plays a major role is upward water transport in stem.
2. The tension in water column in the xylem is negative (- Ψp) dute to transpirational pull.

Question 17.
How does ABA bring about the closure of stomata under water stress conditions? [Mar. 2020]
Answer:

1. Abscisic acid (ABA) is a natural anti-transpirant.
2. This drives the K⁺ ions out of guard cells under water stress conditions making them close.

Short Answer Type Questions

Question 1.
Define and explain water potential. [Mar. 2019, ’18]
Answer:
Water potential (Ψ_w) is a relative term to understand water movement. It represents the difference between chemical potential of water in a system and that of a pure water. Water molecules possess kinetic energy. The greater the concentration of water in a system, the greater is its kinetic energy or water potential. Hence, pure water will have greatest water potential. The water potential of pure water at standard temperatures which are not under any pressure, is taken to be zero. Solute potential (Ψ_s ) and pressure potential (Ψ_p) are the two main components that determine water potential.

Solute potential :
All solutions have a lower water potential than pure water. The magnitude of this lowering due to dissolution of a solute is called solute potential denoted by Ψ_s. It is always negative. The more the solute molecules, the lower is the Ψ_s. For a solution at atmospheric pressure water potential Ψ_w = solute potential Ψ_s.

Pressure potential :
The pressure which developed due to entry of water in a cell is called pressure potential. It is always positive, it is denoted as Ψ_p.

Water potential Ψ_w of a cell is affected by both solute potential and pressure potential.
Ψ_w = Ψ_s + Ψ_p

Question 2.
Write short notes on facilitated diffusion.
Answer:

1. Membrane protein provides sites at which certain molecules can cross the membrane. This process is called facilitated diffusion.
2. Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria, allowing molecules up to the size of small proteins to pass through.
3. Concentration gradient must already be present for molecules to diffuse even if facilitated by the proteins.
4. In’facilitated diffusion, special proteins help to move substance across the membrane without expenditure of ATP energy.
5. Facilitated diffusion cannot cause net transport of molecules from a low to a high concentration this would require input of energy.
6. Transport rate reaches a maximum when all of the protein transporters are being used.
7. Facilitated diffusion is very specific, it allows cell to select substance for uptake.
8. It is sensitive to inhibitors which react with protein side chain.

Question 3.
What is meant by plasmolysis? How is it practically useful to us?
Answer:
Shrinkage of protoplast when the cell is placed in hypertonic solution is called plasmolysis. It occurs when water moves out of the cell and the cell membrane of the cell shrinks away from the cell wall.

The salting of pickles and preserving of fish and meat in salt are good examples of practical application of this phenomenon.

Question 4.
How does ascent of sap occur in tall trees?
Answer:
In tall trees ascent of sap occurs due to transpiration. It depends mainly on the following physical properties of water.

1. Cohesion – mutual attraction between water molecules.
2. Adhesion – attraction of water molecules to polar surfaces (such as the surface of tracheary elements)
3. Transpiration pull – driving force for upward movement of water.

These properties give water high tensile strength i.e., an ability to resist a pulling force and high capillarity i.e-., the ability to rise in thin tubes.

Photosynthesis needs water and it is supplied by xylem vessels from roots to leaf veins. Transpiration pull is created due to evaporation of water through the stomata.

Question 5.
Stomata are turgor operated valves. Explain.
Answer:

1. The cause of opening or closing of the stomata is a change in the turgidity of the guard cells.
2. Guard cells are thick towards inner side and thin towards outer side.
3. When turgidity increases within the two guard cells, the thin outer walls bulge out and force the inner walls into a crescent shape, stoma opens.
4. When the guard cells lose turgor due to water loss (or water stress) the elastic inner walls regain their original shape, the guard cells becomes flaccid and the stoma closes.

Question 6.
Explain pressure flow hypothesis of translocation of sugars in plants.
Answer:
Translocation of sugars from source to sink can be explained by pressure flow or mass flow hypothesis. As glucose is prepared at the source (by photosynthesis) it is converted to sucrose.

The sugar is then moved in the form of sucrose into the companion cells and then into the living phloem sieve tube cells by active transport. This process of loading at the source produces a hypertonic condition in the phloem. Water in the adjacent xylem moves into the phloem by osmosis. it into the energy starch or cellulose. As sugars are removed, the osmotic pressure decreases and water moves out of the phloem and reaches xytem.

This can be shown by following illustration.

Question 7.
“Transpiration is a necessary evil” – Explain. [Mar. 2020, May, Mar. 2017, ’14]
Answer:
Beneficial and harmful effects of transpiration are below :

Beneficial effects of transpiration :

1. It creates transpiration pull for absorption and transportation in plants.
2. It supplies water for photosynthesis.
3. It transports minerals from the soil to all parts of the plant.
4. It cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling.
5. It maintains the shape and structure of the plants by keeping cells turgid.

Harmful effects of transpiration :

1. Photosynthesis is limited by available water which is depleted by transpiration.
2. The humidity of rain forests is largely due to cycling of water from the root to the leaf to the atmosphere and back to the soil.

Even though large amount of water absorbed is lost during transpiration, it is beneficial to plant in many ways. Hence it is considered as “necessary evil”.

Question 8.
A gardener forgot to water a potted plant for a day in summer. What will happen to the plant? Do you think it is reversible? Explain.
Answer:
Plants show wilting.

Yes, it is reversible by watering.
As the potted plant was not water for a day, it will show incipient wilting or temporary wilting. During incipient wilting there is no external symptoms of wilting but the mesophyll cells have lost sufficient water due to transpiration being higher“than the availability of water.

Temporary wilting is the temporary drooping down of leaves and young shoots due to loss of turgidity during noon. It can be corrected only after rate of transpiration decreases accompanied by replenishment of water around root hairs.

Question 9.
Explain the type of molecular movement which is highly selective and requires special membrane proteins, but does not require any metabolic energy.
Answer:
Facilitated diffusion is the type of molecular movement membrane which is highly selective and requires special membrane protein but does not require any metabolic energy.

The random movement of molecules takes place from high concentration to low concentration till equilibrium reaches is called simple diffusion. Membrane provides special proteins to move substances across the membrane by diffusion. This process is called Facilitated diffusion. When all these protein transporters are used transport rate reaches maximum.

Facilitated diffusion is very specific, it allows the cell to select substances for uptake. It is sensitive to inhibitors which react with protein side chain. These special proteins form channels which may be always open or controlled.

Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria, and some bacteria allowing molecules.

Extra cellular molecules are bound to the transport protein. The transport protein then rotates and releases the molecules inside the cell.

Question 10.
How does most of the water move within a healthy plant body and by which path? [May 2014]
Answer:
Water is absorbed along with mineral solutes by the root hairs purely by diffusion. Once water is absorbed by the root hair, it moves by two distinct path ways.
a) Apoplast pathway
b) Symplast pathway

a) Apoplast pathway :
Apoplast is the continuous system of cell walls and intercellular spaces in plant tissues. In apoplast pathway, water movement is through the cell wall without crossing membranes.

b) Symplast pathway :
Symplast comprises of network of cytoplasm of all cells inter-connected by plasmodesmata. The movement of water occurs from one cell to another through plasmodesmata.

Most of the water flows in the cortex occurs via the apoplast. Since the cortical cells are loosely packed and hence offer no resistance. Endodermis radial walls are thickened with casparian strip. Passage cells allows water by symplast, through pericycle it reaches xylem.

Question 11.
Transpiration and photosynthesis – a compromise. Explain.
Answer:

1. Transpiration has more than one purpose : it
   a) creates transpiration pull for absorption and transportation in plants.
   b) supplies water for photosynthesis.
   c) transports minerals from the soil to ail parts of the plant.
   d) cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling.
   e) maintains the shape and structure of the plants by keeping cells turgid.
2. An actively photosynthesising plant has an insatiable need for water.
3. Photosynthesis is limited by available water which can be swifty depleted by transpiration.

Question 12.
Do different species of plants growing in the same area show the same rate of transpiration at a particular time? Justify your answer.
Answer:
No. Different species of plant growing in the same area does not show the same rate of transpiration at a particular time. Plants take up huge amount of water daily but most of it is lost to the air through evaporation from the leaves he., transpiration.

A mature corn plant absorbs almost three litres of water in a day while a mustard plant absorbs water equal to its own weight in about 5 hours.

Long Answer Type Questions

Question 1.
Explain how plants absorb water.
Answer:
Water is absorbed along with mineral solutes by the root hairs purely by diffusion. Once water is absorbed by the root hair, it moves by two distinct pathways.
a) Apoplast pathway
b) Symplast pathway

a) Apoplast pathway :
it is “the system of adjacent cell walls that is continuous throughout plant, except at the casparian strips of endodermis in roots.” Apoplastic movement of water takes place through intercellular spaces and cell walls. Movement through apoplast does not include crossing cell membrane. It is dependent on the gradient. It does not provide any barrier to water movement. The water movement is through mass flow. When the water evaporates into intercellular spaces or atmosphere, a tension develops in continuous stream of water in apoplast. Therefore mass flow of water takes place as a result of adhesive and cohesive properties of water.

b) Symplast system :
It is “the system of interconnected protoplasts.” The adjacent cells are connected through cytoplasmic strands that extend through plasmodesmata. In it, water travels through the cells – their cytoplasm; intercellular movement is through plasmodesmata. Water enters the cells only through the cell membrane, so the movement is relatively slower. The movement is again down a potential gradient. It may cytoplasmic streaming.

Most of the water flow in roots takes place by apoplast as cortical cells are loosely packed. They offer no resistance to water movement. Endodermis is impervious to water due to casparian strip. Water moves through symplast and again crosses a membrane to reach the cells of the xylem. In the endodermis, the movement of water is symplastic. This is the only way water and other solutes can enter the vascular cylinder.

Question 2.
Define transpiration. Explain the structure and mechanism of opening and closing of stomata.
Answer:
Transpiration is defined as the loss of water in the form of vapour from the living tissues of aerial parts of the plants.

Structure of Stomata :
Each stoma is composed of two bean shaped cells known as guard cells. In grasses, the guard cells are dumb-bell shaped. The outer walls of guard cells are thin and inner walls are thickened. The guard cells possess chloroplasts and regulate the opening and closing of stomata. Sometimes, a few epidermal cells, in the vicinity of the guard cells become specialised in their shape and size and are known as subsidiary cells. The stomatal aperture, guard cells and the surrounding subsidiary cells are together called stomatal apparatus.

Opening and closing of stomata :
The stoma is the turgor operated valve in the epidermis.

1. The immediate cause of opening or closing of the stomata is a change in the turgidity of the guard cells.
2. The inner wall of each guard cell towards the pore or stomatal aperture is thick and elastic.
3. When turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape,
4. The opening of the stoma is also aided by the orientation of the microfibrils in the cell walls of the guard cells.
5. Cellulose microfibrils are oriented radially rather than longitudinally, making it easier for the stoma to open.
6. When the guard cells lose turgor due to water loss the elastic inner walls regain their original shape, the guard cells becomes flaccid and the stoma closes.

Levitt proposed K⁺ pump theory to explain the mechanism of opening and closing of photoactive stomata.

Opening of stomata :

1. According to this theory accumulation of K⁺ ions into the guard cells from the subsidiary cells occurs in the presence of light.
2. This coupled with efflux of protons leads to increase in pH of the guard cells.
3. Accumulation of K⁺ ions into the guard cells is associated with passive influx of Cl^– ions thereby decreasing the water potential of the guard cells.
4. Water thereby enters the guard cells making them turgid.
5. As the outer walls are thin and elastic, the guard cells expand outwardly, leaving a minute pore at the centre open.

Closing of stomata :

1. At night, in the absence of light, the K⁺ and Cl^– ions move out of the guard cells due to which the water potential of guard cells increases and water starts moving out of them leading to closure of stomata.
2. Under water stress conditions, abscisic acid (ABA) a natural anti-transpirant drives the K⁺ ions out of guard cells making them close.
3. In Succulent plants, the water potential gradient established due to accumulation of organic acids at night makes the guard cells become turgid, hence stomata opens at night.

Intext Question Answers

Question 1.
Differentiate uphill and downhill transport.
Answer:
Active transport uses energy to pump molecules against a concentration gradient. That means from a low concentration to a high concentration. It is called uphill.

Passive transport does not use energy. It transport from high concentration to low concentration. It is called downhill.

Question 2.
Compare facilitated diffusion and simple diffusion.
Answer:
The random movement of individual molecules from a region of higher concentration to a region of lower concentration is called simple diffusion. Facilitated diffusion is a type of molecular movement which is highly selective and required special membrane proteins. These proteins form channels in the membrane for molecules to pass through. These channels may be open always or controlled.

Question 3.
What happens when two solutions of different concentrations are separated by egg membrane? State the reason.
Answer:
When two solutions of different concentrations are separated by egg membrane osmosis takes place. Osmosis occurs spontaneously in response to a driving force. The net direction and rate of osmosis depends on both the pressure gradient and concentration gradient.

Question 4.
Compare imbibing capacities of pea and wheat seeds.
Answer:
Proteins have a very high imbibing capacities followed by starch and cellulose the least. That is why proteinaceous pea seeds swell move on imbibition than starchy wheat seeds.

Question 5.
In general in a plant which path of water movement is more and why?
Answer:
Major proportion of water flow in the root cortex occurs via the apoplast since the cortical cells are loosely packed and hence offer no resistance to water movement. However, the inner boundary of the cortex, the endodermis, is impervious to water because of a band of suberised matrix called the casparian strip. Water flow is by symplast, through pericycle and ultimately reaches xylem.

Question 6.
Why do Pinus seeds fail to germinate in the absence of mycorrhiza?
Answer:
A mycorrhizae is a symbiotic association of a fungus with a root system. The fungus provides minerals and water to the roots, in turn the roots provide sugars and N – containing compounds to the mycorrhizae. Some plants have an obligate association with the mycorrhizae. So pinus seeds fail to germinate in the absence of mycorrhizae.

Question 7.
Which structures do you think the Pinus plant does not possess due to which its seeds fail to germinate?
Answer:
Mycorrhizae.

Question 8.
What do you think is the driving force for ascent of sap?
Answer:
Transpiration pull.

Question 9.
Why do stomata close under water stress conditions?
Answer:
Under water stress conditions, abscisic acid (ABA) a natural anti-transpirant drives the K+ ions out of guard cells making them close.

Question 10.
How are stomata distributed in a typical monocot plant?
Answer:
Stomata are distributed .equally, both in upper epidermis and lower epidermis in typical monocot plant.

Question 11.
In what form the sugars are transported through phloem?
Answer:
Sucrose

Question 12.
The inward movement of water into a plant begins either as symplast or apoplast. How does it conclude before entering into xylem?
Answer:
Symplast

Question 13.
Why does the root endodermis transports ions in one direction only?
Answer:
Endodermis have casparian strips on their radial walls. Hence the movement of water through the root layers is symplastic in endodermis. This is the only way water and other solutes can enter vascular cylinder. So in endodermis transportation of ions occurs in one direction.

Question 14.
If a ring of bark is removed from an actively growing plant, what will happen and why?
Answer:
In the absence of downward movement of food the portion of the bark above the ring on the stem becomes swollen after a few weeks. This shows that phloem is the tissue responsible for translocation of food and that transport takes place in one direction i.e., towards the roots.

Question 15.
A flowering plant is planted in an earthen pot and watered. Urea is added to make the plant grow faster, but after some time the plant dies. Why?
Answer:
As urea is added the concentration of water in the pot increases. Instead of root hairs absorbing water, the water in the plant comes out due to exosmosis. The cells get plasmolyzed. After some time the plant dies.