Cbse Class 12 Biology Revision Notes Chapter 5

Q.1 Who rediscovered Mendel’s laws of heredity?

Ans

De Vries, Carl Correns and Tschermak.

Q.2 What does monogenic inheritance deal with?

Ans

Quantitative traits.

Q.3 State III^rd law of inheritance given by Mendel. Explain it by taking an example of a dihybrid cross.

Ans

Law of independent Assortment: The law states that when two pairs of trait are combined in a hybrid, segregation of one pair of character is independent of the other pair of character.
Dihybrid cross: Seed colour and Seed shape-

Phenotypic ratio- round yellow: round green: wrinkled yellow: wrinkled green

9 : 3 : 3 : 1

Independent assortment of genes explain that RrYy plant is capable of producing four different types of gametes with equal probability that is 25% each type RY, Ry, rY, ry.

If we consider the individual monohybrid crosses separately we get the monohybrid ratio of 3:1 for both of them. This implies that the 2 pairs of characters under consideration have assorted themselves independently.

Q.4 Haemophilia victims are mostly men, very rarely women are affected. Why?

Ans

Haemophilia is a sex linked recessive disease. Females are affected by the disease only when mutant allele responsible for haemophilia is present on both the sex chromosomes. But in males, allele is present only on one X chromosome; corresponding allele on Y chromosome is absent. If one mutant allele is present in males, this will result in haemophilia in the individual.

Female Male
X^H X^H – Normal X^HY – Normal

X^H X^h – Carrier, Phenotypically normal X^hY – Haemophilia

X^h X^h – Haemophilia

Q.5 Explain the phenomenon of co-dominance by taking example of blood groups in human beings.

Ans

Co- dominance:-Co- dominance is the type of inheritance pattern in which F1 generation resembles both the parents, e.g,
(a) In human beings ABO blood groups are controlled by gene I.
b) The gene I has these alleles I^A I^B produce different types of sugar at the surface of RBC. I does not produce any sugar.
c) Since there are three different alleles, there are six different combinations of the genotypes of the human ABO blood types.

Q.6 Justify the fact that “In human beings, sex of the child is determined by father, not by mother”.

Ans

In males, two types of gametes are produced, 50% of the sperm carry X-chromosomes and 50% of the sperm carry Y-chromosomes. Females are homogamete and produce only one kind of gamete with X-Chromosome. In case, when the ovum fertilises with a sperm carrying X-Chromosome, the zygote develops into a female (XX) and the fertilisation of ovum with sperm carrying Y-Chromosome, results into a male offspring.

Q.7 “In incomplete dominance inheritance pattern, genotypic and phenotypic ratios in F2 generations are same”. Explain.

Ans

In the case of incomplete dominance, one allele is incompletely dominant over the other and is incapable of suppressing it. The phenotypic as well as genotypic results, both show same ratio.

Q.8 Define test cross. What is its significance?

Ans

Test cross: Test cross can be defined as cross between an organism of an unknown genotype and a homozygous recessive individual.
Significance: It reveals the genotype of an organism showing a dominant phenotype.

Q.9 What are Polyploids? How are they produced?

Ans

Organisms containing one or more extra sets of chromosomes in the cells, are called as polyploids. Failure of cytokinesis after telophase stage of cell division results in polyploidy.

Q.10 Explain the law of dominance using a monohybrid cross.

Ans

The law of dominance, given by Mendel states that in a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation.

Q.11 What are Mendel’s monohybrid and dihybrid phenotypic ratios?

Ans

Monohybrid ratio 3:1
Dihybrid ratio 9:3:3:1

Q.12 What is the cause of Down’s syndrome?

Ans

Presence of an additional copy of the 21^st chromosome (trisomy of 21).

Q.13 How does sickle cell anemia occur?

Ans

When Glutamic acid is substituted by Valine at the sixth position of the beta globin chain of the haemoglobin.

Q.14 Define dominance?

Ans

Interaction between two variant forms (i.e., alleles) of a single gene, in which one allele hides the expression of the other.

Q.15 What was the basis on which Mendel formulated the law of purity of gametes?

Ans

Monohybrid crosses

Q.16 Give one example of co-dominance.

Ans

ABO blood groups in human beings.

Q.17 Define mutation.

Ans

Mutation occurs when DNA is damaged or changed in such a way that it alters the genetic message carried.

Q.18 Which sex-determining mechanism does exist in humans?

Ans

XY Type

Q.19 Why was the term linkage coined by Morgan?

Ans

Morgan coined the term ‘Linkage’ to describe the physical association of the genes.

Q.20 Who gave the chromosomal theory of inheritance?

Ans

Sutton & Boveri

Q.21 A cross between two plants heterozygous for a single locus was made. The progeny contained the following:
(i) Round seeds, large starch grains : 1
(ii) Round seeds, intermediate starch grains : 2
(iii) Wrinkled seeds, small starch grains : 1
State the phenomenon exhibited by the above result? Work out the genotype of the parents and offspring using a punnet square.

Ans

The gene controls two traits, a phenomenon called pleiotropy.

The gene is completely dominant for one trait, the seed shape, while it shows incomplete dominance for the size of starch grains.

Parent: Bb X Bb

Gamete: B, b B, b

Q.22 (a) How is the child affected if it has developed from the zygote produced by an XX–egg fertilised by a Y-carrying sperm? What is the term given to this abnormality?
(b) What proportion of individuals produced in the progeny of a cross between two individuals with genotype TtSs will be TtSs and ttss respectively.

Ans

(a)The zygote will exhibit a genotype XXY, the child will be a male. The child will show a number of feminised characters. The term given to this abnormality is Klinefelter syndrome.

(b) Parent: Father X Mother

Gamete: TS, Ts, tS, ts TS, Ts, tS, ts

Therefore, the ratio of TtSs will be 4/16.

And, the ratio of ttss will be 1/16.

Q.23 Huntington’s disease, a disease of the nervous system, is autosomal dominant. The pedigree below shows the inheritance of the disease in three generations of a family. Observe the pedigree carefully and answer the questions that follow:

(a) What is the probable genotype of individual D? How do you know?

(b) What is the probability that individual N will not have Huntington’s disease?

Ans

(a) Individual D is an affected person because individual K has inherited the dominant allele from him.

Now the question arises whether individual D is homozygous (HH) or heterozygous (Hh) for the dominant allele. For this, we have to look at the cross between his parents, i.e., individual A and B:

Parent: Hh X hh

(Individual A) (Individual B)

Gamete: H,h h,h

Therefore, genotype of individual D is Hh.

(b) The genotype of individual N entirely depends on the genotype of Individual H and I.

Genotype of individual H: Individual O is homozygous (HH) for the dominant trait points out that both of his parents are affected individuals. Now, looking at the genotype of individual A and B (parents of individual H), the genotype of any of their affected offspring must be Hh. Thus, the genotype of individual H is Hh.

Genotype of individual I: Individual O is homozygous (HH) for the dominant trait points out that both of his parents are affected individuals. Therefore, individual I can be homozygous (HH) or heterozygous (Hh) for the trait.

So, the probability that individual N will not express the trait depends on the genotype of her parents.

In case individual I is homozygous (HH) for the trait:

Parent: Hh X HH

(Individual H) (Individual I)

Gamete: H,h H,H

In such case, all the offspring will express the trait or the probability that individual N will not express the trait is 0%.

In case individual I is heterozygous (Hh) for the trait:

Parent: Hh X Hh

(Individual H) (Individual I)

Gamete: H,h H,h

In such case, 1/4^th of the offspring will not express the trait or the probability that individual N will not express the trait is 25%.

Q.24 Is the trait that is segregating in the below given pedigree due to a dominant or a recessive allele?

Ans

Both affected individuals have two unaffected parents, which is not in agreement with the hypothesis that the trait is due to a dominant allele. Thus, the trait emerges as a recessive allele.

Q.25 A woman has an uncommon defect of the eyelids called Ptosis, which prevents her from opening her eyes completely. This condition is caused by a dominant allele, ‘P’. The woman’s father had Ptosis, but her mother had normal eyelids. Her father’s mother had normal eyelids.
(a) What are the genotypes of the woman, her father, and her mother?
(b) What proportion of the woman’s children will have Ptosis if she marries a man with normal eyelids?

Ans

(a)

Parent: Pp X pp

(Affected father) (Normal mother)

Gamete: P,p p,p

Pp, pp
(Offsprings)

Therefore, woman’s genotype will be Pp.

(b)

Hence, 50% of the woman’s children will have Ptosis if she marries a man with normal eyelids.

Q.26 The pedigree below shows the inheritance of a dominant trait (R).

What is the possibility that the offspring of the following matings will show the trait:
(a) G X I
(b) H X J

Ans

(a)

Hence, none (0%) of the offspring resulting from the G X I mating will show the trait.

(b)

Hence, 50% of the offspring resulting from the G X I mating will show the trait.

Q.27 The pedigree below shows the inheritance of a recessive trait (r).

What is the chance that the couple G and H of F₂ generation will have an affected child?

Ans
There are 50% chances that the couple G and H of F2 generation will have an affected child.

Q.28 In shorthorn cattle, the genotype RR causes a red coat, the genotype rr causes a white coat and the genotype Rr causes a roan coat. A breeder has red, white and roan cows and bulls.
What phenotypes might be expected from the following matings and in what proportions?
(a) red X red
(b) red X roan
(c) red X white
(d) roan X roan

Ans

(a)

Parent: RR X RR

(Red coat) (Red coat)

Gamete: R R

Hence, all (100%) the offspring will bear red coat as both the parents are homozygous for the dominant allele (RR).

(b) Parent: RR X Rr

(Red coat) (Roan coat)

Gamete: R R,r

Hence, half (50%) of the offspring will bear red coat and the other half will bear roan coat, as one of the parents is homozygous for the dominant allele (RR) and other one is heterozygous for the trait (Rr).

(c) Parent: RR X rr

(Red coat) (White coat)

Gamete: R r

Hence, all (100%) the offspring will bear roan coat, as one of the parents is homozygous for the dominant allele (RR) and the other one is homozygous for the recessive allele (rr).

(d) Parent: Rr X Rr

(Roan coat) (Roan coat)

Gamete: R,r R,r

Hence, 1/4^th (25%) of the offspring will bear red coat, 1/4^th (25%) of

the offspring will bear white coat and 1/2 (50%) of the offspring will

bear roan coat. Because both the parents are heterozygous for the

trait (Rr).

Q.29 (a) Name the law that explains the expression of only one of the parental characters in the F1 generation of a monohybrid cross?
(b) Not all characters show true dominance. What are the two other possible types of dominance? Give an example of each.
(c) A male child was born with 47 chromosomes. Write any two possible combinations of chromosomal abnormalities and write one important symptom of each?

Ans

(a) Law of dominance

(b) The other two possible types of dominance are:

(i) Incomplete dominance, e.g. – Flower colour in snapdragon

(ii) Codominance, e.g. – Blood group AB

(c) 47 chromosomes in a male child can be due to any one of the following:

(i) Klinefelter’s Syndrome = 22 Pairs autosome + XXY

– The individual is a male with feminine characters like breast development.

(ii) Down’s syndrome = Trisomy of 21st chromosome.

Symptoms:

– Furrowed tongue and partially open mouth.

– Broad palm with characteristic palm crest.

Q.30 Wolves have been seen with black coats and blue eyes. If normal coat colour (N) is dominant to black (n) and brown eyes (B) are dominant to blue (b). Suppose the breeding male and female are black with blue eyes and normal coloured with brown eyes respectively and female is also heterozygous for both traits. How many of the offspring (assume 16) living in the pack will have each of the following genotypes?
a) 1)nnBB 2)Nnbb
b)What percent of the offspring will be normal coloured with blue eyes?
c)What percent of the offspring will be brown with brown eyes?

Ans

a) Male-nnbb

Female-NnBb

1) nnBB-0 2) Nnbb-4

b) Normal coloured with blue eyes – Nnbb – 25% (4/16×100)

c) Brown with brown eyes – NnBb – 25% (4/16×100)

Q.31

(a) A colour blind man marries a woman with normal vision whose father was colour blind. Work out a cross to demonstrate the genotype of the new couple and their future sons?(b) Answer the following questions with reference to the given pedigree.

i) Is the trait autosomal dominant, autosomal recessive or sex-linked? Justify your answer.
ii) Give the genotypes of the parents (individual 1 and 2).
iii) Give the genotype of the daughter in the first generation and the son and the daughters in the second generation.

Ans

(a) Since the father of the normal women is colour blind, her genotype will be X^cX (carrier)

Colour blind man X Normal woman

X^cY X^cX

Gamete: X^c, Y X^c, X

50% of their sons will be colour blind.

(b) (i) The trait is sex linked (recessive) as it has appeared in male child in the first generation pointing at the possibility that the female parent is a carrier as his father is normal.

(ii) Genotype of individual 1 => Male => XY

Genotype of individual 2 => Female => X^cX (Carrier)

(iii)
Genotype of individual 3 in generation I => X^cY
Genotype of individual 4 in generation I => X^cX

Genotype of individual 6 in generation II => X^cX Genotype of individual 7 in generation II => X^cX^c

Q.32 (a) Dominance is not an autonomous feature of a gene or the product it codes for. It depends on the gene product and the production of a particular phenotype from the gene product. Justify the statement.
(b) In humans, “unattached” earlobes are dominant over “attached” earlobes. “Widows peak” hairline is dominant over “non-widows peak” hairline. Use E and e for the earlobe phenotype alleles, and W and w for the hairline phenotype alleles.A female and a male, both with genotype EeWw have a child. What is the probability it will be a boy, and have attached earlobes and a widows peak hairline?

Ans

(a)

Dominance depends much on the gene product and the production of a particular phenotype from the product. There are cases where one gene controls more than one phenotype, e.g. size of starch grains and the shape of seed.

(b)

Parent: Father X Mother

(EeWw) (EeWw)

Gamete: EW, Ew, eW, ew EW, Ew, eW, ew

Probability of having attached ear lobes and a widows peak hairline will be 3/16.

As the probability of having a boy is 50%, the probability of having a boy with attached earlobes and a widows peak hairline will be 50% of the actual probability of required combination = 3/16 X 1/2

= 3/32

Q.33 What would be the phenotype ratio of pea plants obtained on crossing two Tt plants, if the gene for tall (T) plants was incompletely dominant over the gene for short (t) plants. What would be the result of crossing two Tt plants?

Ans

1/4 would be tall, 1/2 intermediate height and 1/4 short. The heterozygous offspring (Tt) would be of intermediate height.

Q.34 A genetic cross of red flowered snapdragons with pure white flowered variety, resulted in offspring with pink flowers. When the plants were self-crossed, the resulting plants had a phenotypic ratio of 1 red: 2 pink: 1 white. Give the most likely explanation.

Ans

Heterozygous plants have a different phenotype than either inbred parent because of incomplete dominance of the dominant allele.

The features of crosses involving incomplete dominance are intermediate phenotype of heterozygous individuals and parental phenotypes reappear in F2 when heterozygotes are crossed.

Q.35 In wild, a type of male lizard courts females by bobbing his head up and down while displaying a colorful throat patch. Now, suppose that males prefer to mate with lizards who bob their heads fast (F) and have red throat patches (R) to females that are slow in bobbing and have yellow throats. A male lizard heterozygous for head bobbing and homozygous dominant for the red throat patch mates with a female who is also heterozygous for head bobbing but is homozygous recessive for yellow throat patches.
a) How many of the F1 offspring have the preferred fast bobbing/red throat (assume 16 young)?
b) What percentage of the offspring will lack mates because they have both slow head bobbing and yellow throats?

Ans

a)

Male – FfRR

Female – Ffrr

Fast bobbing red throat – 12

b) Slow head bobbing and yellow throats – ffrr – 0%

Q.36 Normal spots (X^N) on a leopard are a dominant, sex-linked trait compared to dark spots. Suppose as a Biologist, you are involved in the leopard breeding program. One year you cross a male with dark spots and a female with normal spots. She delivers four cubs, out of which two are male and two female. One each of the male and female cubs have normal spots and one each have dark spots.
a)What could be genotype of the mother?
b)Suppose a few years later, you cross the female cub that has normal spots with a male that also has normal spots. How many of each genotype will be found in the cubs (assume 4)?

Ans

a)X^NXⁿ
b)Male-X^NY

X^NX^N: X^NXⁿ: XⁿXⁿ: X^NY : XⁿY:

1, 1, 0, 1, 1

Q.37 A variety of wild beetles have been observed to lay their eggs in dead animals and then they their eggs in the ground until they hatch. Assuming that the preference for fresh meat (F) and tendency to bury the meat shallow (S) are dominant traits.
a) What will be the genotype of the offspring if a female carrion beetle homozygous dominant for both traits mates with a male homozygous recessive for both traits.
b) What will be the expected genotypic ratio of the F2 generation (FFSS : FFSs : FFss : FfSS : FfSs : Ffss : ffSS : ffSs : ffss)?

Ans

a) Female – FFSS

Male – ffss

Phenotype – Fresh meat/shallow

Genotype – FfSs

b) FfSs x FfSs

FFSS : FFSs : FFss : FfSS : FfSs : Ffss : ffSS : ffSs : ffss
1:2:1:2:4:2:1:2:1

Q.38 A bison herd in the dry grasslands has begun to show a genetic defect. Some of the males exhibit “rabbit hock” in which the knee of the back leg is malformed slightly. If it is due to sex linked recessive gene and the herd bull who is normal (X^N) mates with a cow that is a carrier for rabbit hock, what are his chances of producing a normal son?

Ans

Bull – X^NY

Cow – X^NXⁿ

Probability of a normal son is 50%.

Q.39 A woman with normal vision gives birth to a daughter with red-green colour-blindness. Knowing that colour-blindness is a sex-linked recessive gene,
a) what is father’s genotype?
b)The woman marries a man with normal vision. What is the probability they will have sons who are red-green colour-blind
c)What is the probability the above couple will have daughters who are red-green colour-blind?

Ans

a) Women-X^bX

Father- X^bY

b) 50%

c) 50%

Q.40 Define polygenic inheritance.

Ans

Inheritance of a trait which is controlled by two or more sets of alleles.

Q.41 How is the inheritance of human skin colour controlled by polygenes?

Ans

The inheritance of human skin colour is controlled by polygenes. Suppose the trait is controlled by three genes, namely A, B and C.

The dark-skinned and fair-skinned human beings are mated and further the intermediate skin coloured individuals expected at F1 are mated to obtain F2 progeny.

Q.42 Define pleiotropy.

Ans

It is defined as a phenomenon in which a single gene may produce more than one effect or control several phenotypes.

Q.43 State the two types of brood cells in honey bees.

Ans

Depending upon their size, brood cells in the honey bees are of two types:
a) Smaller cells: These cells develop into workers which are females.
b) Larger cells: These cells develop into drones which are males.

Q.44 Where are sperms stored inside the body of the queen honey bee?

Ans

Sperms are stored in the seminal receptacle in the queen’s body.

Q.45 What is colour blindness?

Ans

It is a recessive sex-linked trait in which the eyes fail to distinguish between two different colours, e.g., red and green colours.

Q.46 When does colour blindness appear in females?

Ans

Colour blindness appears in females only when both the sex chromosomes carry the recessive alleles.

Q.47 What is the impact of colour blindness in a patient’s day-to-day life?

Ans

Colour blind people can carry out their normal work, but they cannot distinguish mostly between red and green colours. Moreover, they can follow and understand the traffic light due to the fixed position of the red light at the top and the green light at the bottom.

Q.48 What is thalassaemia? Give any two biochemical abnormalities of the disease.

Ans

Thalassaemia is an inherited autosomal recessive blood disorder which is a type of genetic defect caused due to genetic mutation.
Two biochemical abnormalities of thalassaemia are:
a) Defect in the synthesis of globin polypeptide.
b) Abnormal haemoglobin molecules are formed, which thereby lead to an excessive degradation of red blood cells.

Q.49

Observe the below given graph showing a particular type of inheritance observed in human population and answer the following questions:

1. Deduce the phenotypic ratio occurring in the population.
2. Interpret the data displayed in the graph and identify the type of inheritance shown by this characteristic of human population.
3. Try to draw a rough distribution curve formed by the given data and based on that analyse the type of variation shown by this characteristic of human population.
4. Calculate the sum of phenotypes and genotypes in F₂ generation if skin colour character is controlled by 4 pairs of polygenes.

[1+1+1+2= 5 Marks]

Ans

1. Phenotypic ratio: 1: 6: 15: 20: 15:6:1
2. From the graph, it can be clearly interpreted that extreme phenotypes occur rarely, while the intermediates occur frequently. Polygenic inheritance is shown by skin colour of human population.
3. Rough distribution curve formed by the given data: Bell-shaped curve

Continuous variation is shown by skin colour of human population because it has all possible intermediates between the extremes.

No. of phenotypes for polygenes = 2n+1, where n = 4

So, (2X4+1) = 8+1 = 9

No. of genotypes for polygenes = 3ⁿ, where n = 4

So, 3⁴ = 81

Sum of no. of phenotypes and genotypes = 9+ 81 = 90.

Q.50 Observe the genetically controlled pathway for flower colour in a plant ‘X’.

In the given pathway,

• the dominant allele, M, codes for an enzyme necessary for the conversion of C_o into C₁, while its recessive allele, m, codes for a defective enzyme,
• the dominant allele, N, codes for an enzyme necessary for the conversion of C₁ into C₂, while its recessive allele, n, codes for a defective enzyme,
• the dominant allele, O, codes for an enzyme necessary for the conversion of C₂ into C₃, while its recessive allele, o, codes for a defective enzyme and,
• the dominant allele, P, produces a polypeptide that inhibits the conversion of C₂ into C₃, while its recessive allele, p, produces a defective polypeptide that does not inhibit this reaction.

Study the given data and carry out the mathematical calculations to predict the proportion of plants that will produce:

1. pink flowers,
2. red flowers and,
3. lemon yellow flowers

in F₂ generation if plants of genotype ‘MM nn OO PP’ and ‘mm NN oo pp’ are crossed keeping in mind that the flower colour is determined exclusively by the four genes (M, N, O and P) that assort independently at the time of gamete formation.

[3 Marks]

Ans

1. Proportion of plants that will produce pink flowers = (3/4)⁴ + [(3/4)² × (1/4)]

= (81/256) + [(9/16) X (1/4)]

= (81/256) + (9/64)

= (81+36)/ 256

= 117/256

2. Proportion of plants that will produce red flowers = (3/4)³ × (1/4)

= (27/ 64) × (1/4)

= 27/256

3. Proportion of plants that will produce lemon yellow flowers = 1 – [(117/256) + (27/256)]

= 1 – [(117+ 27)/256]

=1 − 144/256

= (256-144)/ 256

= 112/256

Q.51 In humans the hair type is a codominant trait. Straight and curly hairs are codominant alleles and the heterozygote has wavy hair. Sonam and Rajan both have wavy hair. What are the chances that their children will have:

1. Straight hair
2. Curly hair
3. Wavy hair

Ans

Genotype of wavy hair = H^SH^C

Genotype of straight hair = H^S

Genotype of curly hair = H^C

Therefore;

The chances that their children will have:

1. Straight hair = 1:4
2. Curly hair = 1:4
3. Wavy hair = 1:2

Q.52 Sonam is blood group B and her husband is blood group A. Their daughter is blood group O. What is the genotype of Sonam and her husband?

Ans

The alleles for blood groups can be represented as I^A, I^B and i.

Possible genotype of Blood group A = I^AI^A or I^Ai

Possible genotype of Blood group B = I^BI^B or I^Bi

Genotype of Blood group O or the daughter = ii

Daughter must have received one recessive allele from both the parents:

Hence, the genotype of parents will be:

Sonam’s genotype (Blood group A) = I^Ai

Husband’s genotype (Blood group B) = I^Bi

Q.53 The below given pedigree chart shows the occurrence of vitamin D-resistant rickets in a family. Observe it carefully to answer the following questions (Blue=vitamin D-resistant rickets):

1. Is the trait given in the pedigree segregating due to a dominant or a recessive allele?
2. Is the trait sex linked?
3. What are the chances that person ‘X’ will suffer from the disease?

Ans

1. The disease is caused due to a dominant allele as the:

• Affected parents have affected children
• Unaffected parents only have unaffected children

2. Yes, the trait is sex linked as all the daughters of generation I are affected and all the sons are unaffected
3. There are 100% chances that person ‘X’ will suffer from the disease, as both the parents are affected and the disease is caused due to the dominant allele.

Q.54 The below given pedigree chart shows the ABO blood groups of three generation. Observe it carefully to answer the following questions:

1. Deduce the genotype of each parent.
2. What could be the possible blood groups of person III 3? Give the percentage chance of each group.

Ans

a)

AB individual is IAIB

O individual is ii

B individual is IBi

b) A or B or O or AB; 25% chance of each

Q.55 Rolling of tongue is a genetically controlled autosomal dominant character. Represented as Roller = RR / Rr; Non-roller = rr].

a) A child who can roll the tongue has one brother who cannot role his tongue and two sisters who can roll their tongue. If both the parents can roll their tongue, the genotypes of their parents would be ____.

1. RR x RR
2. Rr x Rr
3. RR x rr
4. rr x rr

b) In a group of 60 students, 45 can roll their tongue and 15 are non-rollers. In the above context, calculate the percentage of dominant and recessive characters.

[2+3=5 marks]

Ans

1. ii. Rr x Rr
2. Total number of students= 60

Tongue rollers (Dominant Character) = 45

Non–rollers (Recessive Character) = 15

Percentage of tongue rollers or Dominant Character

= 45/60 x 100 =75 %

Percentage of Non–rollers or Recessive Character = 15/60 x 100 = 25 %

Q.56 The below given pedigree chart shows the occurrence of Albinism in a family. Observe it carefully to answer the following questions (Green=Normal; White=Albino):

1. Is the trait given in the pedigree segregating due to a dominant or a recessive allele?
2. Is the trait sex linked?
3. As per the laws of inheritance, what is the chance of a child of these parents to have Albinism? Is the ratio observed in pedigree chart in consent with the result obtained?

Ans

1. Both parents are normal and two of the children are albino, hence, albinism is caused by recessive allele and normal pigmentation is an expression of dominant allele.
2. The trait does not show sex linked inheritance as both daughter and son are suffering from the disease.
3. The probability of a child of these parents to have Albinism is 1:4. The ratio observed as per the pedigree chart is 1:2. It is different from the result obtained; but still is in consent with the result as the theoretical ratio of 1:4 could only be observed if the parents had a very large number of children.

Q.57 (a) Radiation effects of atom bomb explosions have been affecting generations in Japan. Analysing the above statements, give your interpretation.

(b) Is the trait given in the pedigree segregating due to a dominant or a recessive allele?

[2+1=3 marks]

Ans

1. A change that affects the body cell is not inherited. However, a change in the gamete is inherited. The radiation effects of atom bomb explosions have been affecting generations as these effects are inheritable; because the genes of the germ cells or gametes have been altered.
2. Both affected individuals have two unaffected parents, which is not in agreement with the hypothesis that the trait is due to a dominant allele. Thus, the trait emerges as a recessive allele.

Q.58 Observe the flow-chart of a monohybrid cross

in a Clitoria plant and write the answers for A, B, C, D:

Character : Colour of the flower

Parents : Blue flowered x White flowered

Ans

A. Bb

B. Selfing

C. 3:1

D. 1:2:1

Q.59 In watermelons, solid green colour (G) is dominant over the striped one (g). A farmer planted watermelon seeds, and noticed that all the new melons were striped. Will the farmer be able to procure solid green coloured watermelons if he interbreeds the new watermelon plants?

Ans

No, the farmer will not be able to procure solid green coloured watermelons if he interbreeds them.

In order to get solid green coloured watermelons, a watermelon plant should have genotype either Gg or GG.

However, his watermelon plants expressed the recessive phenotype and thus have genotype “gg”.

This indicates that all the watermelons are expressing a homozygous recessive trait.

Q.60 A geneticist has performed two crosses between pea plants. In first cross, 3 tall and 1 dwarf offspring are obtained. In second cross, all tall offspring are obtained.

Can you deduce the genotypes of A, B, C and D?

Ans

When two heterozygous parents are crossed, the phenotype of the offspring with dominant allele and the offspring with the recessive allele appears in the ration 3:1. Hence, A and B are Tt.

When a heterozygous and homozygous recessive parents are crossed. The predicted outcome for this cross is 1:1 ratio of tall to dwarf plant. Hence C and D are Tt and tt.

Q.61 In shorthorn cattle, the genotype RR causes a red coat, the genotype rr causes a white coat and the genotype Rr causes a roan coat. A breeder has red, white and roan cows and bulls.

What phenotypes might be expected from the following mating and in what proportions?

(a) red X red

(b) red X roan

(c) red X white

Ans

(a) Parents:

RR X RR
(Red coat) (Red coat)

Gametes: R R
Cross:

Hence, all (100%) the offspring will bear red coat as both the parents are homozygous for the dominant allele (RR).

(b) Parents: RR X Rr

(Red coat) (Roan coat)

Gametes: R R, r

Cross:

Hence, half (50%) of the offspring will bear red coat and the other half will bear roan coat, as one of the parents is homozygous for the dominant allele (RR) and other one is heterozygous for the trait (Rr).

(c) Parents: RR X rr

(Red coat) (White coat)

Gametes: R r

Cross:

Hence, all (100%) the offspring will bear roan coat, as one of the parents is homozygous for the dominant allele (RR) and the other one is homozygous for the recessive allele (rr).

Q.62 The given pedigree diagram shows the inheritance of hair colour in humans. Answer the following questions based on the pedigree diagram given below:

a) Which hair colour represents dominant phenotype?

b) Give reason to support your answer.

c) What will be the genotype of 6 and 7?

d) What is the chance of third child from parents 1 and 2 to have brown hair?

e) What is the chance of third child from parents 3 and 4 to have black hair?

Ans

a) Black hair represents the dominant phenotype.

b) In the given pedigree diagram, parents 6 and 7 have black hair but they also have child with brown hair which shows parents 6 and 7 are heterozygous. They carry a recessive allele for brown hair which is not expressed in them but it is expressed in child 10, which has brown hair. This shows homozygous recessive condition.

c) Since, 6 and 7 carry a recessive allele; their genotype must be Bb and Bb.

d) Since parents 1 and 2 have a child with brown hair (5) which is a homozygous recessive, we can say that the parent 1 is homozygous recessive and parent 2 is heterozygous.

The cross is as follows;

B:allele for black hair b: allele for brown hair

Hence the third child from parent 1 and 2 having brown hair is 50%.

e) The chances for having third child with black hair for couple 3 and 4 is 50%.

Q.63 Groups of alleles associated with the lac operon in the order of dominance for each allelic series are as follows:

R (repressor), I^s (super-repressor), I⁺ (inducible), and I⁻ (constitutive), O (operator), O^c (constitutive, cis-dominant) and O⁺ (inducible, cis-dominant); S (structural), z⁺ and y⁺.

Below given are five genotypes:

1. I⁻O^cz⁺y⁺
2. I⁺O⁺z⁺y⁺
3. I^sO⁺z⁺y⁺
4. I⁻O⁺z⁺y⁺
5. I^sO^cz⁺y⁺

1. Which of the above given genotypes will produce ß-galactosidase and ß-galactoside permease if lactose is present?
2. Which of the above genotypes will produce ß-galactosidase and ß-galactoside permease if lactose is absent?

Ans

1. Genotypes (i), (ii), (iv), and (v) will produce ß-galactosidase and ß-galactoside permease if lactose is present as genotype (ii) is wild-type and inducible, while genotypes (i), (iv) and (v) are constitutive.

Since genotype (iii) has a super-repressor (I^s), it is non-inducible at normal level of lactose.

Therefore, all the genotypes except (iii) will produce ß-galactosidase and ß-galactoside permease in the presence of lactose.

2. Genotypes (i), (iv), and (v) will produce ß-galactosidase and ß-galactoside permease if lactose is absent as repressor cannot bind to O^c in genotypes (i) and (v) and no repressor is made in genotype (iv). So, operon is constitutive in both the circumstances.

Q.64 In a lily plant, the synthesis of purple pigment is a two step process, depicted roughly by the below given pathway.

Analyse the given information and answer the following questions:

1. Predict the phenotype of this plant if it is homozygous for no mutation of gene M.
2. Predict the phenotype of this plant if it is homozygous for no mutation of gene N.
3. Predict the phenotype of this plant if it is homozygous for no mutation of both the genes M and N.
4. Predict the genotypes of the three strains in parts a, b, and c.
5. Assuming the law of independent assortment of characters, predict the ratio in F₂ generation if you cross the plants from parts a and b?

Ans

1. Colourless
2. Pink
3. Colourless
4. Genotype of a: m/m, N/N

Genotype of b: M/M, n/n

Genotype of c: m/m, n/n

5. 9 (purple): 4 (colourless): 3 (pink)

Q.65 Analyse the data illustrated on the table given below and answer the following questions:

1. Calculate the recombination frequency of species A to E.
2. In which of the given species, the chances of crossing over are least? Justify your answer.
3. What will be the distance between the genes in prokaryotic species A and C? How could you deduce your answer?

[2+2+1 = 5 Marks]

Ans

1. Recombination frequency = [(No. of recombinants)/ (Total no. of offspring)] X 100

• For species ‘A’,

Recombination frequency = [64/1600] X 100

= 7%

• For species ‘B’,

Recombination frequency = [77/1100] X 100

= 4%

• For species ‘C’,

Recombination frequency = [55/500] X 100

=11%

• For species ‘D’,

Recombination frequency = [90/3000] X 100

=3%

• For species ‘E’,

Recombination frequency = [36/1200] X 100

=3%

2. The chances of crossing over are least in species ‘D’ and ‘E’ because recombination frequency is minimum in these species. The lesser is the recombination frequency, the lesser are the chances of crossing over.
3. The distance between the genes in prokaryotic species A and C is 7 and 11 map unit respectively. The recombination frequency is directly proportional to the distance between the genes and has the same value, but the unit is different.

Q.66 Emma followed the Mendelian genetics and carried out a cross in which the phenotypic ratio came out to be 15:1 in F₂ generation.

1. Use the given abstract information and predict the condition in which this phenotypic ratio can occur.
2. This phenotypic ratio explicates which cross, monohybrid or dihybrid?
3. Take an example of species which produces this phenotypic ratio in F₂ generation and draw a Punnett square to illustrate the same.

[1+1+3 = 5 Marks]

Ans

1. When the dominant alleles of two gene loci produce the same phenotype, whether inherited together or discretely, the phenotypic ratio 9:3:3:1 gets modified into 15:1.
2. This phenotypic ratio explicates dihybrid cross.
3. The capsule of shepherd’s purse plant occurs in two different shapes, triangular and top-shaped.

When a species with triangular-shaped capsule is crossed with a species with top-shaped capsule, those with triangular-shaped capsules occur in F₁ generation.

When the F₁ progenies are self fertilised, they produce the plants with both triangular and top-shaped capsules in F₂ generation in the ratio of 15:1. This happened because two independently assorting genes (T and S) have influenced the shape of the capsule in the same manner.

So, all the genotypes possessing dominant alleles of both or either of two genes, T and S, would produce the plants with triangular-shaped capsules, while those with genotype ttss would produce the plants with top-shaped capsules.

Q.67 Below given graph illustrates the occurrence of a specific process that inhibits transcription in the people of different age groups.

Based on the given information, answer the following questions:

1. Fill the Y-axis of the graph.
2. Interpret the data illustrated in the graph.
3. Cytosine in DNA can be converted into methylcytosine by addition of a methyl (CH₃) group. This change happens under what condition?

Ans

1. The specific process that inhibits transcription is methylation. So, graph is plotted with age group at X-axis and percentage of methylation level at Y-axis.
2. From the data, it is quite clear that the percentage of methylation is maximum at birth and decreases with the growing age.
3. Cytosine in DNA can be converted into methylcytosine by addition of a methyl (CH₃) group. This enzymatic reaction occurs only where guanine is present at the 3’ side of the cytosine in the base sequence.