Top 14 Problems on Genetics (With Solution)

Here is a list of top fourteen problems on genetics along with its relevant solution.

Problem 1:

Albinism is recessive to normal body pigmentation in man. It is an autosomal trait. If a homozygous normal man marries an albino girl, what would be the phenotypic and genotypic ratios in F₂ generation from this marriage?

Solution:

Since the two traits (albinism & normal body colour) are present in autosomes, they would behave as Mendelian monohybrid cross. Since albinism is recessive it might be denoted with “cc” alleles and normal body pigmentation by “CC” genes because both are homozygous.

Now see chart below:

Conclusions:

(a) The marriage would produce, phenotypically 3 normal children and 1 albino child.

(b) The marriage would produce, genotypically, 1 normal homozygous; 2 normal heterozygous; and 1 albino homozygous child.

Problem 2:

Albinism, the total lack of pigment is due to a recessive gene. A man and woman plan to marry and wish to know the probability of their having an albino child. What advice would you give to them if ?

(a) Both are normally pigmented, but each has one albino parent.

(b) The man is an albino.

(c) The man is an albino and woman’s family includes no albino for at least three generations.

Solution:

Albinism is due to an autosomal gene and is expressed in homozygous condition. As such, the problem shall be solved on the Mendel’s principles of Dominance and Segregation.

(A) Since both man and woman now have normal pigmentation but one of their parents was albino, the genotype of the two individuals shall be heterozygous “Aa” (if we denote albinism with “a” and normal pigmentation with “A”).

If they marry, the chances of having an albino child are 25% as explained below in the chart:

The Genotypes of the two individuals shall be Aa and Aa:

Conclusion:

Out of four possible combinations only one would result in producing an albino child. Hence chances are 25%

(B) If man is albino and woman is normally pigmented their genotypes would possibly be “aa” (homozygous recessive for albinism) for man and for the woman either the genotype shall be “AA” or “Aa”, because both cases shall produce the phenotype normal pigmented.

The two crosses would give under noted results as shown in chart:

Conclusion:

The possibility of an albino child would be 50%. (ii) If man is albino and women is homozygous for normal pigment the possibility of albino child will be nil as shown in the chart.

(C) If the man is albino and women’s family includes no albino for three generations the genotypes of man and woman would be homozygous (“aa” for albino and “AA” for normal pigmentation) and shall produce children having only normal pigmentation.

Conclusion:

All normal pigmented (Heterozygous) children.

Problem 3:

In man brown eyes (B) are dominant to blue (b) and dark hair (R) are dominant to red hair (r). A man with brown eyes and red hair marries a woman with blue eye and dark hair. They have two children, of whom one has brown eyes and red hair. Give the genotypes of the parents and children.

Solution:

The genotype of the two parents may possibly be as follows:

For Brown eye and Red hair the genotype may be BBrr or Bbrr and for blue eye and dark hair it could be either bb RR or bbRr.

(a) We can’t accept both parents being homozygous for both characters because than all children would have brown eye and dark hair as per Law of Dominance, which is not the result in problem above.

(b) We can’t accept that man is heterozygous for brown eye colour and homozygous for red hair and woman homozygous for both characters, because in that case all the children would have dark hair, which is also not true as per the results given in problem above.

(c) The only possibility now left is that either the man is homozygous for both characters (eye colour as well as hair colour) with genotype “BBrr” and woman is heterozygous for dark-hair but homozygous for blue eyes with genotype “bb Rr” or (ii) man is heterozygous for brown colour with genotype “Bb rr” and woman is heterozygous for dark hair with genotype “bb Rr”.

In the first case there are only two possibilities as shown in chart below:

Conclusions:

Since there are only two children and one of them is having brown eye and red hair the other would be having Brown eye and dark hair and their genotypes would be Bb rr and Bb Rr respectively. The genotypes of the parents in this case would be Bb rr (father) and bb Rr (mother).

(ii) However, if we accept the (ii) case the marriage would result in producing four types of children as shown in chart below:

Conclusion:

In this case the phenotypes of children would be brown eye and dark hair; brown eye and red hair; blue eye and dark hair and blue eye and red hair’. But, since in the problem it is specifically mentioned that the couple had two children the result of solution (i) would be more appropriate.

Problem 4:

A brown eyed man marries a blue eyed woman and they have eight children, all brown eyed. What are the genotypes of all the individuals in the family?

Solution:

The brown eye colour is a dominant trait because it has expressed in F, children and shall be denoted by B, whereas blue eye should be denoted by b. Since all the eight children in F₁ are brown eyed, this is only possible as per Law of Dominance, if the parent with dominant character (brown eye) is homozygous having, BB genotype.

Conclusion:

If now we accept this hypothesis the genotypes of parents would be BB for father, bb for mother and Bb for all children.

Problem 5:

A blue eyed man, whose both parents were brown eyed, marries a brown-eyed woman. They had one child, who is blue eyed. What are the genotypes of all the individuals in problem mentioned above?

Solution:

Accepting that brown eye colour as dominant and blue recessive we proceed as follows:

(a) The blue eyed man, whose both parents were brown eyed, should have the genotype bb because, as per law of dominance, blue colour will only express in homozygous condition.

(b) Both parents of this blue eyed man should have the genotype Bb because, as per law of dominance, a recessive character will only become homozygous when both the parents are heterozygous for that particular character.

(c) The blue eyed man with bb genotype marries a brown eyed woman and produced blue eyed child, who would again be having genotype bb: as per the direction of Law of dominance described in (a) above. If both father and child are having genotypes bb the brown eyed mother, automatically, should have the heterozygous genotype Bb.

Conclusions:

The genotypes of all individuals are as under:

Blue eyed man = bb

Parents of blue eyed =Bb and Bb man

Brown eyed woman = Bb

blue eyed child = bb

Problem 6:

What are the chances that the first child of a marriage of two heterozygous brown eyed parents will be blue eyed? If the first child is brown eyed what are the chances that the second child would be blue eyed?

Solution:

Accepting that brown eye colour is dominant over blue eye colour we proceed as under:

(a) From a marriage of parents heterozygous for brown eye colour with genotype Bb, two types of children’s could be produced i.e., either brown eyed or blue eyed in a ratio of 3: 1 (3 brown eyed: 1 blue eyed), as per Law of Segregation.

Conclusions:

(a)The chances of having a blue eyed child are 25%.

(b) If the first child is brown eyed the chances of second child to be blue eyed would be 33%.

Problem 7:

In a case of disputed parentage, the mother of an illegitimate child has blood group N, the child has MN, one suspected father has N and other MN. How would you decide the case?

Solution:

In case of blood groups it is established that blood groups M and N are homozygous and are denoted by genotype M^M M^M and M^N M^N respectively. The blood group MN however, is heterozygous and is denoted by genotype M^MM^N.

Accepting these facts we proceed as follows:

(a) Since, the mother of illegitimate child is N she should have the homozygous genotype M^N M^N and would produce gametes of only one type i.e. having M^N genes.

(b) The Child is, since, MN he would be having the heterozygous genotype M^M M^N and must have received M^M gene from one of the suspected father because the other gene M^N must have come from his mother.

(e) Now, we have two persons in question one of whom is having blood group N and the genotype M^N M^N because blood group N is homozygous. The other person in question is having blood group MN and the genotype in this case would be M^M M^N because blood group MN is heterozygous.

(d) Keeping the genotypes of the two suspected persons in view if we try to analyse the possibility of passage of M^N gene to the child we would arrive on the conclusion that the person with group N could not have contributed M^M gene and therefore the only possibility left is the other person with blood group MN and genotype M^M M^N who can contribute an M^M gene.

Conclusion:

The father of the child must be the person with MN blood group.

Sex-Linked Inheritance Exercises:

Problem 8:

Of what type will be the children with reference to colour blindness, when a man is colour-blind and his wife is normal?

Solution:

The cause of the colour blindness is the presence of recessive (c) gene on the X chromosome.

Because man is colour-blind (X^c Y) and his wife is normal (X^CX^C), the following will be the results of the marriage as shown in chart below:

Following will be the results after fertilization:

(i) X^C X^c i.e., normal but carrier daughter.

(ii) X^c Y i.e. normal son.

Conclusions:

No child will be colour blind.

Problem 9:

When a haemophilic male is married to a heterozygous carrier female, what haemophilic proportion will be present in children of each sex?

Solution:

Since, Haemophilia is a disease that causes delayed clotting of blood, It is duo to a recessive gene h’, located on X chromosome, see chart below:

Conclusions:

One haemophilic daughter

One carrier daughter

One haemophilic son

One normal son.

Problem 10:

When a haemophilic male is married to a homozygous non-haemophilic female what will be the result of this marriage?

Solution:

See chart below:

Conclusions:

A ratio of 2 carrier daughter and 2 normal son will be produced.

Problem 11:

Of what type will be the children with reference to colour blindness, when a woman is colour-blind and her husband is normal?

Solution:

See Chart below:

Conclusions:

In such cases all sons would be colour-blind and of the daughters both would be normal but carrier.

Problem 12:

A woman has normal vision but her father was colour-blind. She marries a man who is colour blind. Find out the probability of the first child being colour blind if it is (a) Son or (b) Daughter

Solution:

Colour blindness is a recessive sex-linked character and gene is located in X^c chromosome. As such it expresses in phenotype in males in homozygous condition but in females it would only express in phenotype when homozygous (present in both’ X^c X^c chromosomes).

Now:

The woman with normal vision had colour-blind father meaning thereby that she is a carrier or heterozygous with genotype “X^c X^C“.

She marries the man, who is colour blind and would have the genotype “X^CY”.

Their marriage would result in as shown in chart below:

Conclusions:

In this problem there is a slight twist that the probability of first colour blind child if male and if female is asked. The answer should be 25% for male child or female child, but if we do not emphasize on the word if than probability of first child being colour blind would be 50%.

Problem 13:

A girl of normal vision, whose father was colour blind, marries a man of normal vision, whose father was also colour blind. What type of vision can be expected in their offspring’s?

Solution:

Knowing fully well that gene for colour blindness is recessive and is located in X^c chromosome we proceed to solve the problem as follows:

(a) Girl has normal vision but her father was colour blind. It means the father, although homozygous, would behave as homozygous in case of daughters because the single X^c chromosome with gene for colour blindness would be transferred to the daughters. Since the daughter still has normal vision it means she is a carrier and carries no colour blind gene in her other X^C chromosome and therefore her genotype shall be “X^cX^C“.

(b) The man also has normal vision and his father was also colour blind. Contrary to the aforesaid explanation in (a) the genotype of man should be “X^CY”. This is because he would have received Y chromosome from his colour blind father and, since he has normal vision he should have the X^C chromosome free of the gene for colour blindness.

(c) The marriage of this girl and this man would produce offspring’s as shown in chart below:

Conclusions:

All daughters shall have normal vision, but 50% of the son’s would have colour blind vision.

Problem 14:

A colour blind man marries a woman of normal vision. They have sons and daughters all of normal vision and all of them married persons of normal vision. Where among the grand children would colour blindness be expected to appear? If there are cousin marriages among these grand children, where among their offspring’s would colour blindness be expected to appear? (All persons with normal vision mentioned are homozygous).

Solution:

Knowing fully well that gene for colour blindness is recessive and is located in X^c chromosome we proceed to solve this problem as follows:

(a) The colour blind man must be having genotype X^CY and the woman of normal vision, at first step, should be homozygous for normal vision and must be having genotype X^CX^C.

(b) The sons and daughters from the marriage of (a) are all having normal vision, which means that all sons are having an X^C chromosome for normal vision with genotype X^CY and all daughters must be carriers because they would have received the X^c chromosome for colour blindness from their father in (a) and their genotype should be “X^c X^C“.

(c) All these sons (X^CY) and daughters (X^c X^C) married (X^c X^C & X^c Y) as mentioned in the end of the problem.

Now, the colour blindness should be expected to appear only in the grand male children produced from this marriage of carrier daughters as shown below in the chart:

Conclusions:

Colour blindness would appear in 50% of the sons (male grand children from the marriage of daughters of (b).

(e) If cousin marriages take place between these children of (d) the colour-blindness would appear in male children from the marriage of normal carrier daughter (X^c X^C) from (ii) and normal male children (sons) of (i) X^CY only as explained below:

Result:

Colorblindness would appear only in 50% male grand children from the cousin marriage between sons of (i) and daughters of (ii).