Solutions For An Introduction to Genetic Analysis Chapter 3 Problem 48P

a) The most likely inheritance of these phenotypes appears to be due to dominant allele. Two affected parents have given a progeny of 3 affected people. The inheritance of dwarfism appears to be due to a recessive allele as affected persons have unaffected parents. Both these traits are not sex-linked and hence, appear to be autosomally inherited.

b) The genotypes of all members in generation III can be deduced after assigning symbols to each of the genes.

• Presence of cataract – A

• Absence of cataract – a

• Absence of dwarfism – B

• Presence of dwarfism - b

The genotypes of the members in generation III are:

1. a/a; B/b

2. a/a; B/b

3. A/a ; B/-

4. a/a ; B/-

5. A/a ; B/b

6. A/a ; B/b

7. a/a ; B/-

8. a/a ; B/-

9. a/a ; b/b

c) A hypothetical mating between IV-1 and IV-5 would be the cross between the following genotypes:

a/a; b/b x A/- ; B/-

The probability of the first child being a dwarf with cataracts can be determined by checking out the genotypes of the parents. Since the genotype of IV-1 is homozygous recessive, we need to check the probability of the genotype of IV-5.

The probability that IV-5 is heterozygous for dwarfism is 2/3. The probability that she has the b allele will be 1/2.

Hence, the total probability that IV-5 has the b allele and will pass it to her child will be:

The probability that IV-5 is homozygous for cataracts would be 1/3. The probability that she is heterozygous for cataracts would be 2/3.

If she is homozygous, she will pass the cataract allele to her child. If she is heterozygous, the probability of passing it to her child is 1/2.

The probability that the first child is a dwarf with cataracts would be that the child should inherit the A and b alleles from its mother.

The probability of inheriting the b allele:

The probability of not inheriting the a allele:

The total probability would be:

The probability that the first child is a dwarf with cataracts would be 2/9.

The probability of having a phenotypically normal child would be calculated in this way:

The probability that the mother donates the a allele:

The probability that the mother donates the B allele:

The total probability would be:

Therefore, the probability of having a phenotypically normal child would be 2/9.