Question

6. Hemochromatosis is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following? (a) All three children are of normal phenotype. (b) One or more of the three children have the disease. (c) All three children have the disease. (d) At least one child is phenotypically normal. (Note: It will help to remember that the probabilities of all possible outcomes always add up to 1.) 7. The genotype of F, individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? (a) aabbccdd (b) AaBbCcDd (c) AABBCCDD (d) AaBBccDd (e) AaBBCCdd

Answer 1

6.

Since both parents are carriers, they will both pass on the normal allele 50% of the time and the disease allele 50% of the time.

a)

All three children have the normal phenotype.

Since a child will have the normal phenotype when they are Homozygous for the normal allele or heterozygous.

Thus, probability of a child being phenotypically normal
= Probability of a child with both normal alleles + probability of a heterozygous child
= (0.5 * 0.5) + 2 * (0.5 * 0.5) = 0.25 + 0.5 = 0.75

Thus, probability of al children being phenotypically normal
= Probability of a Normal Child * Probability of a Normal Child * Probability of a Normal Child
= 0.75 * 0.75 * 0.75 = 0.421875

b)

The Probability of one or more children having the disease = 1 - No children have the disease = 1 - 0.421875
    = 0.578125

c)

Probability of all three children having the disease
= Probability that all 3 children are homozygous recessive for the disease allele
= 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5
= 0.015625

d)

The Probability that at least one child is phenotypically normal
= 1 - Probability of all three children having the disease
= 1 - 0.015625 = 0.984375

7.

The F1 progeny have the genotype AaBbCcDd. Each allele has a 50% chance of being transmitted from either parent.

Since an individual can be homozygous only when both parents contribute the same allele. Thus, the probability of a Homozygous genotype is (0.5 * 0.5).

However, Heterozygotes are formed when one parent passes on the Dominant allele and the other parent passes on the recessive allele. This can happen in two ways. Either Parent 1 passes on the dominant allele (A) and parent 2 passes on the recessive allele (a) or Parent 1 passes on the recessive allele (a) and parent 2 passes on the dominant allele (A). Thus the probability of a heterozygote is 2 * (0.5 * 0.5).

a)

Probability of F2 offspring with genotype aabbccdd
= (0.5 * 0.5) * (0.5 * 0.5) * (0.5 * 0.5) * (0.5 * 0.5) = 0.00390625

This is because each allele being inherited from a parent if 0.5. Since the events are independent, they are multiplied.

b)

Probability of F2 offspring with genotype AaBbCcDd
= 2 * (0.5 * 0.5) * 2 * (0.5 * 0.5) * 2 * (0.5 * 0.5) * 2 * (0.5 * 0.5) = 0.0625

c)

Probability of F2 offspring with genotype AABBCCDD
= (0.5 * 0.5) * (0.5 * 0.5) * (0.5 * 0.5) * (0.5 * 0.5) = 0.00390625

d)

Probability of F2 offspring with genotype AaBBccDd
= 2 * (0.5 * 0.5) * (0.5 * 0.5) * (0.5 * 0.5) * 2 * (0.5 * 0.5) = 0.015625

e)

Probability of F2 offspring with genotype AaBBCCdd
= 2 * (0.5 * 0.5) * (0.5 * 0.5) * (0.5 * 0.5) * (0.5 * 0.5) = 0.0078125