Question

Sickle cell anemia and albinism are both recessive traits in humans. Imagine that a couple, already pregnant with twins, has just learned that they are both heterozygous for both of these traits.As the couple's genetic counselor, the couple asks you the following questions about how their carrier status will affect their offspring.Part AIf the couple has fraternal twins, what is the probability that both children will be unaffected by both conditions?Part BIf the couple has fraternal twins, what is the probability that both of the couple's children will have both sickle cell anemia and albinism?Part CWhat is the probability that one of the fraternal twins is a carrier of either, but not both, of the conditions?(Hint: You will need to use both the product law and the sum law to answer this question.)Part DIf the couple has fraternal twins, what is the probability of having two phenotypically normal children, one being a carrier of only the sickle cell anemia recessive allele, and the other being a carrier of only the recessive allele for albinism?(Hint: You will need to use both the product law and the sum law to answer this question.)

Answer 1

A. 81/256  B.1/256 C 4/9     D 1/4

Explanation:

A. Both the parents are heterozygous for sickle cell anaemia, the probability of gene passing to one child is

1/2A*1/2= 1/4 A

The probability of albinism in one of the child of heterozygous parents:

1/2*1/2= 1/4 A

Now from the above data the probability of a child not having albinism the diseases will be known by:

1-1/4 A

= 3/4 A

Similarly for sickle cell anemia the probability of not occurring the disease is also 3/4 S

Now applying the product rule for getting the probability of both the diseases not occurring.

3/4 A*3/4 S = 9/16

Here the two fraternal twins are in question, so the probability of both the children to be unaffected will be 9/16*9/16

=81/256

B. The chances for each child to inherit defective gene from each parent in case of Albinism .is A 1/2*1/2  = 1/4 A

Similarly same with sickle cell anaemia

1/4 S

applying the product law, we can determine the probability of the occurrence of both the diseases in one child

1/4*1/4

1/16

Hence in two children that are fraternal twins, the probability is

1/16*1/16

= 1/256

C. From the data, we can see that  the probability of twin to be a carrier is 2/3

the chance of occurrence of both the diseases in one child is

2/3*2/3

=4/9

The chances of not carrying the gene of either disease is1/3*1/3

=1/9

Thus, we can know the probability that if it happens to be fraternal twins the chances of diseases

1-4/9+1/9

=4/9

D. we know that the probability of one fraternal twin carrying a gene for either disease:

chances of carrying the gene is 3/4

chances of not occurrence of gene 1/3

So, in case of fraternal twins 3/4*3/4= 9/16

                                               1/3*1/3 = 1/9

So,1- 9/16+1/9

    = 1/4