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 A

If the couple has fraternal twins, what is the probability that both children will be unaffected by both conditions?

Part B

If 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 C

What 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 D

If 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

Answer:

A. 81/256

B. 1/256

C. 4/9

D. 1/4

Explanation:

A. Given,

Both parents to be heterozygous.

Therefore, the chances for one child to inherit a defective gene from each parent

= Â ½ * ½ =Â ¼.

In the same vein, the probability of being affected by albinism

=Probability of possessing a defective gene from each parent

= Â ½ * Â ½

= Â ¼

Now, The probability of not being affected is represented using the formulae

= 1- the probability ofbeing affected

= 1-¼ Â

= Â ¾.

Thus, the probability of not being affected by both albinism and sickle cell anemia

= Â ¾ * ¾

= Â (3 * 3)/(4 * 4)

= Â 9/16.

So,

The probability of neither twins to be affected

= 9/16 * 9/16

= (9 * 9)/(16 * 16)

=81/256.

B. We obtain the chances for a child to possess one defective gene from each parent

= Â ½ * ½

= Â ¼.

In the same vein, the probability of being affected by albinism

=Probability of inheriting one defective gene from each parent

= Â ½ * Â ½

= Â ¼

Now, the probability of being affected by both albinism and sickle cellanemia

= Â ¼ * Â ¼

= 1/16

Therefore, the probability of both twins to be affected

= 1/16 * 1/16

= 1/256.

C. Take, the probability of twin 1 to be a carrier for albinism to be = 2/3 (note, the homozygous recessive is affected, not the carrier).

We can say that the chances of twin 1 to be a carrier for both the diseases

= 2/3 * 2/3

= 4/9.

In like vein, the chances of carrying neither recessive allele

= 1/3 * 1/3

= 1/9.

Therefore, the probability that one of the fraternal twins is a carrier of either, but not both,

= 1- (4/9 + 1/9)

= 1 - (5/9)

= 4/9

D. Given, the probability of one fraternal twin carrying a gene for either disease:

Take;

The provability of carrying the gene to be 3/4

= 3/4 * 3/4

= 9/16

Thw chances of non occurrence of gene = 1/3

And for fraternal twins

      = 1/3 * 1/3

= 1/9

Therefore, 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

= 1 - (9/16+1/9)

= 1/4.