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.
