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
