If both husband and wife are known to be carriers of the allele foralbinism, wha
ID: 1617 • Letter: I
Question
If both husband and wife are known to be carriers of the allele foralbinism, what is the chance of the following combinations in afamily of four children: (a) all four unaffected; (b) threeunaffected and one affected; (c) two unaffected and two affected;(d) one affected and three affected?Assuming that the question states that both parents are carriers,I've come to the conclusion that the cross looks like this: Aa xAa
A: Dominant, normal
a: recessive, albinsim
a)all four affected. The solution manual states: (3/4)^4 =81/256
I'm having a hard time grasping this, why would you multiply4x(3/4)64=81/64????
b)3 unaffected, 1 affected. I get it.4(3/4)(3/4)(3/4)(1/4)=27/64
c)2 unaffected and 2 affected. DON'T GET?
the solution manual states: 6x(3/4)^2x(1/4)^2=256
Why not: 4x(3/4)^2 x (1/4)^2 = 9/64 ????????
d)1 unaffected and 3 affected. I get it. 4 x (3/4)x(1/4)^3=3/64
I've got the answers. If somebody understands the concepts andcould explain parts a. and c. that'd befantastic!
Explanation / Answer
For a) This is because the rule of probability. When you want to calculatethe probability that 4 events happen in the same time, you have tomultiply the probability by themselves according to the number ofevents. You cannot times the number of events. Example: Two are affected (given probability = 3/4) P(Two are affected) = (3/4) x (3/4) P(Three are affected) = (3/4) x (3/4) x (3/4) P(Four are affected) = (3/4) x (3/4) x (3/4) x (3/4) P(n are affected) = (3/4)n For (c) This is because binomial probability The formula is nCr (probability of success)r x(probability of event)1-r (where r is the number of success event, n is the total event) So, finding the probability of 2 unaffected and 2 affected. Let thesuccess event be the number of affected Probaility of affected = 3/4 Total event = 4 (4 children) r = 2 (number of affected) Applied the formula: 4C2 (3/4)2 x (1/4)2 =0.2109 Hope this helps!
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