Hemophilia is an X-linked recessive disorder in humans. If a heterozygous woman has children with an unaffected man, what are the odds of the following combinations of children?

a. An affected son

b. Four unaffected offspring in a row

c. An unaffected daughter or son

d. Two out of five offspring that are affected

a. 1/4 chance of occurring

b. 81/256

c. 3/4

d. 0.26

Explanation:

X-linked recessive inheritance is a pattern of inheritance in which a mutation in a gene on the X chromosome causes the phenotype to always get expressed in males (they have one X and one Y chromosome with mutation on the X) and in females who are homozygous for the gene mutation (mutation on both X chromosomes).

Indicating the mutation with h and the normal allele with H. Thus, if a heterozygous woman (XHXh) and an unaffected man (XHY). we have:

(XHXh) x (XHY)

XHXH XHXh XHY XhY

normal female normal male affected male

0.5 0.25 0.25

1/2 1/4 1/4

a. 1/4

b. Each unaffected offspring has 3/4 chance of occuring: four unffected - 3/4 x 3/4 x 3/4 x 3/4 = 81/256

c. 1/2 + 1/4 = 3/4

d. The probability of an unaffected offspring is 3/4. Here, the binomial expansion equation (where x = 2, n = 5, p = 1/4, and q = 3/4). The answer is 0.26, or 26%, of the time. probability of affected is 1/4. Thus, using this

P (x) = (⁵₂) pˣ (1 - p) ⁿ⁻ˣ =

= (⁵₂) (1/4) ² (3/4) ³ =

= 5!/3!2! (0.0625) (0.4219) =

= 10 (0.026)

= 0.26

a) 1/4b) 81/256c) 3/4d) 0.26

Explanation:

Given

Heterozygous female = X*X (X * is mutated gene);

unaffected male = XY;

so children are X*X, X*Y, XX, XY

now

a) affected son; genotype (X*Y), (1/4) 1 son is affected from 2 sons and from four total children.

b) Four unaffected offspring in a row = 81/256

c) 3/4 (X*X, XX, XY) are unaffected children,

d) The probability of an affected offspring is 0.25, and here the probability of an unaffected offspring is 3/4.

now use the binomial expansion equation for this;

So, two out of five offspring that are affected = 0.26, or 26%.