Question

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

Answer 1

Answer:

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

Answer 2

Answer:

a) 1/4

b) 81/256

c) 3/4

d) 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%.