Respuesta :
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:
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%.
