![PCB 3063L ProBof or aAs 1:2:1 10. In humans, the rh factor is inherited as a dominant gene (R). Individuals with this llele are referred to as rh positive. Two (2) heterozygous individuals (Rr) decided to reproduce. What is the probability of four (4) rh positivechildrenzu- a. b. What is the probability of four (4) rh negative boys? [1] nuy prop r 0.02% nue rc. What is the probability of two (2) rh positive boys and two (2) rh negative girls [] oro . 22% d. If they have five (5) children, what is the probability that the first two (2) will be rh negative and the last three (3) will be rh positive? [11 posifvè: 34 2.6%. e. What is the probability of three (3) rh positive children out of eight (8) children? [11 You will need to use the binomial expansion formula for this problem: - s as 31(R-3) I0320x 042Vas x.DOb4ausbas 220 or275%](http://img.homeworklib.com/questions/426411b0-1fcb-11eb-9fc8-b788cdcbf410.png?x-oss-process=image/resize,w_560)
Homework Answers
rh positive (R) > rh negative (r)
rh positive genotypes = RR, Rr, rR
rh negative genotypes = rr
Genotypes of the parents:
Heterozygous rh positive = Rr
As per Mendelian Laws of Inheritance (Segregation, Independent
Assortment, and Dominance):
Probability of an rh positive child = 3/4
Probability of an rh negative child = 1/4
Probability of a boy = Probability of a girl = 1/2
ANSWER
10.
Answer a.
P of 4 rh+ children = (3/4)4 =
31.64%
Answer
b.
P of 4 rh- boys = ((1/4) * (1/2))4 =
0.0244%
Answer
c.
P of 2 rh+ boys & 2 rh- girls =
4C2? (3/4 * 1/2)2 * (1/4 *
1/2)2 = 1.318%
The order in
which the kids are born also matters hence the
4C2
Answer
d.
P of first 2 to be rh- & last 3 rh+ =
(1/4)2 * (3/4)3 =
2.63%
Answer
e.
P of 3 rh+ children out of 8 children =
?8C3? (3/4)3 * (1/4)5 =
2.307%
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one question:
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The region on the Y chromosome where the male-determining gene
is found
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is found
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