GENETICS: Please explain the following. 1 Assume baldness is determined by one g
ID: 133567 • Letter: G
Question
GENETICS: Please explain the following.
1 Assume baldness is determined by one gene and is recessive. If two heterozygous parents have six childrer What is the probability that all 6 will be normal- What is the probability that all 6 will be bald- What is the probability that at least one is normal What is the probability that at least one is bald- 2 An X-linked dominant allele causes hypophosphatemia in humans. A man with hypophosphatemia marries a normal woman. What proportion of their sons will have the disease? What proportion of their daughters will have the disease?Explanation / Answer
Answer:
1).
B_ = Normal
bb = Bald
Bb x Bb ----Parents
B
b
B
BB (normal)
Bb (normal)
b
Bb (normal)
bb (bald)
Normal = ¾
Bald = ¼
The probability that all 6 will be normal = ¾ * ¾ * ¾ * ¾ * ¾ * ¾ = 729 / 4096 = 18%
The probability that all 6 will be bald = ¼ * ¼ * ¼ * ¼ * ¼ * ¼ = 1 / 4096 = 0.02%
The probability that atleast one is normal = ¾ * ¾ * ¾ * ¾ * ¾ * ¼ = 243 / 4096 = 5.93%
The probability that atleast one is bald = ¼ * ¼ * ¼ * ¼ * ¼ * ¾ = 3 / 4096 = 0.07%
2).
Man( XH Y) x Woman (Xh Xh)---Parents
Xh
XH
XH Xh (daughter with disease)
Y
Xh Y (normal son)
No son will have the disease.
100% of their daughters will have the disease.
B
b
B
BB (normal)
Bb (normal)
b
Bb (normal)
bb (bald)
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