Question

In: Biology

In humans, hypotrichosis (sparse body hair) is a rare autosomal recessive condition. Let h = hypotrichosis...

In humans, hypotrichosis (sparse body hair) is a rare autosomal recessive condition.

Let h = hypotrichosis and H = Normal.

Family A consists of two phenotypically normal parents with five children: the first two have hypotrichosis, and the rest are normal.

Family B consists of a phenotypically normal man and a hypotrichotic woman with three children: one has hypotrichosis, and the other two are normal.

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Family C is formed when a normal child from family A marries a normal child from family B.

e.What is the probability that their first offspring is hypotrichotic?

f.If they have four children, what is the probability that they will all be normal?

Please answer E and F. Thank you!

Expert Solution

Answer:

Family A: Both parents are phenotypically normal. They have normal children and those with hypotrichosis.
This indicates that the recessive allele is present in both the parents, which when passed on to the offsprings will cause the condition to appear.
Therefore family A are phenotypically normal parents and are genotypically heterozygous (Hh). Pedigree is shown here along with probable genotypes.:

Family B: also has hypotrichosis in children. Father is phenotypically normal and his wife has hypotrichosis. Hence genotypically father can only be a heterozygous carrier (Hh). The mother suffers from hypotrichosis hence homozygous recessive (HH). Pedigree is shown here along with genotypes:

Family C: has two phenotypically normal individuals, one from each family A & B. There will be two genotypic possibilities here as shown by two punnet squares (Case I and Case II) below:

Answer e):
The probability that their first offspring is hypotrichotic:

As seen from punnet square above,
In case I - There is a 0% probability, that the first child will be hypotrichotic.
In case II - There is a 25% probability that the first child is hypotrichotic.

Answer f):
The probability that all of their 4 children are normal:

In case I - as seen from the punnet square above, there is a 100% probability, that all 4 children will be normal.

In case II - as seen from punnet square above, the probability can be calculated as -

as one in four children will have hypotrichosis, the chances of a normal child are 3/4 for each child.
Therefore the probability that all four will be normal (without hypotrichosis) will be cumulative
Probability = (3/4) x (3/4) x (3/4) x (3/4) = 81 / 256 = 0.316 OR 31.6% chances that all children will be normal.

gladiator answered 2 days ago

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