Respuesta :
Answer:
The correct answer is 2.56 %.
Explanation:
It is given that in a population, which is in equilibrium only 4 % will demonstrate a recessive trait. Thus, on the basis of Hardy-Weinberg law of equilibrium, q^2 = 4% or 0.04, that is, q = 0.2 and p will be 1-0.2 = 0.8.
It is mentioned that both the parents do not possess the trait, therefore, the parents would be exhibiting either of the two possible genotypes, that is, AA or Aa. However, to have an offspring with the trait, the genotype of both the parents would need to be Aa. The probability of the offspring to possess the recessive trait can be determined with the help of And rule of probability,
P [Aa] * P[Aa] = [0.2*0.8] * [0.2*0.8] = 0.0256 or 2.56%.
The probability that two individuals were chosen from the population at random will have an offspring that expresses the trait is 0.16.
- We know as per the Hardy-Weinberg equilibrium:
[tex]p^2+2pq+q^2 = 1[/tex]
- 4% of individuals in a population at Hardy-Weinberg equilibrium express recessive traits.
- Two copies of an abnormal gene will be required for a recessive trait to express.
So we have:
[tex]q^2 = 4%\\q^2 = 0.04\\q=0.2\\Also, we know,p+q=1[/tex]
Hence
[tex]p=1-q\\p=1-0.2\\p=0.8[/tex]
- the probability that a man who expresses the trait and a woman who does not will have a child that does express the trait is given by:
Probability (child expresses trait) = P (mother is a carrier) x P (Child has both abnormal alleles)
- the mother can be a carrier or normal. But for the child to acquire the trait, the mother has to be a carrier or heterozygous (2pq) for the trait. So
2pq = 0.32.
- if we cross a male showing the trait with a heterozygous female, the chance that the child will express the trait is
= A/2
Putting all this analysis together, the probability that the child will express the trait is:
[tex]0.32 \times \frac{1}{2}\\= 0.16[/tex]
Thus, the probability that two individuals were chosen from the population at random will have an offspring that expresses the trait is 0.16.
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