If 4 percent of a population in equilibrium expresses a recessive trait, what is the probability that the offspring of two individuals who do not express the trait will express it?

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%.  

explanation:

b,bacteria convert nitrates

into nitrogen gas.

d,plants and animals take in

and use nitrogen compounds.

e,bacteria break down waste    

products and return nitrogen  

to the soil.

a,bacteria convert atmospheric

nitrogen to ammonia.

c,bacteria convert ammonia

to nitrites and nitrates.