Let a and a represent dominant and recessive alleles whos respective frequencies are p and q in a given interbreeding population at equilibrium (with p+q=1). a) if 16 percent of the individuals in the population have recessive phenotypes, what percentage of the total number of recessive genes exist in the heterozygous condition? b) if 1.0 percent of the individuals were homozygous recessive, what percentage of the recessive genes would occur in heterozygotes?

A) %

B) %

Explanation:

A) Given

Frequency of individuals in the population having recessive phenotypes %

Frequency of individuals in the population having recessive phenotypes is represented by

Thus frequency of recessive genes is equal to

frequency of dominant genes is equal to

Now,

% of genes exist in the heterozygous condition

B.

Frequency of individuals in the population having recessive phenotypes %

Frequency of individuals in the population having recessive phenotypes is represented by

Thus frequency of recessive genes is equal to

frequency of dominant genes is equal to

Now,

% of genes exist in the heterozygous condition

thats not a question

answer-

the statement that f(x) = |x+a| + b has exactly one x-intercept is sometimes correct.

solution-

it solely depends on b, whether the function will have one or two or zero x intercept.

this plot of the given function, f(x) = |x+a| + b will be the basic absolute value graph i.e v shape, with vertex translated to (-a, b), instead of origin.

1-   if b is zero, the graph will have exactly one x-intercept, at x= -a

2- if b is positive, the whole graph will be above the x-axis, hence it will have no x-intercepts.

3- if b is negative, the graph will be below the x-axis, hence it will have two x-intercepts.

∴ the statement is sometimes true.