A small proportion of humans have resistance to strains of HIV, the virus that causes AIDS. Their immunity is caused by a deletion in gene CCR5, a co-receptor on white blood cells needed for HIV infection. The proportion of people with resistance (genotype aa) to HIV is approximately 8%. If these alleles are in Hardy Weinberg equilibrium, what proportion of people are carriers for the HIV resistant allele?

The correct answer is "40.6%". A further explanation is provided below.

Explanation:

The given values are:

The proportion of people with resistance is,

aa = 8%

or,

    = 0.08 (q²)

q² = 0.08

q   = √0.08

    = 0.28284

Now,

⇒

⇒      

hence,

For the HIV resistant allele, the proportion of people will be:

=

=

On substituting the values, we get

=

=

= (%)

hey, i believe your answer would be d. sugar ribose and the base uracil. i believe this is the answer, because rna consists of sugar ribose, adenine, guanine, cytosine, and uracil.

your first option is incorrect, because that's dna. where it consists of sugar deoxyribose, adenine, guanine, cytosine and thymine.

your second option is also incorrect, this is because sugar deoxyribose is in dna, and uracil is in rna. these two do not make sense together.

your third option being incorrect, because sugar ribose is in rna, but not the base of thymine. rna consists of uracil, not thymine.

lastly your fourth option being correct! this answer is correct, because both sugar ribose, and the base uracil are in rna! ~

answer is d. sugar ribose and the base uracil.

this is a good question! i'll do some research and come back when i have the answer!