Question

In studying the gene pool of a population, you find that 15% of the population is homozygous recessive (tt) for a trait. How would you use this information to solve for the frequency of the dominant T allele?

A. Square root of 0.15

B. 1 – Square root of 0.15

C. 15 ÷ 100

D. 0.152

Answer 1

You can use hardy-weinberg: \(\sf \color{Red}{p^2+pq+q^2}\)
and \(\sf \color{blue}{p+q=1}\)
the total must equal 1, or 100%.

where p = DOMINANT "T", and q = RECESSIVE alllele, "t"

Is any of this sounding familiar?

Answer 2

Just the dominant recessive part

Answer 3

Is it C?

Answer 4

If you know that \(\sf \color{red}{q^2=0.15}\), then you know that \(\sf \color{red}{q^2 = \sqrt{q^2}=q = 0.38}\)

Answer 5

and if p + q, MUST equal 1, then to find p, you do 1-0.38 = p = 0.61

Answer 6

So, as you can see. It is NOT \(C\)

Answer 7

oh..

Answer 8

Do you follow me?

Answer 9

Not at all. Sorry

Answer 10

Where are you lost?

Answer 11

all of it really

Answer 12

Haha. Ok.
Well, let's see hardy-weinberg equation is defined as: \(\sf \color{red}{p^2+2pq+q^2}\), which as you remember from math, is simply factored out form of \(\sf \color{red}{(p+q)^2}\)

\(\sf \color{red}{p = homozygous~DOMINANT~ALLELE}\)
\(\sf \color{BLUE}{q = homozygous~RECESSIVE~ALLELE}\)

These are all frequencies. So for example, you have here is 15% frequency, of what they say \(\sf \color{red}{homozygouz~recessive}\) - tt
that means that it represents \(\sf \color{blue}{q}\)
but, hardy-weinberg shows that you have \(\sf \color{blue}{q^2}\). Remember that you have \(\sf \color{red}{p^2+2pq+q^2}\)
How do you solve for ONLY \(\sf \color{blue}{q}\)? You take the square root.
Also, the total amount should equal 1!

\(\sf \color{blue}{p+q=1}\)

So, that means if you already have \(\sf \color{blue}{q}\), and you want to get \(\sf \color{red}{p}\), you do \(\sf \color{purple}{1-\sqrt{q^2}=p}\)

I hope this makes a little more sense. It's hard to exaplain in few words online.