Question

The Hardy-Weinberg equation is an expression showing the frequencies of 100 percent of the alleles for a specific gene in a population. If biologists can show that either p or q is changing over a period of time, what is occurring in the population?

Answer 1

A. The population is not evolving.
B. The population is decreasing in size.
C. The population is evolving.
D. The population is stabilizing.

Answer 2

p+q = 1

Answer 3

i think it's C

Answer 4

am i right?

Answer 5

or D

Answer 6

Hai Sweet Friend, @faisalalif1999 WelcoMe To OpeN studY !!

In the simplest case of a single locus with two alleles: the dominant allele is denoted A and the recessive a and their frequencies are denoted by p and q;
freq(A)=p; freq(a)=q; p + q = 1.

If the population is in equilibrium, then we will have
freq(AA)=p2 for the AA homozygotes in the population
freq(aa)=q2 for the aa homozygotes
and freq(Aa)=2pq for the heterozygotes.

The overall equation for the Hardy-Weinberg equilibrium is expressed in this way:
p2 + 2pq + q2 = 1


Hope, if U Are Satisfied with this Answer, Please Close This Question !

Thank U !! Keep In Touch, with Open study !!

Answer 7

@Koikkara you just defined the equation

Answer 8

which I already knew

Answer 9

What I'm not understanding is this question

Answer 10

yes, I believe you are right with C.

I would've excluded A, B and D to come to your answer.

Since the parameters change over time, you can exclude A. We know nothing about pop size, so B can't be right.

And D can also be excluded, because we know nothing about the frequency of the change. If a population is stabilizing, it's still evolving in some sort. But the only thing we KNOW for sure after reading the question is that some sort of evolution is taking place here.

Answer 11

oh, thank you :)