Question

A randomly mating population has the dominant allele “B” and the recessive allele “b” at the frequencies 0.7 and 0.3 respectively. A sudden cataclysmic event causes all the homozygous recessive individuals to die. However, the population simply carries on with life and gets busy mating. What was the proportion of homozygous recessive individuals before the cataclysm, and what is the proportion of homozygous recessive individuals born in the first generation after the cataclysm?

Answer 1

Do you know the Castle-Hardy-Weinburg equation?

Answer 2

You will need to use it.

Answer 3

is it the p^2 +2pq + q^2=1 equation

Answer 4

how would i start this question im a bit confused :S

Answer 5

i know were trying to find q^2

Answer 6

OK. You are given the frequencies p = 0.70 and q = 0.30 which you use in the CHW equation.

The first thing you probably want to do at this point is find the frequencies of the different genotypes. That is:

freq(pp) = 0.7^2 = 0.49
freq(2pq) = 2*0.7*0.3 = 0.42
freq(qq) = 0.3^2 = 0.09.

Those are the frequencies before all the qq individuals die off.

Next you want to calculate the frequencies of q and p after the die off.

Answer 7

Do you have ideas about how to go about that?

Answer 8

i wish i did lol

Answer 9

OK, so after all the recessive homozygous individuals (qq) are dead, the only ones left are homozygous dominant (pp) and heterozygous (2pq) individuals.

So your CHW equation, which gives the frequencies of each genotype in the population goes from this:

p^2 + 2pq + q^2 = 1

to this:

p^2 + 2pq = 1.

Answer 10

The key is that removing all the qq individuals has changed the relative frequencies of the p and q alleles - there are more ps relative to fewer qs in the entire population.

Answer 11

I hope you dont mind @blues. @jatt3092 asked for my help on the chat, so I'm just going to throw in a clue to help out. 'q' represents the allele frequency for the reccesive gene but it says that all the homo recessive individs die. But you can still have a recessive allele with a dominant individual. That would make them heterozygous, so go from there.

Answer 12

Thanks Shay, the frequency equation after the die off (p^2 + 2pq = 1) plainly has the recessive allele (q) in it and the remark that the die off is merely a reduction of the recessive allele frequency relative to the dominant allele frequency has that clearly established.

Out of curiosity, how would you go about solving the rest of the prob? : D

Answer 13

im curious as well :D

Answer 14

I'm actually not sure. q^2 should stay the same right? because the frequency of the population remains the same I think even if the allele is reduced. Or wait, can you multiply the q^2 by the total population to find the number of homo recessive individs? Then subtract that from the original total population which would be the new population because all the bb individs died. Then go from there, finding a new frequency for the p? I'm not too sure, but this might work.

Answer 15

Or no, the new population would be the 2 pq and since the p^2, not q^2 remains the same, you can use math to find the rest I think

Answer 16

That is, in a nutshell, what needs to be done next. Well done. :D

You have the frequencies of the homozygous dominant and heterozygous individuals before the die off. Next you need to calculate their frequencies after the die off.

It is easiest just to normalize by dividing by the frequency of remaining individuals.

Frequency surviving = pp + 2pq.
frequency surviving = 0.49 + 0.42 = 0.91.

Find the post die of frequencies of homozygous dominant individuals and heterozygous individuals by dividing those frequencies by 0.91.

So after the die off the frequencies of homozygous dominant and heterozygous individuals is this:

pp = 0.49 / 0.91 = 0.5384
2pq = 0.42 / 0.91 = 0.46154.

What happened to the person who we're supposedly tutoring?

Answer 17

Hahaha... They got scared away I think. lol. I dont know? Maybe their helping someone else?

Answer 18

Bugger. In that case, I shall wait until they return to finish the prob.

Answer 19

Okay, I'm going to go have a look to see if I got any more help on my question. @blues you any good with essays by any chance? Because I desperately need help with my question. @jatt3092, If you need any more help, just leave me a message or tag me in a comment. :)

Answer 20

yeah sorry guys I was busy another problem my apologizes..but yeah shay whats ur essay question i could help u out.

Answer 21

Hey Jatt, do you know how to get from the renormalized proportions of homozygous dominant individuals and heterozygous individuals to new values for p and q?

Answer 22

@jatt3092, its is posted right here: http://openstudy.com/users/shay88#/updates/4f81f0e8e4b0505bf08380d4
I will let you figure out @blue's question on your own. Just think about all the stuff that you've been taught and the formulas.

Answer 23

dont' we have to figure out the proportion of homozygous recessive inviduals?

Answer 24

so basically once we find the new question you just have to square it and u got the answer

Answer 25

q value* not question

Answer 26

Yes, exactly. You need to find new values for p and q for the post die off population. Once you have found q, you have to square it to get the number of homozygous recessive individuals in the new population.

But before that you have to solve for p and q...

Answer 27

Well that's for before the catalysmic. You have to find the new q value using the p^2 value we found. Do you have any ideas of how to go about this?

Answer 28

no :(

Answer 29

OK Jatt, when we renormalized the population of surviving heterozygotes and homozygous dominant individuals, we found the proportion of pp and qp individuals in the new population. That is what the new pp = 0.5384 and new 2pq = 0.46154 are all about.

Is that clear?

Answer 30

Use another formula given to you for the ALLELE frequencies, instead of the one given for the population frequencies that we just used.

Answer 31

Shay, I'm sorry but I don't know which way you're going with that last.

Answer 32

Well we have p^2 so you square that to get p. Right? You following so far?

Answer 33

yes im clear about those

Answer 34

Then we need q for the recessive allele frequency.

Answer 35

blue is the answer 0.0729

Answer 36

The equation for population frequencies is p^2 + 2pq = q^2
The equation for allele frequencies though is different.

Answer 37

Its not by my calculations jatt

Answer 38

So essentially those are two new equations for p and q. We have:

p^2 = 0.5384
2pq = 0.4615.

We can solve those very easily for the new allele frequencies p and q, can't we?

p = sqrt(0.5384) = 0.7338

Once we have that we can get q from the second equation:

q = 0.4615 / (2*p)
q = 0.4615 / (2*0.7338)
q = 0.3145.

And then to get the frequency of homozygous recessive (qq) individuals in the new population is trivial. We just compute q^2.

q ^ 2 = 0.3145 ^ 2 = 0.0989.

Clear?

Answer 39

Well there is an easier way to do it using the allele frequency equation which is p + q = 1. So once you've got p = 0.7338, you get q by manipulating that, and getting q = 1- p. And then squaring that value.

Answer 40

Or you can do it that way if you please. ;D

Answer 41

So I get 0.0729. My mistake, you were right @jatt3092

Answer 42

blue u got a different answer :S

Answer 43

Or no 0.0709. Sorry Hold on I'm just going to go grab an actual calculator instead of the computer one and check this out.

Answer 44

I double checked my math and I actually have to get going. Post it and I'll catch up with it. :D

Answer 45

k thanks you :)

Answer 46

I got 0.2662 for my q value, so that might be the problem.

Answer 47

nah blues did it right

Answer 48

Yeah I cant figure out what I did wrong, but when you add blues values together for p^2+ 2pq + q^2 it comes up for 1. So ignore me a little bit. and go with what blues did.