Consider a gene with two alleles, C and M. The table below...

Consider a gene with two alleles, C and M. The table below describes fitness for different genotypes in two populations. Fitness CC CM MM Population 1 1.0 1.0 0.6 Population 2 0.9 0.9 1.0 Assume that both populations begin with frequencies of 0.5 for each allele, population size is infinite, and there is no migration between populations. Based on the table, how would you expect the frequency of C to change over time in population 2? Frequency of C will increase. Frequency of C will decrease. Frequency of C will stay the same. Frequency of C will change randomly.

Transcript

00:01 Very long question.

00:02 And i invite you to look at the table as we go through this.

00:05 I'm not going to redraw the table.

00:07 I'm just going to go through this very quickly and make a few comments and a few notes as we go along the way here.

00:15 But there's a lot to cover.

00:16 So when you look at these questions, do you have to write whether they're in hardy -weimberg or not? so the first one, you can see everybody has a big a allele.

00:26 So p is 1, q is 0.

00:29 So it's in hardy -wyne equilibrium because you wouldn't expect there to be any heterozygots or homozygots recessive.

00:35 For the second one, the answer is no, it's not in hardy -weinberg equilibrium.

00:40 And the reason is that there are heterozygous there, and you'd expect when they mate and have offspring, that some of those offspring would be homozygotes.

00:49 For three, answers, yes, it is in hardy -weinberg equilibrium.

00:54 And again, for the same reason as number one is, for four.

01:01 It's not in hardy -winberg equilibrium.

01:04 And you can simply do the basic math here to get p and q.

01:07 It's a little bit tedious, but not terrible.

01:15 For five, again, not in hardy -winberg equilibrium.

01:19 And if you look at those numbers, we would expect there to be more heterozygotes, and we're not getting that.

01:25 So, again, these numbers here would be 0 .375 and 0 .625.

01:33 All right.

01:35 It's for six.

01:39 This was also yes, as i lose track on my notes.

01:48 This one was in hardy -wyneberg equilibrium.

01:51 And you can, whenever they're in equilibrium, all you have to do to get p's and q is to take the square root of, say, the percentage proportion of homazigous recessives.

02:02 That'll give you q, and then you can just solve for p.

02:04 So that's pretty straightforward.

02:06 For seven, this is not in hardy -wringberg equilibrium.

02:10 If you look at it, it's just such a strange set of numbers.

02:13 Something is going on and solving for it.

02:17 You're going to have half the alleles will be big a and half will be a little a.

02:24 Eight is yes.

02:27 It is in hardy -oenberg equilibrium.

02:29 And again, it's easy when they're hardy -wiring your equilibrium because you just have to take the square root, say the number of homozygous or a proportion of homozygous recessive.

02:37 To get the values.

02:40 9 is yes, and it's simply the reverse of the number 8.

02:47 So that's pretty straightforward.

02:49 And then finally, number 10 has a tiny numbers involved with it, but it is in hardy -weinberg equilibrium.

02:56 And that's 0 .993 and 0 .007.

03:02 So that's a and b here in the questions.

03:06 C is asking about question six...

Question

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