**Hardy-Weinberg problem set**

Hardy-Weinberg Theorem states that if a population is NOT evolving then the frequencies of the alleles in the population will remain stable across generations - it is in equilibrium.

We can use the Hardy-Weinberg equation to make predictions about the relative frequency of the different alleles (as well as the associated genotypes), even if there is population growth, as long as the five conditions we discussed in class hold true. Think about what those five conditions are.........

**no mutations, no natural selection, no gene flow,
large population size, random mating**

Alternatively, Hardy-Weinberg equation can be a means to determine if a population is indeed evolving - that the allele frequencies are changing and therefore NOT at equilibrium.

The trick to using the Hardy-Weinberg equation to help evaluate the frequency of a particular allele frequency at time X, or to make a prediction about the frequency of a particular genotype/phenotype in future generations, is to go through the math is a step-by-step manner.

For the problems below assume all genes have only two alleles and there is a simple dominant recessive relationship.

-------------------------------------------------------------------------------------------

*1. If 98 out of 200 individuals in a population express the recessive
phenotype, what percent of the population would you predict would be heterozygotes?
*

(a) I have given you information on the frequency of the homozygous recessive
(or q*2*). So start by determining q*2* and then solving for q.

**q 2 = (98/200) = 0.49 (or 49%)**

**q = square root of 0.49 = 0.7 (70%)**

(b) Now that you have q, you can solve for p. Remember there are only two alleles in the population, so if you add the frequency of the two alleles, you have accounted for all possibilities and it must equal 1. So p + q = 1.

**p = 1-q **

**p = 1 - 0.7 = 0.3 (30%)**

(c) Now what is the formula for heterozygotes? Think back to the Hardy-Weinberg equation -- it is dealing with the genotypes of individuals in the population.

**p 2 + 2pq + q2 = 1**

**frequency of homozygous dominant + frequency
of heterozygotes + frequency of homozygous recessive = 1 **

**so.....2pq = frequency of heterozygotes**

**frequency of heterozygotes = 2 (0.3)(0.7) = 0.42
or 42%**

(d) Now that you have figured out the % of heterozygotes, can you figure out the % of homozygous dominant? Does the % of homozygous dominant, heterozygotes and homozygous recessive individuals add up to 100%? If not, you have made an error. Those are the only three genotypes possible with only two alleles and a simple dominant and recessive relationship.

**p 2 = (0.3)(0.3) = 0.09 (or 9%)**

**p 2 + 2pq + q2 = 1**

**0.09 + 0.42 + 0.49 = 1.0**

*2. Your original population of 200 was hit by a tidal wave and
100 organisms were wiped out, leaving 36 homozygous recessive out of the 100
survivors. If we assume that all individuals were equally likely to be wiped
out, how did the tidal wave affect the predicted frequencies of the alleles
in the population? *

Again, start with the frequency you know -- homozygous recessive. Follow the same step-by-step procedure as above.

What is the frequency of homozygous recessive?

**q 2 = (36/100) = 0.36**

**q = square root of 0.36 = 0.6**

What is the predicted frequency of heterozygotes?

**frequency of heterozygotes = 2pq**

**p = 1 - 0.6 = 0.4**

**frequency of heterozygotes = 2 (0.4)(0.6) = 0.48**

What is the predicted frequency of homozygous dominant?

**p 2 = (0.4)(0.4) = 0.16**

**Double check: **

**p 2 + 2pq + q2 = 1**

**0.16 + 0.48 + 0.36 = 1.0**

Given that the allele frequencies did change as the result of the tidal wave, we would say that microevolution has occurred. What do we call the phenomenon that caused this evolution?

**the drastic reduction in size of a population
due to some chance event is called a bottleneck event - particularly
when the original gene pool (allele frequencies) is no longer represented in
the surviving population **

**since there is now a small population, chances
are likely that it will be subjected to genetic drift and continue
to shift away from the original allele frequency (pre-tidal wave)**

------------------------

*3. Lets say that brown fur coloring is dominant to gray fur coloring
in mice. If you have 168 brown mice in a population of 200 mice........*

**Again start with what you know -- if 168 out
of 200 are brown, that means 32 mice must be gray. There are only two phenotypes
(three genotypes). So the first step is to determine q 2. **

**Why don't you start with p 2 since I
given you the frequency of the dominant phenotype? **

**Think about it -- you have two DIFFERENT genotypes
that can give you the dominant phenotype. Homozygous dominant and heterozygotes.
So the frequency of the dominant phenotype would be: **

**frequency of dominant phenotype
= p 2 + 2pq**

**You can't solve for p, if both p and q are unknown.
So solve for q first --- **

**q 2 = (32/200) = 0.16**

**q = 0.4**

**p = 1 - (0.4) = 0.6**

What is the predicted frequency of heterozygotes?

**2pq = 2 (0.6)(0.4) = 0.48**

What is the predicted frequency of homozygous dominant?

**p 2 = (0.6)(0.6) = 0..36**

What is the predicted frequency of homozygous recessive?

**q 2 = (0.4)(0.4)= 0.16**

p2+ 2pq + q2= 1

0.16 + 0.48 + 0.36 = 1.0

------------------------** **

4. **If 81% of a population is homozygous recessive for a given trait..........
Note questions are out of order in how you should figure out the answers --
I have indicated the order in which I would answer the questions. **

**(STEP 2)** What is the predicted
frequency of homozygous dominant?

p2 = p * p or 0.1 * 0.1 = 0.01

**(STEP 3)** What is the predicted
frequency of heterozygotes?

2pq = 2(0.1)(0.9) = 0.18

**(STEP 1) **What is the frequency
of the dominant and recessive alleles in the population?

In order to figure this out you need to know what p (frequency of the dominant allele) -

You know q2 = .81 (frequency of homozygous recessive)

q = square root of 0.81 or 0.9 = the frequency of the recessive allele

p = 1-q or 1-0.9 = 0.1 - the frequency of the dominant allele

**(STEP 4)**
DOUBLE CHECK:

p2 + 2pq + q2 = 1

0.01 + 0.18 + 0.81 = 1

------------------------** **

5. **If 51% of the population carries at least one copy of the recessive
allele....... Note questions are out of order in how you
should figure out the answers -- I have indicated the order in which I would
answer the questions. **

**(STEP 2)** What is the predicted
frequency of individuals in the population that express the dominant phenotype?

the predicted frequency of individuals who express the dominant phenotype are:

frequency of homozygous dominant + the frequency of heterozyotes
**or **

**p2 + 2pq **

**(0.7 * 0.7) + (2(0.7)(0.3) **

**0.49 + 0.42 = 0.91 **

**(STEP 3)** What is the predicted
frequency of individuals in the population that express the recessive phenotype?

frequency of individuals with the recessive phenotype
is the frequency of homozgyous recessive **or q2 **

**q2 = q * q or 0.3 * 0.3 = 0.09 **

**Note the above answer and this one equals 1 which
is a good double check **

**p2+2pq+q2 = 1**

**0.49 + 0.42 + 0.09 = 1**

**(STEP 1)** -
the information in the question gives you the % of individuals who are homozygous
dominant, but not who expresses the dominant phenotype (which would be homozygous
dominant and heterozygotes).

So need to solve for q and p and then answer the questions above:

p2 (frequency of homozygous dominant) would be equal to 100%-51% or 49% or 0.49

p = the square root of 0.49 or 0.7

q = 1-p or 1-0.7 = 0.3