Hardy Weinberg Answers (Alt)

Hardy-Weinberg Assignment

Here are the answers to the problem set for this week on the Hardy Weinberg assignment.  I’ve provided explanations for each answer, so read through them carefully to understand how to solve problems such as these.

 

1.   A population of pikas has the allelic frequencies of H=.6 and h=.4 in the parental generation.

a.   Calculate the values of p, q, p+q, p², 2pq and q².

b.   What do each of the above values represent?  (answer in parenthesis)

p = .6 (Dominant allelic frequency – H)

q = .4 (recessive allelic frequency – h)

p + q = 1 (This represents the frequency of the population that has the dominant or recessive alleles, i.e, all of them -- H + h)

p² = (.6) x (.6) =  .36  (homozygous dominant phenotypic frequency – HH)

2pq = 2 x (.6) x (.4) =  .48  (heterozygous phenotypic frequency – Hh)

q² = (.4) x (.4) = .16  (homozygous recessive phenotypic frequency – hh)

2.   Suppose this population of pikas now have allelic frequencies of H = .5 and h = .5 in the parental generation.

a.   Using a Punnett Square with the appropriate p and q values, predict the genotypic frequencies of the offspring this population will produce in the next generation.

Entire Population Reproduces (p+q)2	p = .5	q = .5
p = .5	p x p = p2 = .25	pq = .25
q = .5	qp = .25	q x q = q2 = .25

This represents the entire population (p+q) reproducing with itself (with p representing the dominant allele H and q representing the recessive allele h)  This population cross is represented by (p+q)(p+q) or (p+q)².  The Punnett Square gives the genotypic frequencies for pp, pq, qp, and qq.  You can combine pq and qp into one value 2pq.  So the genotypic frequencies in the next generation are predicted to be:

p² = .25 (represents the predicted frequency of HH)

2pq = .5 (represents the predicted frequency of Hh)

q² = .25 (represents the predicted frequency of hh)

This is assuming NO evolutionary change is happening with respect to this trait.

b.   What should be the phenotypic frequencies (high call or low call) of the offspring?

Dominant phenotype (HH and Hh) =  p² + 2pq = (.25) + (.5) = .75

Recessive phenotype (hh) = q² = .25

c.   Suppose you then count the number of high call and low call pikes offspring and find that the numbers match your predictions.  Does this mean this population is evolving (changing) with regard to this trait?  Why or why not?

Because the predicted frequencies match the actual frequencies in the next generation, that means that nothing was causing the values to change from the expected, i.e., no selection force was influencing this trait.  This means the population is not evolving or changing with regard to this particular trait.

3.   You discover a new population of pikas but you don’t know their allelic frequencies.  You do a careful count of the pikas and find that there are 100 total pikas with 84 high call pikas and 16 low call pikas.

a.   What is the frequency of the recessive phenotype (q²) in this population?

The frequency of recessive phenotype is number of low call pikas divided by the total population:

q² = 16/100 = .16

b.   What is the recessive allelic frequency (q) of this population?

Since you know q2, you can figure out q by taking the square root: 

q = √q² = √.16 = .4

c.   What is the dominant allelic frequency (p) of his population?

If you know q, then you can figure out p since p+q = 1:        

p= 1-q = 1 – (.4) = .6

d.   Given these p and q values, how many of the high call pikas are heterozygous (2pq)?

2pq = 2 x (.4) x (.6) = .48

4.   For the population in question three, imagine that the homesite of the pikas experiences heavy spring rains which washes away much of loose earth, leaving behind more solid rock exposed on the surface.  After the population reproduces, you do a count of the offspring and find that there are a total of 100 pikas with 75 high call pikas and 25 low call pikas.

a.   What are the p and q values of this offspring population?

q² = 25/100 = .25

q = √q² = √.25 = .5

p = 1 – q = 1 – (.5) = .5

b.   You determine that this population has experienced evolutionary change.  How can you use the values in (part a) can you use to justify this conclusion?

The p value has decreased from .6 to .5 and the q value has increased from .4 to .5.  The low call pikas are increasing in the population and the high call are decreasing.  This is a change in the genetic frequency in the population which is evolutionary change.

c.   Which of the four forces of evolution (natural selection, migration, genetic drift or mutation) is mostly like the cause of this evolutionary change, based upon the scenario?

This change was caused by environmental change, namely the loss of top soil and the exposure of rock which made the low call expression more advantageous.  This is the basis for evolutionary force of natural selection.

5.   Continuing with the population of pikas in question 4, using their p & q values.  The following spring, new population of pikas comes in and joins this existing population.  You determine that this new combined population now has 200 pikas total, with 128 high call and 72 low call pikas.

a.   What are the new p and q values of this population (notice that the total is NOT 100)?

q² = 72/200 = .36

q = √q² = √.36 = .6

p = 1 – q = 1 – (.6) = .4

b.   Has this population experienced evolutionary change?

Yes.  From the values in question 4, the value of q has increased and the value of p has decreased.  The low call pikas are increasing and the high call pikas continue to decrease.

c.   Which of the four forces of evolution is most likely the cause of this change?

This change was caused by the influx of new pikas (which apparently had a higher percentage of low call pikas).  This is evolutionary change brought on by migration.