Module 8 Population Genetics

4.8.3 page 2

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In addition to giving us the conditions for genetic equilibrium and micro-evolution, Hardy and Weinberg also gave us a mathematical formula that allows us to go back and forth between determining allele frequencies and genotype frequencies. The symbols used are p and q. (For these examples we’ll use A as the dominant allele and a as the recessive allele.)

 

p = frequency of the dominant allele = f(A) =     the # of the dominant alleles         
                                                              the total # of alleles in the population
           
q = frequency of the recessive allele = f(a) =     the # of the recessive allele          
                                                              the total # of alleles in the population

If(A) + f(a)  =  all the alleles in the population, so:

p     +   q     =   1

 

Follow the logic on the Punnett square on p. 682 and you will see how Hardy and Weinberg derived the following equation for determining the frequency of each genotype in the population. (For this example in the Punnett square, their assumption is that 0.70 or 70% of the alleles in the gene pool are dominant: f(B) = 0.70, and that 0.30 or 30% of the alleles in the gene pool are recessive:
f(b) = 0.30 or 30%)

 

The Hardy Weinberg equation we use to predict the frequency of genotypes in a population is:

 

 p2      +       2pq      +      q2      =     1
f(BB)   +     f(Bb)     +     f(bb)    =    the whole population 

 

where:            

p2 = the frequency of the homozygous genotype in the population

 p2 = f(AA)

 

2pq = the frequency of the heterozygous genotype in the population

2pq = f(Aa)

 

q2 = the frequency of the homozygous recessive genotype in the population
 q2 = f(aa)

 

Read 

 

Read carefully through the sample problems on p. 682-3 and work them out yourself using pen and paper. You will need a calculator.

 

Here is another sample problem if you need a little more help:

 

Assume that in grizzly bear populations, brown-tipped fur is dominant to silver-tipped fur.

For the purposes of this problem assume A indicates the brown tipped allele, and a indicates the silver–tip allele.

 

For each question, first of all, decide which of the following you’ve been given in the question. Then decide which of the following you want to find.

 

p                   q                     p2                  2pq                  q2
f(A)               f(a)                  f(AA)               f(Aa)               f(aa)

1. If 45 of 75 grizzly bears have the recessive phenotype of silver-tipped fur, what is the frequency of the recessive allele in this population of grizzlies?
   
   
   • Given: 45/75 or 0.6 of the bears have the recessive phenotype, and therefore the genotype of aa. The frequency of aa is q2, so you’re given: q2 = 0.6
   • Want: The frequency of the recessive allele, so you want: q
     
     
   • Solve: square root  q2 to get q = 0.774
     
     Answer: 0.77 or 77% of the alleles in the gene pool are recessive.
     Note; remember to read the questions carefully, especially on exams. Does the question require frequency, or % frequency?
2. What percentage of the population of bears are heterozygotes?
   
   
   • Given: 45/75 or 0.6 of the bears have the recessive phenotype, and therefore the genotype of aa. So, you’re given: q2 = 0.6
   • Want: the percentage of bears of the heterozygous genotype (Aa) in the population, so you want: 2pq
   • Solve: Square root q2 to get q.  q = 0.77
               p + q = 1, so 1 - q = p.   p = 0.23
               2pq = 2 (0.23) (0.77)   2pq = 0.35
     
     Answer: 35% of the bears in this population are heterozygotes.
3. How many bears are pure-breeding for the brown-tipped phenotype?
   
   
   • Given: 45/75 or 0.6 of the bears have the recessive phenotype, and therefore the genotype of aa. So, you’re given: q2 = 0.6
   • Want: Number of bears pure-breeding for the brown-tipped phenotype or the AA genotype, so you want: p2
     
     
   • Solve:
              The square root of q2 gives you q.    q = 0.77
               p + q = 1, so 1 - q = p                   p = 0.23
               p2 = f(AA) = 0.05
     
     Answer: 0.05 or 5% of the 75 bears are pure-breeding for brown-tipped hair. Therefore 6 of the 75 bears are pure–breeding brown tipped hair.
4. If 10 years ago, only 24 of 77 bears had silver-tipped fur, has evolution occurred? Justify your answer.
   
   By definition, evolution has occurred if the allele frequencies have changed, so we have to find p or q for these two dates and compare them.
   
   Solve:
   • Bears with silver-tipped fur are those with aa genotype or q2
   • q2 10 years ago = 24/77 = 0.31      Now q2 = 45/75 = 0.6
   • q ten years ago = 0.096                Now q is 0.36
     
     Answer: Yes, evolution has occurred because the allele frequencies have changed.

Here is a question that goes the other direction, from genotype frequency to allele frequency. It uses a fictitious organism called dworps. Dworps can be tall (dominant) or short (recessive). If 50% of dworps are tall, what’s the frequency of the tall allele in the gene pool? (A = tall, a = short)

• Given: 0.5 are tall = f(AA) + f(Aa)
• So, 0.5 are short = f(aa), which is q2
• Want: p = f(A), and you have q2.
• Solve: square root of q2 = q = 0.71, So p = 1 - q= 0.29
  
  Answer: the frequency of the dominant allele (p) = 0.29