Evolution – Biology 4250 Hardy-Weinberg Problems
SHOW ALL YOUR WORK ! ! ! Round answers to the nearest two significant digits past the decimal point. Unless otherwise specified, assume populations are in a fictitious H-W equilibrium.
1. In a population with 2 alleles for a particular locus (D and d), the frequency of the D allele is 0.55.
a) What is the frequency of the d allele? 1 - .55 = .45
b) What is the frequency of homozygous dominant individuals in the population? .552 = .3025
c) What is the frequency of homozygous recessive individuals in the population? .452 = .2025
d) What is the frequency of heterozygotes in the population?
1 - (.3025 + .2025) = .495
2. The fraggles are a population of mythical, mouselike creatures that live in undergrown tunnels and chambers beneath a large vegetable garden that supplies their food. Of the fraggles currently in this population, 372 have green fur and 182* have gray fur. Green fur is controlled by a dominant allele F and gray fur by a recessive allele f.
a) What is the frequency of the gray allele f? √182/554 = .573
b) What is the frequency of the green allele F? 1 - .573 = .427
c) How many fraggles are heterozygous (Ff)? 2(.427)(.573) x 554 = 271 Ff
d) How many fraggles are homozygous recessive (ff)? 182* (listed in the problem!)
e) How many fraggles are homozygous dominant (FF)? 554 - (271 + 182), so 101 FF
3. In a population that is in Hardy-Weinberg equilibrium, 37% of the
individuals exhibit the recessive trait (ss).
q = freq. s = √.37 = .608
a) What is the frequency of the dominant allele (S) in the population? 1 - .608 = .392
b) What percent of the population possesses the dominant
allele (S)? 63% (everyone who
isn't ss possesses an S)
4. In a large, randomly mating population with no appreciable forces working to change gene frequencies, the frequency of homozygous recessive individuals for the characteristic of extra-long eyelashes is 60 per 1000. What percent of this population carries this very desirable trait but displays the dominant phenotype of short eyelashes?
q = √60/1000 = .245
Carriers (heterozygotes) = 2(.245)(.755) = .37
p = .755 37%
5. Coat color in sheep is determined by a single gene. Allele B, for white wool, is dominant over allele b, for black wool. We have followed a population of sheep for two years. Below are the statistics we have compiled.
Year 1 Year 2
White sheep 1567 1723
Black sheep 429 496
Total number of individuals 1996 2219
a) Determine the frequency of both alleles (B & b) in year 1.
429/1996 = .215 . . . q = freq b = √.215 = .464 p = freq B = .536
b) Determine the frequency of both alleles (B & b) in year 2.
496/2219 = .224 . . . q = freq b = √.224 = .473 p = freq B = .527
c) Is this population in Hardy-Weinberg equilibrium? Why or Why not?
would accept either:
"Yes, it is in virtual H-W equilibrium, because there was no appreciable change in frequencies from year 1 to year 2." or "No, because there was at least a little change from year 1 to year 2."
d) If the allelic frequencies for a particular gene in a population remain constant from year to year, what does this mean about the evolution of wool color in this population of sheep?
It means that there is no evolution taking place in fur color (at the moment).
6.In a particular species of flower, C1 codes for red flowers, C2 codes for white, with the heterozygous individuals being pink.
a. If the frequency of pink individuals in the population was .7126, would
you be able to estimate the frequencies of the individual alleles in the
population? Why or why not?
No, because the number of heterozygous individuals have two variables in the H-W equations (p & q).
b. If the frequency of red individuals in the population was .329, what would the estimated frequency of pink and white individuals be in this same population?
Freq C1 is
√.329 = .574
Freq C2 is = .426
So, freq of white = .4262 = .181 and freq of pink is .490 (= 2 x .426 x .574)
7. Suppose the number of red, pink, and white individuals in another population of flowers was 555, 555, and 555 respectively.
a. Could this population be said to be in H-W equilibrium? SHOW YOUR WORK!
555 red 1110 freq C1 = .5
555 pink 555 555 freq C2 = .5
555 white 1110
Total: 1665 1665 1665
Predicted C1C1 frequency = .52
frequency = .52 = .25
.25 x 1665 = 416 red
.5 x 1665 = 833 pink
.25 x 1665 = 416 white
to the observed, the expected are nowhere close to the observed, so this
population certainly could not
be considered to be in H-W equilibrium.
b. Which flowers could be said to be at an apparent selective advantage?
Both red and white, as the observed for both (555) is more than the expected (416)