Ecology – 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.
30 points total.
1. In a population with 2 alleles for a particular locus (D and d), the
frequency of the D allele is 0.64.
(4 pts.)
a) What is the frequency of the d allele?
1 - .64 = .36
b) What is the frequency of homozygous dominant individuals in the population?
.642 = .41
c) What is the frequency of homozygous recessive individuals in the population?
.362 = .13
d) What is the frequency of heterozygotes in the population?
1 - .13 - .41 = .46
2. Green elephants result from the cross between blue elephants and yellow
elephants. In a population
of elephants, 187 are blue, 927 are green and 533 are yellow. Could it be said
that the population is in
genetic (HW) equilibrium? If not, who seems to have a selective advantage? Can
you give a reasonable
explanation as to why that (those) particular color(s) would be selected for? (9 pts.)
C1 - blue C2 - yellow
Observed:
C1
C2
Blue 187 2 x 187 = 374
Freq C1 = 1301/3294 = .395
Green 927 927 927
Freq C2 = 1993/3294 = .605
Yellow 533 ____
2 x 533 = 1066
1301
1993
Expected:
1647 Total
Total alleles
Blue = (.395)2 x 1647 = 257
3294
Green = 1647 - 257 - 603 = 787
Yellow = (.605)2 x 1647 = 603
Not in H-W equilibrium; observed does not equal
expected.
Green appears to be at a selective advantage (more
green observed than expected).
Mate choice -- yellow prefers blue and vice versa, and
green prefers green (also accepted camouflage).
3. In a population that is in Hardy-Weinberg equilibrium, 20% of the individuals
exhibit the recessive trait (ss).
(2 pts.)
√.20 = .447
a) What is the frequency of the dominant allele (S) in the population?
freq S = 1 - .447 = .553
b) What percent of the population possesses the dominant allele (S)?
Remember, BOTH SS and Ss posses the dominant allele, so the answer is everyone
not
homozygous recessive, which was given as 20%. So
the answer is 80%
4. The frequency of children homozygous for a recessive lethal allele is about
1/25,000. What proportion
of the population (assuming equilibrium) are carriers of the lethal allele? (2
pts.)
q = √1/25000 =
√.00004
= .0063
Carriers (heterozygotes) = 2(.0063)(.9937) = .0125,
p = .9937 or 1.25%, or 125 out of 10000 (313 out of 25000)
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. (8 pts.)
Year 1
Year 2
White sheep
564
497
Black sheep
157
193
Total # of
721
690
a) Determine the frequency of both alleles (B & b) in year 1.
157/721 = .217
= q2 . . .
q = freq b = √.217
= .47 p = freq B = .53
b) Determine the frequency of both alleles (B & b) in year 2.
193/690 = .28
= q2 . . . q = freq b = √.28
= .53 p = freq B = .47
c) Is this population in Hardy-Weinberg equilibrium? Explain why or why not.
It
is not in H-W equilibrium, because there was a
change in frequencies from year 1 to year 2.
In this case, the change is 6%,
which is HUGE in one year.
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 that trait (at the moment)
6. The formation of methylmercaptan from ingested asparagus is a recessive
trait. To find out if
you are a methylmercaptan forming individual, simply eat a serving of asparagus
and wait
approximately 20 minutes before urinating. You will be able to detect a very
distinctive odor if
you have inherited the ability to form methylmercaptan. What percent of the
population would be
homozygous recessive for this trait if it is known that 32% are homozygous
dominant? Understand
that this trait has no known selective value. (2 pts.)
p2
= .32 p = freq M = √32.
= .57
q = freq m = .43
Homozygous
recessive = .432 = .18, so 18%
7. 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 .5163, 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 .629, what would
the estimated frequency of
pink and white individuals be in this same population?
Freq C' is
√.629 = .79
Freq C" is = .21
So, freq of white = .212 =
.044 and freq of pink is .327 (1 - [.629 +.044] or 2 x .21 x .79)
8. Suppose the number of red, pink and white individuals in another population
of flowers was 433, 396,
and 671 respectively. C1 - red
C2 - white
a. Could this population be said to be in H-W equilibrium? SHOW YOUR WORK!
Observed:
C1
C2
Red 433 2 x 433 =
866
Freq C1 = 1262/3000 = .42
Pink 396 396
396
Freq C2 = 1738/3000 = .58
White 671 ____
2 x 671 = 1342
1262
1738
Expected:
1500 Total
Total alleles
Red = (.42)2 x 1500 = 264
3000
Pink = 1500 - 264 - 505 = 731
White = (.58)2 x 1500 = 505
Not in H-W equilibrium; observed does not equal
expected.
b. Which flowers could be said to be at an apparent selective advantage?
Red and white both appear to be at a selective advantage (more red/white observed than expected).