MG. Ch. 26

Question 1

If the frequency of the M allele in the human MN blood group system is 0.65 in a population at equilibrium, What is the frequency of the N allele?

Answer 1

.35

M+N = 1 (So 1 - .65 = .35)

Question 2

If a recessive disease is found in 50 out of 100,000 individuals, what is the frequency of the heterozygote carriers for this disease?

Answer 2

.043

1) calculate q squared = 50/100000 = .0005
2. ) Calculate p (.978)
3) plug into heterozygote equation

Question 3

In a population of birds in Africa, it was observed that birds with small or large beaks could efficiently crack and eat small or large seeds, respectively. Birds with intermediate beaks had trouble with both types of seeds. What type of selection would be expected to occur in this population if small and large seeds were the only types of food available to these birds?

Answer 3

Disruptive
Birds with an intermediate beak phenotype are at a disadvantage in this population and will be selected against because they are less fit.

Question 4

inbreeding often occurs in small populations, thereby increasing the chance for

Answer 4

homozygosity

Question 5

With increased homozygoity comes a ______ chance that recessive alleles will be expressed

Answer 5

Increased

Question 6

A mainland and an island are both inhabited by brown and white mice. Allele A1 codes for brown coat color; allele A2 codes for white coat color. A1 is dominant to A2 .

Mice do not swim between the mainland and the island, and there are no shipping routes to the island, thus mice cannot hitch a ride on ships. During storms, however, some mice get “rafted” to the island on logs.

The island contains 240 mice. The frequency of the A1 allele is 0.85; the frequency of the A2 allele is 0.15.

On the mainland, the frequency of the A1 allele is 0.55; the frequency of the A2 allele is 0.45.

During a severe storm, 60 mice from the mainland are rafted to the island. What are the new frequencies of A1 and A2 on the island after the storm?

Answer 6

the frequency of A1 = .790
the frequency of A2 = .210

1.) determine the portion of population that is made of migrants (m) vs. island (1-m).
Original island = 240 + 60 migrants = 300 total population and m = 60/300 = .2 & 1-m = 240/300 = .8
2.) now calculate new population frequency
(Pn = (1-m)(A1 frequency island) + (m)(A1 frequency main))
(Pn = (.8)(.85) + (.2)(.55) = .79)
3.) 1 - .79 = q(A2) = .21
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Alternative:
1.) determine the number of current allele copies
(.85)x(240) = 204 current allele copies for A1
(.15)x(24) = 36 current allele copies for A2
2.) determine how many alleles were migrant
(.55)x(60) = 33 A1 alleles
(.45)x(60) = 27 A2 alleles

3.) divide total number of alleles by population
(204+33/300 = .79)

Question 7

Consider a population of frogs. Allele B1 codes for spotted frogs; allele B2 codes for non-spotted frogs. B1 is dominant to B2 .
The allele frequencies on the mainland are B1 = 0.75 and B2 = 0.25.
The allele frequencies on the island are B1 = 0.55 and B2 = 0.45. There are 170 frogs on the island.

During a storm, 30 non-spotted frogs come to the island by floating on debris. What are the new frequencies of B1 and B2 on the island?

Answer 7

B1 = .470
B2 = .530

1) determine genotype before the storm
2.) add population after storm
3.) Determine freq. of alleles after the storm:
[(2 x 1st allele) + hetero allele]/ Total alleles of population