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Population Genetics and Evolution Practice Test (page 3)

By — McGraw-Hill Professional
Updated on Aug 23, 2011

Testing a Locus for Equilibrium Questions

  1. Apair of codominant alleles governs coat colors in Shorthorn cattle: CRCR is red, CRCW is roan, and CWCW is white. A sample of a cattle population revealed the following phenotypes: 180 red, 240 roan, and 80 white.   (a) What is the frequency of the CR allele?   (b) What is the frequency of the CW allele?   (c) Does the sample indicate that the population is in equilibrium?   (d) What is the chi-square value?   (e) How many degrees of freedom exist? ( f) What is the probability that the deviation of the observed from the expected values is due to chance?
  2. A blood group system in sheep, known as the XZ system, is governed by a pair of codominant alleles (X and XZ). A large flock of Ram bouillet sheep was blood-grouped and found to contain 113 X/X, 68 X/XZ, and 14 XZ/XZ.   (a) What are the allelic frequencies?   (b) Is this population conforming to the equilibrium expectations?   (c) What is the chi-square value?   (d) How many degrees of freedom exist?   (e) What is the probability of the observed deviation being due to chance?
  3. The frequency of the T allele in a human population = 0.8, and a sample of 200 yields 90% tasters of a chemical called PTC (Problem 9.22) (T-) and 10% nontasters (tt).   (a) Does the sample conform to the equilibrium expectations?   (b) What is the chi-square value?   (c) How many degrees of freedom exist?   (d) What is the probability that the observed deviation is due to chance?
  4. In poultry, the autosomal gene FB produces black feather color and its codominant allele FW produces splashed-white. The heterozygous condition produces Blue Andalusian. A splashed-white hen is mated to a black rooster and the F2 was found to contain 95 black, 220 blue, and 85 splashed-white.   (a) What F2 ratio is expected?   (b) What is the chisquare value?   (c) Howmany degrees of freedom exist?   (d) What is the probability that the observed deviation is due to chance?   (e) May the observations be considered to conform to the equilibrium expectations?

Answers

Vocabulary

  1. random mating or panmixis
  2. gene pool
  3. Hardy-Weinberg rule
  4. equilibrium
  5. Mendelian population
  6. genetic drift
  7. disequilibrium
  8. race or subspecies
  9. reproductive isolation
  10. evolution

Multiple-Choice

  1. e(035)
  2. b(455%)
  3. a
  4. d
  5. a
  6. b
  7. e(0286)
  8. e(029)
  9. a
  10. a

Hardy-Weinberg Equilibrium

  1. 0.5
  2. All individuals are AaBb.
  3. (a) pm = pf = 0:6, qm = qf = 0:4   (b) AA = 0:36, Aa = 0:48, aa = 0:16
  4. (a) 1/7 = 0:143   (b) 0.1008
  5. (a) 0.25   (b) 1/2 AABB : 1/2 aabb or 1/2 aaBB : 1/2 AAbb
  6. AB = 0:12, Ab = 0:08, aB = 0:48, ab = 0:32

Calculating Gene Frequencies

Autosomal Loci with Two Alleles

Codominant Autosomal Alleles

  1. CD = 0:9, CG = 0:1
  2. (a) LM = 75:5%, LN = 24:5%   (b) 210

Dominant and Recessive Autosomal Alleles

  1. t = 0:56
  2. (a) C = 0:30   (b) a = 0:58
  3. Y = 0:7, y = 0:3
  4. R = 0:613, r = 0:387
  5. 0.5; all individuals are heterozygous carriers of the lethal allele.

Sex-Influenced Traits

  1. 9775
  2. (a) 50%   (b) 75%

Autosomal Loci with Multiple Alleles

  1. (a) IA = 0:05, IB = 0:25, i = 0:70   (b) A = 44:6%, B = 21:6%, AB = 14:4%, O = 19:4%
  2. G r = 0:1, gi = 0:6, g = 0:
  3. 64% wild type, 20% dark bay, 7% seal brown, 9% black

Sex-Linked Loci

  1. (a) 1/100   (b) 1/10,000
  2. (a) 0.96   (b) 99.84%
  3. w = 0:19

Testing a Locus for Equilibrium

  1. (a) 0.6   (b) 0.4   (c) Yes   (d) 0   (e) 1   (f) 1
  2. (a) X = 0:75, XZ = 0:25   (b) Yes   (c) 0.7   (d) 1   (e) 0.3–0.5
  3. (a) No   (b) 18.75   (c) 1   (d) < 0:001
  4. (a) 1/4 black : 1/2 blue : 1/4 splashed-white   (b) 4.50   (c) 1   (d) 0.01–0.05   (e) No
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