BIO 340                                              Problem Set #5                        ANSWER KEY

 

1.   a.  The genetic variance, Vg, is the variance of the 7 line means.  The overall mean is 481.  The variance of the line means is 9179, which is calculated as follows:

      [(642-481)2 + 547-481)2 + É + (351-481)2] / (7 Ð 1)

      b.  The environmental variance, Ve, is the within-line variance.  There are seven within-line variances, each of which is an estimate of the environmental variance.  Given that we have 7 estimates of the same thing, the most logical thing to do is take their average.  The average within-line variance is 6429.

      c.  The heritability (broad sense) of seed weight is Vg / Vp = Vg / (Vg + Ve)

                                                = 9179 / (9179 + 6429) = 0.588

 

2.   The total sample size is 3 + 44 + 55 = 102.

      a.  Frequency of M allele = #M alleles in sample / total # of alleles in sample.

                                                = [(3 x 2) + 44]/ (102 x 2) = 50/204 = 0.2451

 

            The frequency of the N allele = [44 + (55 x 2)] / 204 = 0.7549

     

      b.  The expected number of MM individuals is (0.2451)2 x 102 = 6.13

            The expected number of MN individuals is 2(0.2451)(0.7549) x 102 = 37.75

            The expected number of NN individuals is (0.7549)2 x 102 = 58.13

 

      c.  Genotype          Observed         Expected          (O-E)2/E

            MM                 3                      6.13                 1.598

            MN                 44                    37.75               1.035

            NN                  55                    58.13               0.169

                                                                        Total    2.802 = chi square

            The degrees of freedom , df = 1.

            Using the table on p. 49 of your text, we see that the probability of this chi square value is between 0.05 and 0.20, if our null hypothesis that this population is in Hardy-Weinberg equilibrium at this locus is true.  Since the probability is greater than 5%, we will NOT reject the null hypothesis.

 

3.   a.  1 in 15, 000 = 1/15,000 = 0.00006667 is the frequency of homozygous recessive individuals.  Let this equal q2, where q is the frequency of the allele that causes PKU.  Taking the square root of 0.00006667 gives us q = 0.008.

      b.  If q = 0.008, then the frequency, p, of the normal allele is 1 Ð 0.008 = 0.992.  The frequency of carriers is 2pq = 2(0.992)(0.008) = 0.016, or about 1.6% of the population.

 

4.   Given that you are type O and your partner is heterozygous type A, the chance that any one child will be type O is 1/2 (i.e., chance is 1/2 that a child will receive the recessive i allele from your partner).  Therefore, the chance that all three children will have type O blood (and that the A allele will be lost) is (1/2)3 = 1/8.

 

5.   For one 10-allele locus, the number of different diploid genotypes is (10 x 11)/2 = 55.  Therefore the total number of possible 13-locus genotypes is (55)13 = 4.21 x 1022 (approximately).  In words, this is about 42.1 trillion billion.  (For comparison, I read recently that there are believed to be about a billion galaxies in the observable universe, each with about a billion stars.  That is ÒonlyÓ about 1018 stars in the entire universe.)

 

6.   a.  The probability that an individual would have the genotype a2 a4 for gene A is 2(0.06)(0.15) = 0.018.  The probability of being b1 b1 at gene B is (0.22)2. = 0.0484.  For all five genes, the Hardy-Weinberg genotype frequencies are:

      Gene          Genotype         Expected frequency

      A               a2 a4                2(0.06)(0.15) = 0.018

      B               b1 b1               (0.22)2. = 0.0484

      C               c2 c6                2(0.03)(0.11) = 0.0066

      D               d3 d8               2(0.08)(0.17) = 0.0272

      E                e1 e5                2(0.35)(0.02) = 0.014

 

      Assuming that these 5 loci are independent of one another, the probability of this particular 5-locus genotype in the population is the product of the probabilities for each locus:

      = (0.018)(0.0484)(0.0066)(0.0272)(0.014) = 2 x 10-9, or about 2 chances in a billion.

      b.  I think I'd feel pretty safe in my conclusion that this suspect left a DNA sample at the crime scene.  The chance of some randomly chosen (innocent) person having a matching genotype is only 2 in a billion.  Note: in practice some "correction" would be made to this calculation to account for the fact that allele frequencies may differ among various human racial and ethnic groups.  For example, the probability of a random match to this particular 5-locus genotype might be higher in certain ethnic groups.