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If the female progeny from Cross #1 have tailless offspring, the Alternatively, if the unknown mutation fails to complement fs, the mutation must be in fs. Offspring) - i.e., the unknown mutation fails to complement torso and therefore the unknown mutation is in torso. If the unknown mutation (called mut in the diagram below) is in torso, the progeny of the cross will also have the same phenotype (tailless Is crossed with the known torso mutant strain or the fs strain. Strand of the template DNA (so the promoter has to be at the oppositeĭo a complementation test. To the target mRNA, so it has to be transcribed off the other The RNA transcribed from the construct has to be complementary Incapable of expressing LDL receptors anyway, depleting the bodyĬonstruct 2. The premise of the resin treatment is that depletion of bile willĬause liver cells to express more LDL receptors so as to increase That the second person is in fact a heterozygote = 0.05 x 0.3 The 30% false negatives (probability = 0.3). That the person is heterozygous (probability = 0.05) but is among That could either mean that the person is homozygous normal, or If one member is tested and not found to have a disease allele, Probability that both members will not be correctly identified = 1 - 0.49 = 0.51 (or 51%).ĥ% (= 0.05, the frequency of heterozygotes in the population). Therefore, the probability that both members in a heterozygote/heterozygoteĬouple will be correctly identified = 0.7 x 0.7 = 0.49. Probability of correct identification of each heterozygote = 0.7. All we need to measure is the number of homozygous recessiveĪnd that lets us calculate the predicted number of the other classes V(1 - p) = up v - vp = up up + vp = v p = v/(u + v) So the change in p = 0 (as is the change in q)- vq - up = 0 vq = up Since q = 1 - p, we can substitute and solve for p. Gain from forward mutation and loss from back mutation: Change in p = vq - up Change in q = up - vqĪt equilibrium, change in p is exactly matched by change in q, Then the change in p would include loss from forward mutationĪnd gain from back mutation likewise, change in q would include Let the frequency of D = p, and the frequency of d = q, forward mutation rate = u, and back mutation rate = v The d allele will be more frequent, as the forward mutation ( D to d) occurs at a higher rate than the back mutation. Therefore, in the next generation the frequency of So the recessive allele makes up 1/3 of the total alleles in the Now the heterozygotes make up 2/3 of the surviving population, Is: Homozygous dominant = p 2 = 0.25 Heterozygotes = 2pq = 0.5 Homozygous recessive = q 2 = 0.25Īfter the viral epidemic, the only cats left are homozygous dominantĪnd heterozygotes: Homozygous dominant = p 2 = 0.25 Heterozygotes = 2pq = 0.5 Homozygous recessive = q 2 = 0.25
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The frequency of the black allele in the next generation willīefore the virus comes through, the frequency of the three genotypes Then only the recessive (black) allele will be left in the population If only black cats are left standing after the virus goes through, However, doing so would ignore the contribution of recessive allelesįrom the heterozygotes in each population. Number of recessive individuals from the two populations and toĭerive q from that - i.e., take the square root of (4 + 54). = 0.22 Frequency of black cats in the next generation = q 2 = (0.22) 2 = 0.0484.Ī potential source of error in this problem is to simply add the In the merged population - Frequency of recessive allele q = ((400 x 0.1) + (600 x 0.3))/1000 In the larger population - Frequency of the recessive phenotype = (q 2) 2 = 54/600 Frequency of the recessive allele = q 2 = 1/10 = 0.3. In the smaller population - Frequency of the recessive phenotype = (q 1) 2 = 4/400 Frequency of the recessive allele = q 1 = 1/10 = 0.1 Answer key to practice problems - Genetics 371B Autumn 1999