Problem set 4 key

Problem set 4 key - Bio 340: Problem Set 4 Due in class...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Bio 340: Problem Set 4 Name: Due in class June 13 th TA: 1) When pink sweet peas were self-pollinated and the seeds were collected and sown, the following flower colors were obtained: Red 36 Pink 56 White 28 Use a χ 2 test to determine whether these results are consistent with the hypothesis that pink flowers are heterozygous for a single pair of color alleles, showing incomplete dominance. (4 points) Incomplete dominance implies the following genotypes and phenotypes: cc = red, Cc = white, CC = pink. If this is true and if pink flowers are crossed (Cc × Cc), we expect a phenotypic ratio of 1:2:1 in the offspring. With 120 offspring, the expected numbers of each phenotype are 30 red, 60 pink, and 30 white. Class Observed Expected (O – E) 2 (O – E) 2 /E Red 36 30 36 1.20 Pink 56 60 16 0.27 White 28 30 4 0.13 χ 2 = sum( (O-E) 2 /E ) = 1.60 Degrees of freedom = 2 Critical value of χ 2 = 5.99 The statistic is less than the critical value, so the null hypothesis is not rejected. These results are consistent with incomplete dominance. 2) Horses can be cremello (a light cream color), chestnut (a brownish color), or palomino (a golden color with white in the horse’s tail and mane). Of these phenotypes, only palominos never breed true. Consider the results of the following crosses: Cremello × palomino ½ cremello, ½ palomino Chestnut × palomino ½ chestnut, ½ palomino Palomino × palomino ¼ chestnut, ½ palomino, ¼ cremello Determine the mode of inheritance by assigning gene symbols and indicating which genotypes yield which phenotypes. (3 points) These coat color phenotypes show incomplete dominance: CC = cremello Cc = palomino cc = chestnut a. Predict the F 1 and F 2 phenotypic proportions of matings between cremello and chestnut horses. (4 points) F 1 generation = 100% palomino F 2 generation = ¼ chestnut, ½ palomino, ¼ cremello. Palominos cannot breed true, because they are always heterozygous. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Bio 340: Problem Set 4 Name: Due in class June 13 th TA: 3) Like the ABO blood group, the MN blood group in humans shows codominance. Homozygotes for the L M allele present the M antigen on the surface of their red blood cells. Homozygotes for the L N allele present the N antigen, and heterozygotes present both antigens. The left side of the following table shows five human matings (1-5), identified by both maternal and paternal phenotypes for ABO and MN blood-group antigen status. Each mating produced one of the five offspring whose blood group phenotypes are shown on the right side (a-e). Match each offspring with one correct set of parents, using each parental set only once. Is there more than one set of correct answers? (5 points) Parental phenotypes Offspring phenotypes Answer 1 A, M × A, N a A, N 1c 2 B, M × B, M b O, N 2d 3 O, N × B, N c O, MN 3b 4 AB, M × O, N d B, M 4e 5 AB, MN × AB, MN e B, MN 5a This is the only answer that works for all five sets of parents. Parents 5 could have
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

Problem set 4 key - Bio 340: Problem Set 4 Due in class...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online