Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 3, Problem 28P
Summary Introduction
To determine: The genotypes of the parents for each of the resulting progenies.
Introduction. The mutation is the change in the
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The pedigree below shows the phenotypes of the ABO blood groups and Rhesus factors [positive (+) and negative (-)] for several members of a family.
I
(B+
AB-
1
2
3
4
II
O-
A+
В-
B-
AB+
A+
1
2
4
5 6
a. What are the ABO blood group genotypes of individuals I-1 and I-2?
b. Which child/ren of individual I-4 can donate blood to him?
c. Which individual in the pedigree can donate blood to all the other individuals in the pedigree?
Among dogs, short hair is dominant to long hair and dark coat color is dominant to white (albino) coat color. Assume that these two coat traits are caused by independently segregating gene pairs. For each of the crosses given below, write the most probable genotype (or genotypes if more than one answer is possible) for the parents. It is important that you select a realistic symbol set and define each symbol below. Assume that for cross (d), you were interested in determining whether fur color follows a 3:1 ratio. Set up (but do not complete the calculations) a Chi-square test for these data [fur color in cross (d)].
In dogs, dark coat color is dominant over albino, andshort hair is dominant over long hair. Assume that theseeffects are caused by two independently assorting genes.Seven crosses were done as shown below, in which D andA stand for the dark and albino phenotypes, respectively,and S and L stand for the short-hair and long-hairphenotypes.Number of progenyParental phenotypes D, S D, L A, S A, La. D, S × D, S 88 31 29 12b. D, S × D, L 19 18 0 0c. D, S × A, S 21 0 20 0d. A, S × A, S 0 0 29 9e. D, L × D, L 0 31 0 11f. D, S × D, S 45 16 0 0g. D, S × D, L 31 30 10 10Write the genotypes of the parents in each cross. Use thesymbols C and c for the dark and albino coat-color allelesand the symbols H and h for the short-hair and long-hairalleles, respectively. Assume parents are homozygousunless there is evidence otherwise.
Chapter 3 Solutions
Introduction to Genetic Analysis
Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 10PCh. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13P
Ch. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 31.1PCh. 3 - Prob. 31.2PCh. 3 - Prob. 31.3PCh. 3 - Prob. 31.4PCh. 3 - Prob. 31.5PCh. 3 - Prob. 31.6PCh. 3 - Prob. 31.7PCh. 3 - Prob. 31.8PCh. 3 - Prob. 31.9PCh. 3 - Prob. 31.10PCh. 3 - Prob. 31.11PCh. 3 - Prob. 31.12PCh. 3 - Prob. 31.13PCh. 3 - Prob. 31.14PCh. 3 - Prob. 31.15PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 57P
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- For the following cross, show the P generation Genotypes and the Phenotypic ratio that would be seen in the F1 and F2. Remember, to produce the F2 generation you want to cross Heterozygotes from the F1. d) Genes 1 and 2 exhibit Epistasis (9:6:1) and Gene 3 is an Autosomal Dominant. In the P generation, the Male is Homozygous Recessive for the Genes showing Epistasis. Use E1, E2 and E3 to represent the Phenotypes shown by Epistasis. Report your results in the following format: P = aabb x AABB, F1 = 100%AaBb (Phenotype), and %3! F2 = 9/16 A_B_ (Phenotype), 3/16 aaB (Phenotype), 3/16 A_bb (Phenotype), 1/16 aabb (Phenotype)arrow_forwardThe pedigree concerns eye phenotypes in Tribolium beetles. The solid symbols represent black eyes, the open symbols represent brown eyes, and the cross symbols (X) represent the "eyeless" phenotype, in which eyes are totally absent. Assume that these three phenotypes are the result of three alleles of a single gene, E, where E' is black, g? is brown, and is eyeless. Assign a genotype to each individual in generation II. 1 2 3 II 4 7 8 II 9 10 | 11 12 IV 13 Grifths, et al, Introduction to Genetic Analysis 12e © 2020 Macmillan Leaning Answer Bank E'? E'E What is the mode of inheritance of the eyeless phenotype? autosomal dominant X-linked recessive X-linked dominant autosomal recessive COarrow_forwardIn goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We’ll use the symbol Bb for the beard allele and B+ for the beardless allele. Another independently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Give the phenotypes and their expected proportions for the following crosses. a. B+ Bb Ww male × B+ Bb Ww female b. B+ Bb Ww male × B+ Bb ww female c. B+ B+ Ww male × Bb Bb Ww female d. B+ Bb Ww male × Bb Bb ww femalearrow_forward
- Yellow guinea pigs crossed with white ones always produce cream-colored offspring. Two cream guinea pigs, when crossed, produce yellow, cream, and white offspring in the ratio of 1 yellow : 2 cream : 1 white. What principle of genetics is involved in this cross? (1 point) 2. The shape of radishes may be long, round, or oval. The following results were obtained in the different possible crosses: a. long x oval gave ½ long and ½ oval b. oval x round gave ½ oval and ½ round c. long x round gave all oval d. oval x oval gave ¼ long, ½ oval, and ¼ round Explain these results. Hint: Show genotypes of each cross) (2 points). a. b. c. d. 3. In human blood types, what are the genotypes of the following parents? (2 points). Phenotypes of ParentsPhenotypes of OffspringGenotypes of parents ABABO A x AB½ 0½ 0_______ x _______ A x AB½ ¼¼0_______ x _______ A x A¾00¼_______ x _______ A x O½00½_______…arrow_forwardIn the pearl-millet plant, color is determined by three alleles at a single locus: Rp1 (red), Rp2 (purple), and rp (green). Red is dominant over purple and green, and purple is dominant over green (Rp1 > Rp2 > rp). Give the expected phenotypes and ratios of offspring produced by the following crosses. a. Rp1/ Rp2 × Rp1/ rp b. Rp1/ rp × Rp2/ rp c. Rp1/ Rp2 × Rp1/ Rp2 d. Rp2/ rp × rp/ rp e. rp/ rp × Rp1/ Rp2arrow_forwardIn roses, purple flower color is determined by the dominant P allele, while pp homozygotes are white. The presence of long stems is determined by the dominant S allele, while ss homozygotes have short stems. Both mutations are completely penetrant. A test cross was performed between a rose plant of unknown genotype with a white flowered, short stemmed rose plant (pp ss) and the following 200 progeny plants were obtained: 84 white flowers, long stems 16 purple flowers, long stems 82 purple flowers, short stems 18 white flowers, short stems Select the statements below that are TRUE. Select 2 correct answer(s) The P and S genes independently assort during meiosis. The map distance between P and S is 17 CM. The genotype of the progeny plants with purple flowers and short stems is PP ss. The map distance between P and S is 83 CM. The homologs in the plant with unknown genotype are p S and Ps. The homologs in the plant with unknown genotype are PS and p s.arrow_forward
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