INTRO TO GEN ANALYSIS W/ACHIEVE ACCESS
12th Edition
ISBN: 9781319423865
Author: Griffiths
Publisher: MAC HIGHER
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Chapter 2, Problem 52P
Summary Introduction
To determine: The implication of the results and the proposed genotypes of all plants.
Introduction: The human blood group is an example of codominance where the alleles for the two antigens A and B are equally dominant. This means that when the two antigens A and B are present the blood group is AB while in the absence of both the antigens the blood group is O.
Summary Introduction
To determine: The possibility of predicting F1 from the original mutant A with original mutant B.
Introduction: When the two or more alleles equally expressed themselves through the
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Two plants with white flowers, each from true-breeding strains, were crossed. All the F1 plants had red flowers. When these F1 plants were intercrossed, they produced an F2 consisting of 177 plants with red flowers and 142 with white flowers.
(a) Propose an explanation for the inheritance of flower color in this plant species.
(b) Propose a biochemical pathway for flower pigmentation and indicate which genes control which steps in this pathway.
The wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long
bristles. Mutant strains have been isolated with either curled wings or short bristles.
The genes representing these two mutant traits are located on separate
chromosomes. Carefully examine the data from the five crosses below. (a) For each
mutation, determine whether it is dominant or recessive. In each case, identify
which crosses support your answer; and (b) define gene symbols and determine the
genotypes of the parents for each cross.
Cross
1. straight, short X
straight, short
2. straight, long X
straight, long
3. curled, long X
straight, short
4. straight, short X
straight, short
5. curled, short X
straight, short
straight
wings,
long
bristles
30
120
40
40
20
Number of Progeny
straight
curled
wings, wings,
short
long
bristles
bristles
90
10
0
40
120
60
40
40
0
20
curled
wings,
short
bristles
30
0
40
0
60
The wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated with either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes. Carefully examine the data from the five crosses below. (a) For each mutation, determine whether it is dominant or recessive. In each case, identify which crosses support your answer; and (b) define gene symbols and determine the genotypes of the parents for each cross.
Chapter 2 Solutions
INTRO TO GEN ANALYSIS W/ACHIEVE ACCESS
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10P
Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 56.1PCh. 2 - Prob. 56.2PCh. 2 - Prob. 56.3PCh. 2 - Prob. 56.4PCh. 2 - Prob. 56.5PCh. 2 - Prob. 56.6PCh. 2 - Prob. 56.7PCh. 2 - Prob. 56.8PCh. 2 - Prob. 56.9PCh. 2 - Prob. 56.10PCh. 2 - Prob. 56.11PCh. 2 - Prob. 56.12PCh. 2 - Prob. 56.13PCh. 2 - Prob. 56.14PCh. 2 - Prob. 56.15PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 1GSCh. 2 - Prob. 2GSCh. 2 - Prob. 3GS
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- A researcher crosses mice with brown eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 6 apricot, short : 30 brown, long : 15 brown, short : 9 apricot, long You know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these results. a) Making use of the appropriate genetic convention for naming alleles, give the genotype of the male parent in this cross. b) What is your null hypothesis for the chi-square test? c) Give the expected number of individuals in the "brown, long" class. d) You obtain a value of 3.47 for the chi-square test. What conclusion can you make from the results of the chi-square test? P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357…arrow_forwardConsider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion. What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+? Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.arrow_forwardMale Drosophila from a true-breeding wild-type stock were irradiated with X-rays and then mated with females from a true-breeding stock carrying the following recessive mutations on the X chromosome: yellow body (y), crossveinless wings (cv), cut wings (ct), singed bristles (sn), and miniature wings (m). These markers are known to map in the order: Recessive alleles: y, cv, ct, sn, m Dominant alleles: y+, cv+, ct+, sn+, m+ y-cv-ct-sn-m у CV ct sn m X-rays х х X ct sn CV у m y+ CV+ ct+ sn+ m+ х X ? Exceptional female: Most of the female progeny of this cross were phenotypically wild type, but one female exhibited ct and sn mutant characteristics. When this exceptional ct sn female was mated with a male from the true-breeding wild-type stock, twice as many females as males appeared among the progeny. a. What is the nature of the X-ray-induced mutation present in the exceptional female? b. Draw the X chromosomes present in the exceptional ct sn female as they would appear during pairing…arrow_forward
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