Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 4, Problem 11P
In the rats identified in Problem
a.
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c.
d.
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Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL.
A. What would be the physical phenotype of these flies?
B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products?
C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker?
D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?
Pigment in mouse fur is only produced when the C allele is present. Individuals of the cc genotype are white. If color is present, it may be determined by the A, a alleles. AA or Aa results in agouti color, while aa results in black coats. (a) What F1 and F2 genotypic and phenotypic ratios are obtained from a cross between AACC and aacc mice? (b) In three crosses between agouti females whose genotypes were unknown and males of the aacc genotype, the following phenotypic ratios were obtained: (1) 8 agouti (2) 9 agouti (3) 4 agouti 8 white 10 black 5 black 10 white What are the genotypes of these female parents?
Three autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progeny
Chapter 4 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 4 - 1. Define and distinguish incomplete penetrance...Ch. 4 -
2. Define and distinguish epistasis and...Ch. 4 - When working on barley plants, two researchers...Ch. 4 - Fifteen bacterial colonies growing on a complete...Ch. 4 - 5. In a type of parakeet known as a “budgie,”...Ch. 4 - 6. The and blood groups are given below for four...Ch. 4 - The wild-type color of horned beetles is black,...Ch. 4 - 8. Two genes interact to produce various...Ch. 4 - Prob. 9PCh. 4 - 10. In rats, gene produces black coat color if the...
Ch. 4 - 11. In the rats identified in Problem, a third...Ch. 4 - Using the information provided in Problems 10 and...Ch. 4 - 13. Total cholesterol in blood is reported as the...Ch. 4 - 14. Flower color in snapdragons results from the...Ch. 4 - 5. A plant line with reduced fertility comes to...Ch. 4 - Prob. 16PCh. 4 - The coat color in mink is controlled by two...Ch. 4 - Prob. 18PCh. 4 - 19. Feather color in parakeets is produced by the...Ch. 4 - Brachydactyly type D is a human autosomal dominant...Ch. 4 - 21. A male and a female mouse are each from...Ch. 4 - Xerodermapigmentosum (XP) is an autosomal...Ch. 4 - 23. Three strains of green-seeded lentil plants...Ch. 4 - Blue flower color is produced in a species of...Ch. 4 - 25. The following crosses are performed between...Ch. 4 - Two pure-breeding strains of summer squash...Ch. 4 - Marfan syndrome is an autosomal dominant disorder...Ch. 4 - 28. Yeast are single-celled eukaryotic organisms...Ch. 4 - Prob. 29PCh. 4 - Dr. Ara B. Dopsis and Dr. C. Ellie Gans are...Ch. 4 - Human ABO blood type is determined by three...Ch. 4 - In rabbits, albinism is an autosomal recessive...Ch. 4 - Dr. O. Sophila, a close friend of Dr. Ara B....Ch. 4 - In a breed of domestic cattle, horns can appear on...Ch. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - 37. Epistatic gene interaction results in a...Ch. 4 - 38. Draw a pedigree containing two parents and...
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- In the fungus Neurospora, a strain that is auxotrophic for thiamine (mutant allele t) was crossed with a strain that isauxotrophic for methionine (mutant allele m). Linear asci were isolated and classified into the following groups: a. Determine the linkage relations of these two genes to their centromere(s) and to each other. Specify distances in map units. b. Draw a diagram to show the origin of the ascus type with only one single representative (second from right).arrow_forwardConsider three genes L, U, and W, for which the count of F2 phenotypes after a 3-point cross is as follows: Phenotype F2 count: L U w 19 L u W 1 l u W 21 L U W 33 l U W 274 l u w 41 l U w 2 L u w 259 Which of the following statements about genes L, U, and W are TRUE? (may be more than one correct ans) A. L, U, and W are each on a different chromosome B. Only U and L are on the same chromosome C. Only U and W are on the same chromosome D. Only W and L are on the same chromosome E. L, U, and W are all on the same chromosomearrow_forwardIn com, male sterility is controlled by maternal cytoplasmic elements. However, the presence of a nuclear fertility restorer gene (F_) restores fertility to male sterile lines. a. What are the crosses male sterile female x FF male? Give the genotypes and phenotypes of the offspring in each cross. Explain.arrow_forward
- In roses, purple flower color is determined by the dominant P allele, while pphomozygotes 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 two statements below that are TRUE. options: 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…arrow_forwardIn a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.arrow_forwardMouse fur color is affected by TWO unlinked loci. GG or Gg mice are agouti. Mice with genotype gg are albino since all pigment production is blocked, regardless of the phenotype at the second locus. At the other locus, the F allele (agouti coat) is dominant to the f allele (black coat). How many black mice would result from a cross between two agouti mice of genotype GgFf? (just write in the numeral)arrow_forward
- The wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated that have either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes. Carefully examine the data from the following five crosses shown below (running across both columns). (a) Identify each mutation as either dominant or recessive. In each case, indicate which crosses support your answer. (b) Assign gene symbols and, for each cross, determine the genotypes of the parents.arrow_forwardDeduce the phenotypic proportions in the progeny of thefollowing crosses of autotetraploids in which the a+/a locus is very close to the centromere. (Assume that the fourhomologous chromosomes of any one type pair randomly two by two and that only one copy of the a+ allele isnecessary for the wild-type phenotype.)a. a+/a+/a/a × a/a/a /ab. a+/a/a /a × a/a/a/ac. a+/a/a /a × a+/a/a /ad. a+/a+/a /a × a+/a/a /aarrow_forwardThe data set attached summarizes F2 numbers from an F1 cross arising from two, true-breeding Drosophila strains (P generation), which differ with respect to two mutant traits. Here are the hypothesis: Leg length - The wild-type and mutant alleles for leg length are incomplete dominant relative to each other. Justification: The data set includes three phenotypic categories for leg length: wild type (long leg), medium leg, and truncated wings. The presence of three distinct phenotypes suggests an incomplete dominance pattern, where the heterozygous individuals exhibit an intermediate leg length phenotype (medium leg). The absence of purebred short-legged individuals supports the idea that the long leg allele is dominant over the short leg allele. This shows that mode of inheritance is incomplete dominance of the alleles relative to each other. Since the data does not mention any specific differences between males and females, we can assume that the mode of inheritance for the trait is…arrow_forward
- The wild-type (W) Abraxas moth has large spots on its wings, but the lacticolor (L) form of this species has very small spots. Crosses were made between strains differing in this character, with the following results:Provide a clear genetic explanation of the results in these two crosses, showing the genotypes of all individual moths.arrow_forwardTwo Drosophila flies that had normal (transparent, long) wings were mated. In the progeny, two new phenotypes appeared, dusky wings (having a semi-opaque appearance) and clipped wings (with squared ends). The progeny were as follows: Females: 179 transparent, long 58 transparent, clipped Males: 92 transparent, long 89 dusky, long 28 transparent, clipped 31 dusky, clipped a) Provide a genetic explanation for these results, showing genotypes of parents and of all progeny classes under your model. b) Design a test for your model.arrow_forwardIn Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F1, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F2 generation, and 1000 offspring were counted, with the results shown in the following table. Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. (a) Using proper nomenclature, determine the genotypes of the P1 and F1 parents. (b) Determine the sequence of the three genes and the map distances between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence. Does it represent positive or negative interference?arrow_forward
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