Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 5, Problem 19PDQ
If the cross described in Problem 18 were made, and if Sb and cu are 8.2 map units apart on chromosome III, and if 1000 offspring were recovered, what would be the outcome of the cross, assuming that equal numbers of males and females were observed?
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Students have asked these similar questions
Could you have generated a chromosome map from the F2 progeny in cross 1 (assuming you crossed F1 males and females to each other to generate the F2)? What is the advantage of using cross 2 instead?
In a cross between a white-eyed female (ww) and a red-eyed male (w+Y), nearly all the progeny were either red-eyed females (w+w) or white-eyed males (wY). However, about 1 in every 2000 F1 flies had an "exceptional phenotype" and was either a white-eyed female or red-eyed male. How did Bridges explain this unexpected result?
A) Crossing over
B) Incomplete cytokinesis
C) Incorrect synapsis
D) Nondisjunction
E) Pseudoautosomal region
In corn, the genes v (virescent seedlings), pr (red aleurone), and bm (brown midrib) are all on chromosome 5, but not necessarily in the order given. The cross:
v+ pr bm/ v pr+ bm+ with v pr bm/ v pr bm produces 1000 progeny with the following phenotypes:
v+ pr bm 226
v pr+ bm+ 229
v+ pr bm+ 153
v pr+ bm 185
v+ pr+ bm 59
v pr bm+ 71
v+ pr+ bm+ 36
v pr bm 41
What is the gene order, and the (b) genetic map of these three genes?
Chapter 5 Solutions
Concepts of Genetics (12th Edition)
Ch. 5 - In a family with one autistic child the risk for...Ch. 5 - Given that the prenatal test can provide only a...Ch. 5 - Consider two hypothetical recessive autosomal...Ch. 5 - With two pairs of genes involved (P/p and Z/z), a...Ch. 5 - In Drosophila, a heterozygous female for the...Ch. 5 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 5 - Review the Chapter Concepts list on page 94. Most...Ch. 5 - Describe the cytological observation that suggests...Ch. 5 - Why does more crossing over occur between two...Ch. 5 - Explain why a 50 percent recovery of...
Ch. 5 - Why are double-crossover events expected less...Ch. 5 - What is the proposed basis for positive...Ch. 5 - What two essential criteria must be met in order...Ch. 5 - The genes dumpy (dp), clot (cl), and apterous (ap)...Ch. 5 - Colored aleurone in the kernels of com is due to...Ch. 5 - In the cross shown here, involving two linked...Ch. 5 - In a series of two-point mapping crosses involving...Ch. 5 - Two different female Drosophila were isolated,...Ch. 5 - In Drosophila, a cross was made between femalesall...Ch. 5 - Another cross in Drosophila involved the...Ch. 5 - In Drosophila, Dichaete (D) is a mutation on...Ch. 5 - Drosophila females homozygous for the third...Ch. 5 - In Drosophila, two mutations, Stubble (Sb) and...Ch. 5 - If the cross described in Problem 18 were made,...Ch. 5 - Are mitotic recombinations and sister chromatid...Ch. 5 - What possible conclusions can be drawn from the...Ch. 5 - An organism of the genotype AaBbCc was testcrossed...Ch. 5 - Based on our discussion of the potential...Ch. 5 - Traditional gene mapping has been applied...Ch. 5 - DNA markers have greatly enhanced the mapping of...Ch. 5 - In a certain plant, fruit is either red or yellow,...Ch. 5 - Two plants in a cross were each heterozygous for...Ch. 5 - A number of humanmouse somatic cell hybrid clones...Ch. 5 - A female of genotype produces 100 meiotic tetrads....Ch. 5 - In laboratory class, a genetics student was...Ch. 5 - Drosophila melanogaster has one pair of sex...Ch. 5 - In Drosophila, a female fly is heterozygous for...Ch. 5 - The gene controlling the Xg blood group alleles...Ch. 5 - Prob. 34ESP
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- To determine the recombination frequency between body color and wing genes in flies, you perform several crosses where you cross an F1 having red body and smooth wings with a yellow-bodied, crinkle-winged fly. You get the following results. What is the distance between the genes for body color and wing surface in map units?arrow_forwardAssume performing only one cross, which produced 274 flies, which consisted of 193 wild-type flies and 81 white-eyed flies. (a) Based upon this information, what is the most likely genotype of the female parent? (b) How to confirm the above answer with only the result from this one cross?arrow_forwardIn tomato the mutant genes o (oblate=flattened fruit), p (peach=hairy fruit) and s (compound inflorescence) were found to be in chromosome 2. The test cross results are: 73 110 + 348 2 2 + 306 p 96 63 a) Construct the linkage map b) What are the genotypes of the homozygous parents used in making the F1 heterozygote? c) Compute for the coefficient of coincidence. O o O O +arrow_forward
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- When true-breeding mice with brown fur and short tails (BBtt)were crossed to true-breeding mice with white fur and long tails(bbTT), all of the F1 offspring had brown fur and long tails. TheF1 offspring were crossed to mice with white fur and short tails.What are the possible phenotypes of the F2 offspring? Which F2offspring are recombinant, and which are nonrecombinant? Whatare the ratios of phenotypes of the F2 offspring if independentassortment is taking place? How are the ratios affected by linkage?arrow_forwardThe radish (Raphanus sativus) has a diploid number (2n) of 18. The cabbage (Brassica oleracia) also has a diploid number (2n) of 18. Why is it that if you cross these two species to produce a hybrid with a total of 18 chromosomes in its somatic cells it is sterile, but if you form an allotetraploid (also known as amphidiploid) from these two species it is fertile?arrow_forwardA PORTION OF THE LINKAGE MAP OF CHROMOSOME 2 IN THE TOMATO IS ILLUSTRATED HERE. ci (compound influorescence) o (oblate) - 15 CM 20 CM p (peach) THE OBLATE PHENOTYPE IS A FLATTENED FRUIT, THE PEACH PHENOTYPE IS HAIRY FRUIT (LIKE A PEACH), AND COMPOUND INFLORESCENCE MEANS CLUSTERED FLOWERS. IGNORE THE PEACH LOCUS. AMONG 1000 GAMETES PRODUCED BY A PLANT OF GENOTYPE O CI /+ +, WHAT TYPES OF GAMETES WOULD BE EXPECTED, AND WHAT NUMBER WOULD BE EXPECTED OF EACH?arrow_forward
- In tomato, the following genes are located on chromosome 2: + tall plant d dwarf plant + uniformly green leaves m mottled green leaves + smooth fruit p pubescent (hairy) fruit Results of the cross +++ / dmp and dmp / dmp were: + + + = 470 + m p = 1 + + p = 14 d + p = 25 d + + = 0 d m p = 441 + m + = 19 d m + = 30 Identify the single and double crossovers among the progeny.arrow_forwardThe following recombinant frequencies were found between genes A, B, C, D, and E. Determine the order of these genes on the chromosome and draw a linkage map of this chromosome with the map units labeled. А, С: 10% А, D: 30% А, Е: 6% В, С: 4% В, D: 16% В, Е: 20% C, D: 20%arrow_forwardin a cross AaBb X aabb, the following progeny were obtained: AaBb=50 aabb=46 Aabb=7 aaBb=6 What is the recombination frequency between the genes A and B? How far apart are they if they were on the same chromosomearrow_forward
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