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 4, Problem 24P
Summary Introduction
To determine: The gene sequence on the chromosome of three loci, map distance between the genes and coefficient of coincidence.
Introduction. Recombination is the process that is exclusive to the meiotic division as it allows the exchange of genetic material between the non-homologous chromosomes. The recombination process is responsible for the shuffling of the characters and producing a zygote that is different from both the parents but has the chromosomes from both the parents.
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SET 2: In corn, the following allelic pairs have been identified in chromosome 3:
+/b
plant color booster vs. non-booster
+/lg
with ligule vs. without ligule
+/v
green plant vs. virescent
A test cross between triple recessive plants and plants heterozygous for the three pairs gave the following
progeny:
+ v lg
305
b + lg
b v lg
+ + lg
b v +
148
58
+ + +
62
+ y +
152
b + +
275
Total
1000
Calculate the parental and recombinant frequencies, the map distances between genes, and the coefficient of
coincidence. Illustrate the gene map or sequence.
In corn, the following genes are linked on chromosome 3: va - variable sterile v - virescent gl - glossy leaves
Two homozygous plants were crossed and produced an all-normal F1. The F1 was test-crossed, progeny phenotypes appeared as follows:
235 virescent
40 variable sterile
48 virescent, glossy7 glossy
270 variable sterile, glossy
4 variable sterile, virescent60 wild
62 variable sterile, virescent, glossy
Show the diagram of the chromosome map for the 3 loci in this format: a--7.00 cM--b--8.00 cM--c--9.00 cM
What is the correct sequence of the three gene loci?
What is the coefficient of coincidence (COC) and the interference in this cross?
In fruit flies, red eyes (pr+_) are dominant to purple eyes (prpr) and normal wings (vg+_) are dominant to vestigial wings (vgvg). The genes are located on the same chromosome. A pure-breeding red-eyed fly with vestigial wings was crossed with a pure-breeding purple-eyed fly with normal wings. All of the F1 progeny had a WT phenotype. The recombination frequency between the two genes is 15%. If an F1 individual were test crossed, what percentage of the progeny would you expect to have the WT phenotype?
Chapter 4 Solutions
Introduction to Genetic Analysis
Ch. 4 - Prob. 1PCh. 4 - Prob. 5PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19P
Ch. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 21.1PCh. 4 - Prob. 21.2PCh. 4 - Prob. 21.3PCh. 4 - Prob. 21.4PCh. 4 - Prob. 21.5PCh. 4 - Prob. 21.6PCh. 4 - Prob. 21.7PCh. 4 - Prob. 21.8PCh. 4 - Prob. 21.9PCh. 4 - Prob. 21.10PCh. 4 - Prob. 21.11PCh. 4 - Prob. 21.12PCh. 4 - Prob. 21.13PCh. 4 - Prob. 21.14PCh. 4 - Prob. 21.15PCh. 4 - Prob. 21.16PCh. 4 - Prob. 21.17PCh. 4 - Prob. 21.18PCh. 4 - Prob. 21.19PCh. 4 - Prob. 21.20PCh. 4 - Prob. 21.21PCh. 4 - Prob. 21.22PCh. 4 - Prob. 21.23PCh. 4 - Prob. 21.24PCh. 4 - Prob. 21.25PCh. 4 - Prob. 21.26PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 38.1PCh. 4 - Prob. 38.2PCh. 4 - Prob. 38.3PCh. 4 - Prob. 38.4PCh. 4 - Prob. 38.5PCh. 4 - Prob. 38.6PCh. 4 - Prob. 38.7PCh. 4 - Prob. 38.8PCh. 4 - Prob. 38.9PCh. 4 - Prob. 38.10PCh. 4 - Prob. 38.11PCh. 4 - Prob. 38.12PCh. 4 - Prob. 38.13PCh. 4 - Prob. 38.14PCh. 4 - Prob. 38.15PCh. 4 - Prob. 38.16PCh. 4 - Prob. 38.17PCh. 4 - Prob. 38.18PCh. 4 - Prob. 38.19PCh. 4 - Prob. 38.20PCh. 4 - Prob. 38.21PCh. 4 - Prob. 38.22PCh. 4 - Prob. 38.23PCh. 4 - Prob. 38.24PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69P
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- 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_forwardSET 2: In corn, the following allelic pairs have been identified in chromosome 3: +/b plant color booster vs. non-booster +/lg with ligule vs. without ligule +/v green plant vs. virescent A test cross between triple recessive plants and plants heterozygous for the three pairs gave the following progeny: + v lg b + lg b v lg + + lg 305 148 58 bv + + + + 62 + v + b + + 152 275 1000 Total Calculate the parental and recombinant frequencies, the map distances between genes, and the coefficient of coincidence. Illustrate the gene map or sequence.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 autotetraploid Chinese primrose (Primula sinensis L.), the gene controlling stigma color is very near the centromere of the chromosome carrying it. The allele G for green stigma is dominant to g for red stigmas. A homozygous green autotetraploid strain is crossed with a homozygous red autotetraploid strain. Each of the F1 GGgg plants would obtain 12 gametes which are 2GG, 8Gg, and 2g. How were these obtained?arrow_forwardWaxy endosperm (wx), virescent seedling (v), and shrunken endosperm (sh) are all recessive mutants in corn. Waxy and shrunken are both on chromosome IX, located 18 map units apart. Virescent is on chromosome V and is not linked to the other loci. A strain of corn that is homozygous for waxy and virescent is crossed to a strain that is homozygous for shrunken endosperm. The F1 were entirely wild type. An F1 individual was then test crossed to a strain that is homozygous for waxy, virescent, and shrunken. Of 1,000 offspring, how many of the offspring will be virescent and shrunken, but not waxy? Using the information from Problem 1, how many of the offspring will be waxy, virescent, and shrunken?arrow_forward+ ec +/Y + + w/Y y ec +/Y + ec +/y ec w ++ w/y ec w у ес +у ес и Determine the order in which the three loci y, ec, and w Occur on the chromosome and prepare a linkage map. 7.22 A cross involving X-linked genes was made between yellow, bar, vermilion female fies and wild males, and the F1 females were crossed with y B v males. The following phenotypes were obtained when 1000 progeny were exam- ined: Dra ord ma the 7.2 546 244 160 50 + + + + Bv y Bv y+ + y+v y B+ and an and and and +B + re + + v ge Determine the order in which the three loci occur on the chromosome and prepare a linkage map. 7.23 Female Drosophila heterozygous for ebony (e"le), scarlet (st*/st), and spineless (ss*/ss) were testcrossed, and the following progeny were obtained: PROGENCY PHENOTYPES NUMBER ir Wild type Ebony Ebony, scarlet Ebony, spineless Ebony, scarlet, spineless Scarlet 67 8. 68 347 78 368 Scarlet, spineless Spineless (a) Are these genes linked? Justify your answer. (b) Write the genes given on a…arrow_forward
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- Three autosomal genes are linked along the same chromosome.The distance between gene A and B is 7 mu, the distance betweenB and C is 11 mu, and the distance between A and C is 4 mu. Anindividual that is AA bb CC was crossed to an individual that is aaBB cc to produce heterozygous F1 offspring. The F1 offspring werethen crossed to homozygous aa bb cc individuals to produce F2offspring.Where would a crossover have to occur to produce an F2 offspringthat was heterozygous for all three genes?arrow_forward. Chromosome 3 of corn carries three loci (b for plant-color booster, v for virescent, and lg for liguleless). A testcross of triple recessives with F1 plants heterozygous forthe three genes yields progeny having the followinggenotypes: 305 + v lg, 275 b + +, 128 b + lg, 112 + v +,74 + + lg, 66 b v +, 22 + + +, and 18 b v lg. Give the genesequence on the chromosome, the map distances between genes, and the coefficient of coincidence.arrow_forwardThree autosomal genes are linked along the same chromosome.The distance between gene A and B is 7 mu, the distance betweenB and C is 11 mu, and the distance between A and C is 4 mu. Anindividual that is AA bb CC was crossed to an individual that is aaBB cc to produce heterozygous F1 offspring. The F1 offspring werethen crossed to homozygous aa bb cc individuals to produce F2offspring.If we assume that no double crossovers occur, what percentageof F2 offspring is likely to be homozygous for all three genes?arrow_forward
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