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 31.9P
Summary Introduction
To explain: The reasons for the alteration of the classic
Introduction: The classic ratio obtained in the F2 generation of a dihybrid cross is 9:3:3:1.
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In fruit flies, you are mapping three genes in a three point cross. The mutants are hairy
body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent
with a homozygous recessive parent and obtain the following results:
Type
Number
h se g.
5
+ se +
450
+ se g
27
++g_
h se +
+ + +
h + g.
h + +
TOTAL
is the gene in the middle and the distance in map units between se and g is
Oh; 16.4
se; 7.1
Oh; 7.1
70
82
7
327
32
1000
se; 16.4
In Drosophila, the brown mutation (bw, chromosome 2, position 104.5) results in brown eyes, while miniature (min, chromosome X, position 36.1) results in wings that are 2/3 the length of wild type. True breeding, wild type females are mated with true breeding males with brown eyes and miniature wings.
Using Drosophila notation, diagram the P1 and F1 crosses.
P1 F1
Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work.
Phenotype
Females
Males
Overall (♀and ♂)
=1 =1 =1
In Drosophila, the allele for red eyes (pt) is wild-type and the allele for purple eyes (p¯) is mutant. The allele for grey body (b+) is wild-type and the allele for black body (b¯) is
mutant. Flies with p*p* b*b* genotypes are mated with flies that have p p- b¯b¯ genotypes. A testcross was then performed in which the F1 offspring with p*p¯ b*b¯ genotypes
were mated with flies with p p¯ b¯b genotypes. Ten-thousand flies were produced from this test cross. The following results were observed:
4,300 red eye, grey body flies
550 red eye, black body flies
4,500 purple eye, black body flies
650 purple eye, grey body flies
Which F2 phenotypes are parental types? Which F2 phenotypes are recombinant types? What is the distance between the gene loci for eye color and body color?
Use the equation for Hardy-Weinberg equilibrium for the following questions. p+q = 1
p2 + 2pq + q2 = 1
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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- In Drosophila melanogaster, vestigial (short) wings (vg) are caused by a recessive mutant gene that independently assorts with a gene pair that influences body hair. Hairy (h ) results in a hairy body. A cross is made between a fly with normal wings and a hairy body and a fly with vestigial wings and a normal body. The phenotypically normal F1 flies were crossed among each other and 1024 F2 flies were reared. What phenotypes would you expect in the F2 and in what actual numbers (not ratio) would you expect to find them?arrow_forwardIn Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?arrow_forwardAnother cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w) and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F1 females were wild type for all three traits, while the F1 males expressed the yellow-body and white eyes traits. The cross was carried to an F2 progeny and only male offspring were tallied. Based on the data shown here, a genetic map was constructed. a) Diagram the genotypes of the F1 parents. b) Construct a map, assuming the white is at locus 1.5 on the X-chromosome. Phenotype Male offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0arrow_forward
- Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w) and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F1 females were wild type for all three traits, while the F1 males expressed the yellow-body and white eyes traits. The cross was carried to an F2 progeny and only male offspring were tallied. Based on the data shown here, a genetic map was constructed. a) Diagram the genotypes of the F1 parents. b) Construct a map, assuming the white is at locus 1.5 on the X-chromosome *******ANSWER PART B NOT PART A!!!! Phenotype Male offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0arrow_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_forwardIn Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1arrow_forward
- Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F1 females were wild type for all three traits, while the F1 males expressed the yellow-body and white-eye traits. The cross was carried to an F2 progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed. Phenotype Male Offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0 (a) Diagram the genotypes of the F1 parents. (b) Construct a map, assuming that white is at locus 1.5 on the X chromosome. (c) Were any double-crossover offspring expected? (d) Could the F2 female offspring be used to construct the map? Why or why not?arrow_forwardThe allele b gives Drosophila flies a black body and b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings and wx+ gives non-waxy, the wild-type phenotype. The allele cn of a third gene gives cinnabar eyes and cn+ gives red, the wild-type phenotype. A female heterozygous for these three genes is testcrossed, and 1000 progeny are classified with the following phenotypes. 382 cinnabar 379 black, waxy 69 waxy, cinnabar 67 black 48 waxy 44 black, cinnabar 5 wild type 6 black, waxy, cinnabar Based on this data, what is the correct map of these genes in terms of order and distance?arrow_forwardIn silkmoths (Bombyx mori), red eyes (re) and white-banded wings (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have wild-type eyes and wild-type wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are wild-type eyes, wild-type wings red eyes, wild-type wings wild-type eyes, white-banded wings red eyes, white-banded wings a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes? b. What is the rate of recombination between the gene for red eyes and the gene for white-banded wings?arrow_forward
- In Drosophila, the dominant Bar mutation (B, chromosome X, position 57) results in thin bar- shaped eyes, while the recessive singed (sn, chromosome X, position 21) results burnt looking bristles. True breeding, wild type females are mated with true breeding males with Bar eyes and singed bristles. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1arrow_forwardIn the fruit fly Drosophila melanogaster, the trait of black body is due to a gene on chromosome 2 and black body b is recessive to wild type body b + . The trait of purple eyes is controlled by a gene that is also on chromosome 2 and purple eyes p is recessive to wild type eyes p + . A true-breeding wild type strain is crossed with a true breeding strain that has black bodies and purple eyes. The F1 generation is then testcrossed to the black body, purple eye strain and 500 progeny are produced as follows: 224 wild type for both body and eye 236 black body and purple eye 18 wild type body and purple eye 22 black body and wild type eye. What is the recombination frequency and genetic map distance between the two genes?arrow_forwardIn Drosophila melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/Xw V/v×Xw/Y v/v, Xw/Xw V/v × XW/Y V/v.arrow_forward
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