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 31.6P
Summary Introduction
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Introduction: The dominant eye color for the Drosophila melanogaster or fruit fly is red (RR), while brown-color is the recessive trait. Similarly, long wings are dominant in Drosophila melanogaster, while short wings are a recessive trait.
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Students have asked these similar questions
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 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?
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
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, 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_forwardIn 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?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
- In 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 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_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, 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_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/XwV/v×Xw/Y v/v, Xw/XwV/v×XW/Y V/v.arrow_forward
- The phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly that is homozygous for normal wings and has a hairy body and a fly with vestigial wings that is homozygous for normal body. The wild-type F1 flies were crossed among each other to produce 1024 F2 offspring. Which phenotypes would you expect among the F2 offspring, and how many of each phenotype would you expect? Group of answer choices 192 wild type, 256 vestigial, 64 hairy, and 192 vestigial and hairy All vestigial and hairy. 576 wild type, 192 vestigial, 192 hairy, and 64 vestigial and hairy All wild type 256 wild type; 256 vestigial, 256 hairy, and 256 vestigial and hairyarrow_forwardIn Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the F1 cross? X X 3+ cu e + X X e + + + + + cu e + O + ■ 3+ X X X X Y Y + + ■ cu cu cu ' + ■ cu ■ ' + e + e e e e e + cu +arrow_forwardIn 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 = 1arrow_forward
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