Introduction To Genetic Analysis
12th Edition
ISBN: 9781319114787
Author: Anthony J.F. Griffiths, John Doebley, Catherine Peichel, David A. Wassarman
Publisher: W. H. Freeman
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Chapter 3, Problem 43.7P
Summary Introduction
To explain: The full
Introduction: Drosophila melanogaster or fruit fly is a heavily used model organism for the study of genetics and gene regulation. It shows a normal red eye color, but some strains are also available which shows brown eye color (recessive). Similarly, wings are normally long, but there are strains with short wings (recessive).
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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, 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?
Chapter 3 Solutions
Introduction To Genetic Analysis
Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10P
Ch. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 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. 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. 43.1PCh. 3 - Prob. 43.2PCh. 3 - Prob. 43.3PCh. 3 - Prob. 43.4PCh. 3 - Prob. 43.5PCh. 3 - Prob. 43.6PCh. 3 - Prob. 43.7PCh. 3 - Prob. 43.8PCh. 3 - Prob. 43.9PCh. 3 - Prob. 43.10PCh. 3 - Prob. 43.11PCh. 3 - Prob. 43.12PCh. 3 - Prob. 43.13PCh. 3 - Prob. 43.14PCh. 3 - Prob. 43.15PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 70PCh. 3 - Prob. 1GSCh. 3 - Prob. 2GSCh. 3 - Prob. 3GS
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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 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_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
- 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 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, 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
- 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_forwardIn an 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. What is the genotype of the F1? Show the types of gametes the F1’s may be expected to form and derive the expected proportion of each. What phenotypic ratio of green to red is expected if: The F1’s are intercrossed? The F1’s are crossed with red plants If the G locus were 50 or more map units from the centromere, what types and proportions of gametes would the F1 be expected to produce? Derive the expected F2 phenotypic ratio.arrow_forwardIn 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_forward
- A mutant sex-linked trait called “notched” (N) is deadly in Drosophila when homozygous in females. Males who have a single N allele will also die. The heterozygous condition (Nn) causes small notches on the wing. The normal condition in both male and females is represented by the allele n. a) Indicate the phenotypes of the F1 generation from the following cross: XNXn x XnY b) Explain why dead females are never found in the F1 generation no matter which parents are crossed. c) Explain why the mating of female XNXn and a male XNy is unlikely.arrow_forwardIn the following cross, imagine that you have a female fly that has two Xs and one Y due to a nondisjunction event in her mother's germ cells. Draw out what the possible gametes are for both the female and the male and also a Punnett square showing the genotypes, phenotypes, and sex of the possible flies as a result of this cross. You do not need to provide the probabilities of each of these. Red-eyed wi C Ở Red-eyed wt XX Y X Y Meiosisarrow_forwardIn Drosophila, a gene controls body color producing either normal body color or the mutant form black body color (bl. A second gene controls wings. The flies either have normal wings or are wingless (wn). A cross is made between a homozygous wild type fly and fly with black body and wingless. A test cross was then performed. The following progeny was observed: Phenotype # Observed Wild Type 405 Normal, wingless 85 Black, normal 100 Black, wingless 410 A. List the genotypes of the original parents. B. List the genotypes of the testcross. C. Are these two genes linked? How do you know? Edit Format Table 12pt v Paragraph v BIUAarrow_forward
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