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In a cross involving Drosophila melanogaster, an F_2 population included 4400 flies with normal wings and...
Drosophilia homozygous for the lobed eye mutation (L) were mated with flies homozygous for a wing...
Drosophilia homozygous for the lobed eye mutation (L) were mated with flies homozygous for a wing mutation known as plexus (px) All the f1 progeny had lobed eyes and wild type wings. The F1 flies were test crossed resulting in the following progeny: Number 18 F2 phenotypes Wild type eyes and wings Lobed eyes Plexus wings Lobed eyes and plexus wings 35 33 14 Write the genotypes of the parental and f1 flies. Use proper drosophilia nomenclature. Be sure to...
Genetic Linkage The six genes listed below are all located on Chromosome 2 of Drosophila melanogaster....
Genetic Linkage The six genes listed below are all located on Chromosome 2 of Drosophila melanogaster. Your goal is to construct a genetic map of Chromosome 2. That is, determine the order of these genes along chromosome 2 and the map distance in centimorgans between each gene. To complete this task, you will be given the results of a variety of two-point test crosses involving these genes. For each test cross you may assume that the female is heterozygous and...
Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to...
Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F1 females...
In Drosophila, the allele dp determines long wings and dp determines short ("dumpy") wings. At a...
In Drosophila, the allele dp determines long wings and dp determines short ("dumpy") wings. At a separate locus e* determines gray body and e determines ebony body. Both loci are autosomal. The original parents were purebred with antagonistic traits, one of the parents had long wings and ebony body. The dihybrid F1 was test crossed with the following result: Long, ebony 151 Long, gray 136 Short, gray 140 Short, ebony 145 Total You want to determine if the data support...
Finally, bow-legs is hypothesized to be X-linked recessive in Drosophila melanogaster. The P1 virgin females were,...
Finally, bow-legs is hypothesized to be X-linked recessive in Drosophila melanogaster. The P1 virgin females were, once again, homozygous wild type but the males were bow-legged. There were 52 wild type males and 67 wild type females in the F1 generation. The F2 generation contained 30 wild type males, 75 wild type females, 40 bow-legged males and no bow-legged females. Does this data support or reject the hypothesis? Use chi square to prove your position.
A dihybrid cross is performed in Drosophila - one parent is heterozygous for both genes (B/b...
A dihybrid cross is performed in Drosophila - one parent is heterozygous for both genes (B/b ; F/f) and is test-crossed with a homozygous recessive type (b/b ; f/f). (where B = black body; b = brown body; F = forked bristles; f = unforked bristles.) The results are: black, forked 230 black, unforked 210 brown, forked 240 brown, unforked 250 How many degrees of freedom would you use in this case? What would be the expected number for each...
recalling that wild type is tan body, red eyes, straight wings and straight antennae, the data...
recalling that wild type is tan body, red eyes, straight wings and
straight antennae, the data is consistent with the logical
hypothesis that mutant trait-pick a mutant-is -autosomal or
sexlinked-, -dominant or recessive-.
using the f2 geb data and assuming we want to fail to reject the
null hypothesis abd support "logical" hypothesis for this mutant
trait, for the chi square goodness of fit test what is the chi
square statistical value? (value should be to the hundreths)
Here is...
LABORATORY 6. MENDELIAN GENETICS: GENERAL REMARKS AND Drosophila 1. Are the genes for the body color,...
LABORATORY 6. MENDELIAN GENETICS: GENERAL REMARKS AND Drosophila 1. Are the genes for the body color, eye color, and wing shape recessive or dominant? Explain your answer. Yellow: _______________ White: _________________ Miniature: ____________ Sepia: _________________ Explanation: _________________________________________ _________________________________________ _________________________________________ _________________________________________ _________________________________________ 2. Where are these genes located? On an autosome(s) or sex chromosome? Explain your answer Yellow: _______________ White: _________________ Miniature: ____________ Sepia: _________________ Explanation: _________________________________________ _________________________________________ 2. Set up a X2 to test if the yellow (y) (body color)...
55 sn car sn cart sn* car sn+Car+ 200 TABLE 5.2 Critical Chi-Square Values Values 0.99...
55 sn car sn cart sn* car sn+Car+ 200 TABLE 5.2 Critical Chi-Square Values Values 0.99 0.90 0.50 0.10 0.05 0.01 0.001 Degrees of Freedom Values w - 0.02 0.45 2.71 0.02 0.21 1.39 4.61 0.11 0.58 2.37 6.25 0.30 1.06 3.36 778 0.55 1.61 4.35 9.24 3.84 6.64 10.83 5.99 9.21 13.82 7.81 11.35 16.27 9.49 13.28 18.47 11.07 15.09 20.52 In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles. The wild-type...
use problem above to answer Complete the following chi-square problems. (Show all work!!) In the cross...
use problem above to answer
Complete the following chi-square problems. (Show all work!!) In the cross AaBb x AaBb O-E (O-E) (O-E)?/E 311 105 1. Phenotype A-B- A-bb aaB- aabb Totals 25 calculated X2= mith, Beatty and Ware 7/14 revised by for Fall 2015, BIOL 251 Genetics Lab Manual, Benedictine Univ page 3 a. What are your null and alternative hypotheses? b. How many degrees of freedom? c. What is your table X2 value? d. What is your region of...
Drosophilia homozygous for the lobed eye mutation (L) were mated with flies homozygous for a wing mutation known as plexus (px) All the f1 progeny had lobed eyes and wild type wings. The F1 flies were test crossed resulting in the following progeny: Number 18 F2 phenotypes Wild type eyes and wings Lobed eyes Plexus wings Lobed eyes and plexus wings 35 33 14 Write the genotypes of the parental and f1 flies. Use proper drosophilia nomenclature. Be sure to...
Genetic Linkage The six genes listed below are all located on Chromosome 2 of Drosophila melanogaster. Your goal is to construct a genetic map of Chromosome 2. That is, determine the order of these genes along chromosome 2 and the map distance in centimorgans between each gene. To complete this task, you will be given the results of a variety of two-point test crosses involving these genes. For each test cross you may assume that the female is heterozygous and...
Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F1 females...
In Drosophila, the allele dp determines long wings and dp determines short ("dumpy") wings. At a separate locus e* determines gray body and e determines ebony body. Both loci are autosomal. The original parents were purebred with antagonistic traits, one of the parents had long wings and ebony body. The dihybrid F1 was test crossed with the following result: Long, ebony 151 Long, gray 136 Short, gray 140 Short, ebony 145 Total You want to determine if the data support...
recalling that wild type is tan body, red eyes, straight wings and
straight antennae, the data is consistent with the logical
hypothesis that mutant trait-pick a mutant-is -autosomal or
sexlinked-, -dominant or recessive-.
using the f2 geb data and assuming we want to fail to reject the
null hypothesis abd support "logical" hypothesis for this mutant
trait, for the chi square goodness of fit test what is the chi
square statistical value? (value should be to the hundreths)
Here is...
55 sn car sn cart sn* car sn+Car+ 200 TABLE 5.2 Critical Chi-Square Values Values 0.99 0.90 0.50 0.10 0.05 0.01 0.001 Degrees of Freedom Values w - 0.02 0.45 2.71 0.02 0.21 1.39 4.61 0.11 0.58 2.37 6.25 0.30 1.06 3.36 778 0.55 1.61 4.35 9.24 3.84 6.64 10.83 5.99 9.21 13.82 7.81 11.35 16.27 9.49 13.28 18.47 11.07 15.09 20.52 In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles. The wild-type...
use problem above to answer
Complete the following chi-square problems. (Show all work!!) In the cross AaBb x AaBb O-E (O-E) (O-E)?/E 311 105 1. Phenotype A-B- A-bb aaB- aabb Totals 25 calculated X2= mith, Beatty and Ware 7/14 revised by for Fall 2015, BIOL 251 Genetics Lab Manual, Benedictine Univ page 3 a. What are your null and alternative hypotheses? b. How many degrees of freedom? c. What is your table X2 value? d. What is your region of...
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