In fruit flies, assume bright red eyes (r) and curly wing (w) are encoded by two...
- In fruit flies, assume bright red eyes (r) and curly wing (w) are encoded by two alleles that are recessive to wild-type alleles (R and W). These two genes are on the same chromosome. A fly homozygous bright red eyes and curly wings is crossed to a homozygous dominant fly. The F1 have a normal eyes and wings. The F1 flies are test crossed to homozygous recessive flies.
Wild-type eyes, wild type wing 418
Bright red eyes, wild type wing 19
Wild-type eyes, curly wing 16
Bright red eyes, curly wing 426
What is the genetic distance between the genes for bright red eyes and curly wings?
I got 39.8 map units
- You can conduct the crossed above and are told that the two genes in question are 32.6 map units apart. What percentage of the offspring above what you expect to have right red eyes and wild type of wings? What action number of flies would you expect to display these two traits?
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In fruit flies, assume bright red eyes (r) and curly wing (w)
are encoded by two...
If an apterous female fruit fly with wild type (+) red eyes is crossed with a...
If an apterous female fruit fly with wild type (+) red eyes is crossed with a white eyed male with normal wild type wings, what is the resting F1 generation? Please show on a punnet square.Also what is the punnet dihybrid cross for the F2 generation. Note: the recessive white eyed allele is X-linked while the apterous (wingless) recessive allele is NOT x-linked. SO to summarize: male with recessive white eyes and with normal wild type wings and a recessive...
39. You have crossed two fruit fly individuals that have long wings and red eyes -...
39. You have crossed two fruit fly individuals that have long wings and red eyes - the wild-type phenotype. In the progeny, the mutant phenotypes called curved wings and lozenge eyes appear as follows: Females 600 long-wing, red eyes 200 curved wing, red eyes Males 300 long-wing red eyes 300 long-wing, lozenge eyes 100 curved wing, red eyes 100 curved wing, lozenge eyes According to these data, the lozenge eye allele is A. sex-linked dominant B. sex-linked recessive C. autosomal...
6. In Cross 1, a yellow eyed, long wing fruit fly from a pure breeding strain...
6. In Cross 1, a yellow eyed, long wing fruit fly from a pure breeding strain is mated to a red eye, short wing fruit fly from a pure breeding strain. All of their offspring (F1) had red eyes and long wings. In Cross 2, one of the F1 offspring is mated to a fly with yellow eyes and short wings, and this cross gave the following F2 population: 194 flies with long wings and red eyes, 796 flies with...
A female fruit fly with vermilion eyes and normal wings is crossed to a male with...
A female fruit fly with vermilion eyes and normal wings is crossed to a male with normal red eyes and cut wings. The F1 progeny consist of females with red eyes and normal wings, and males with vermilion eyes and normal wings. When the F1 progeny are interbred, the F2 consists of two types of females vermilion eyes, normal wings (1/2) and red eyes, normal wings (12), and two types of males-vermilion eyes, normal wings (12) and red eyes, cut...
Question 5 In Drosophila, ebony body (e) and rough eyes (r) are encoded by linked autosomal...
Question 5 In Drosophila, ebony body (e) and rough eyes (r) are encoded by linked autosomal recessive genes on the third chromosome and are completely linked at O map units apart. A homozygous ebony female was crossed to a homozygous rough male. The F1 were all wild ty The F1 females were crossed to ebony, rough males. Give the phenotypes and their proportions in the F2. ebony, rough 20%, wildtype 20%, ebony 30%, rough 30% ebony, rough 30%, wildtype 30%,...
"Ebony body, held-out wings and sepia eyes are ALL RECESSIVE AND ALL ON CHROMOSOME 3. The...
"Ebony body, held-out wings and sepia eyes are ALL RECESSIVE AND ALL ON CHROMOSOME 3. The COEFFICIENT of COINCIDENCE for chromosome 3 is 0.5. A homozygous ebony male is mated to a homozygous held out wing and sepia eyed female. The F1 are wild-type. An F1 female is test crossed to a fly that is homozygous for all three recessive alleles and they produce 1000 progeny. If ebony and held-out are separated by 12 map units, and held-out and sepia...
Question 1, Part 2: A student crossed two flies and collected Fl progeny. The first parental...
Question 1, Part 2: A student crossed two flies and collected Fl progeny. The first parental fly was homozygous for two recessive mutations - yellow body (y) and bent wings (b), while the other parental was homozygous for the wild- type alleles - black body (Y) and straight wings (B). All of the F1 progeny had straight wings and black bodies. He did a testcross between the yy bb parent and the F1 progeny, and observed the following classes in...
The genes for sepia eye color, short bristles, and dark body coloration are on the same...
The genes for sepia eye color, short bristles, and dark body coloration are on the same chromosome of Drosophila melanogaster. Each gene has two alleles: wild type, which is dominant, or mutant, which is recessive. se+ is dominant and causes red eyes; se is recessive and causes sepia eyes sb+ is dominant and causes long bristles; sb is recessive and causes short bristles b+ is dominant and causes gray body coloration; b is recessive and causes dark...
Assortment of genes on same chromosome In the fruit fly Drosophila, there is a dominant gene...
Assortment of genes on same chromosome In the fruit fly Drosophila, there is a dominant gene for normal wings and its recessive allele for vestigial wings. At another gene locus. there is a dominant gene for red eyes and its recessive allele for purple eyes. A female that was heterozygous at both gene loci was mated with a male that is homozygous for both recessive alleles. Knowing this, complete the sentences with the correct terms. 94% crossing over independent assortment...
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...
39. You have crossed two fruit fly individuals that have long wings and red eyes - the wild-type phenotype. In the progeny, the mutant phenotypes called curved wings and lozenge eyes appear as follows: Females 600 long-wing, red eyes 200 curved wing, red eyes Males 300 long-wing red eyes 300 long-wing, lozenge eyes 100 curved wing, red eyes 100 curved wing, lozenge eyes According to these data, the lozenge eye allele is A. sex-linked dominant B. sex-linked recessive C. autosomal...
6. In Cross 1, a yellow eyed, long wing fruit fly from a pure breeding strain is mated to a red eye, short wing fruit fly from a pure breeding strain. All of their offspring (F1) had red eyes and long wings. In Cross 2, one of the F1 offspring is mated to a fly with yellow eyes and short wings, and this cross gave the following F2 population: 194 flies with long wings and red eyes, 796 flies with...
A female fruit fly with vermilion eyes and normal wings is crossed to a male with normal red eyes and cut wings. The F1 progeny consist of females with red eyes and normal wings, and males with vermilion eyes and normal wings. When the F1 progeny are interbred, the F2 consists of two types of females vermilion eyes, normal wings (1/2) and red eyes, normal wings (12), and two types of males-vermilion eyes, normal wings (12) and red eyes, cut...
Question 5 In Drosophila, ebony body (e) and rough eyes (r) are encoded by linked autosomal recessive genes on the third chromosome and are completely linked at O map units apart. A homozygous ebony female was crossed to a homozygous rough male. The F1 were all wild ty The F1 females were crossed to ebony, rough males. Give the phenotypes and their proportions in the F2. ebony, rough 20%, wildtype 20%, ebony 30%, rough 30% ebony, rough 30%, wildtype 30%,...
"Ebony body, held-out wings and sepia eyes are ALL RECESSIVE AND ALL ON CHROMOSOME 3. The COEFFICIENT of COINCIDENCE for chromosome 3 is 0.5. A homozygous ebony male is mated to a homozygous held out wing and sepia eyed female. The F1 are wild-type. An F1 female is test crossed to a fly that is homozygous for all three recessive alleles and they produce 1000 progeny. If ebony and held-out are separated by 12 map units, and held-out and sepia...
Question 1, Part 2: A student crossed two flies and collected Fl progeny. The first parental fly was homozygous for two recessive mutations - yellow body (y) and bent wings (b), while the other parental was homozygous for the wild- type alleles - black body (Y) and straight wings (B). All of the F1 progeny had straight wings and black bodies. He did a testcross between the yy bb parent and the F1 progeny, and observed the following classes in...
Assortment of genes on same chromosome In the fruit fly Drosophila, there is a dominant gene for normal wings and its recessive allele for vestigial wings. At another gene locus. there is a dominant gene for red eyes and its recessive allele for purple eyes. A female that was heterozygous at both gene loci was mated with a male that is homozygous for both recessive alleles. Knowing this, complete the sentences with the correct terms. 94% crossing over independent assortment...
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...
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