Homework Answers
Animals 10,8, 14 and 19 shows recombination between W and marker B.In all these there is a recombinant of WB11 can be seen.To find the linkage distance divide the number of recombinant gametes into the total number of gametes.Therefore cM distance is 4.4cM.
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QUESTION D: In another set of mating experiments, a pink rabbit was mated to a pink...
In another set of mating experiments, a pink rabbit was mated to a pink rabbit and...
In another set of mating experiments, a pink rabbit was mated to a pink rabbit and the progeny tested for RFLPs A and B and for coat color. Since both parents are heterozygous at A, B, and W, each progeny is the result of two meiosis events in which recombination COULD have occurred. a. How many meiosis events are represented in the 22 progeny shown in the table shown below? 1 2 3 4 5 6 7 8 9 10...
draw a southern blot in which genomic DNa frow WW rabbits, Ww rabbits pink, and ww...
draw a southern blot in which genomic DNa frow WW rabbits, Ww
rabbits pink, and ww rabbits white has been digested with EcoRI and
hybridized to probe A
3. You have discovered a coat color gene ("W" gene) in rabbits that has two alleles, W and w. Rabbits that are homozygous for W (WW) are red. white. Heterozygous rabbits (Ww) are pink. Rabbits that are homozygous for w (ww) are You have also identified a DNA sequence of about 100...
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16...
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 10kb Probe А 5kb -UI III TIT II Probe B 11kb 7kb — color pink white red pink red pink pink white pink white red pink red white pink pink red pink white pink red pink c. probes A and B: list the animals that show recombination between these markers and then calculate CM distance between them. Which...
In lizards, the gene for color and the gene for pattern are located on the same...
In lizards, the gene for color and the gene for pattern are located on the same chromosome and exhibit recombination. The lined pattern allele (l) is recessive to the solid unpatterned allele (L), and the aqua color allele (b) is recessive to the blue color allele (B). Geneticists cross a homozygous lined aqua lizard with a homozygous solid blue lizard. Then, they carry out a testcross using the F1 progeny. The parental chromosomes for the testcross are given. Identify all...
The results for the F2 progeny are shown for a F1 cross using the two X-linked...
The results for the F2 progeny are shown for a F1 cross using
the two X-linked markers:
w+ and m+.
As shown in the figure, the F1 cross is between the following
two parents:
- a phenotypically wild-type (red eye, full wing), heterozygous
female: w+m+/wm
- a double mutant male (white eye, miniature wing):
wm/Y
Answer the following questions:
1) In the F1 cross, can you infer the genotype for the phenotype
in the female? Why or why not?
2)...
4. You are studying three genes in corn and performed a three-point cross. The three genes...
4. You are studying three genes in corn and performed a three-point cross. The three genes are green (V) vs Viniscent (V) seedlings, blue (Pr) vs red kernels (pr) and white (B ) or brown (bm) leaf midrib color. The table below list the offspring resulting from a testcross with a female F1 heterozygote. Type of gamete (NR, SCO, DCO Phenotype Progeny Gamete from F1 heterozygote Virescent seedling red 101 kernel Red kernel 175 Red kernel, brown midrib 259 Virescent...
15. You are studying com plants. You were able to determine that the gene for kernel...
15. You are studying com plants. You were able to determine that the gene for kernel color (C-colored and no color) and kernel waxiness (W = waxy and not waxy) are on the same chromosome and are 30 CM apart. You are ultimately interested in determining the potential phenotypic ratio if crossing over occurs. You start by crossing a male who is homozygous dominant for both genes with a female who is homozygous recessive for both genes. The F1 heterozygotes...
PPPPPP Step 5: Determine the phenotype of cach possible allele combination shown in the offspring. In...
PPPPPP Step 5: Determine the phenotype of cach possible allele combination shown in the offspring. In this instance, three of the offspring Chose with the genotypes EE, Ee, and Be) have free earlobes, because the free earlobe allele is dominant to the attached-carlobe allele. One offspring, ec, has attached earlobes. Therefore, the proba- bility of having offspring with free earlobes is 3/4 and with attached earlobes is 1/4. Additional Single-Factor Inheritance Problems (One Trait Followed from One Generation to the...
can u tell me if these answers are correct please!??!!! Choose the best answer for the...
can
u tell me if these answers are correct please!??!!!
Choose the best answer for the following questions. Place your answer on the line. If your answer is not on the line.it does not count 1 Mender's discovery that characteristics are inherited due to the transmission of hereditary factors resulted from his (1) dissections to determine how fertilization occurs in pea plants (2analysis of the offspring produced from many pea plant crosses (3) careful microscopic examinations of genes and chromosomes...
Question 5 (14 marks) The allele cn gives fruit flies cinnabar eyes, whereas the allele cn...
Question 5 (14 marks) The allele cn gives fruit flies cinnabar eyes, whereas the allele cn gives red, the wild-type phenotype. The allele wx of a separate allele gives waxy wings, and wx gives non-waxy wings (wild-type). The allele b of a third gene gives fruit flies a black body, whereas the allele b gives a brown (wild-type) body colour. A geneticist knows these three genes are on the same chromosome, but does not know what order the genes are...
In another set of mating experiments, a pink rabbit was mated to a pink rabbit and the progeny tested for RFLPs A and B and for coat color. Since both parents are heterozygous at A, B, and W, each progeny is the result of two meiosis events in which recombination COULD have occurred. a. How many meiosis events are represented in the 22 progeny shown in the table shown below? 1 2 3 4 5 6 7 8 9 10...
draw a southern blot in which genomic DNa frow WW rabbits, Ww
rabbits pink, and ww rabbits white has been digested with EcoRI and
hybridized to probe A
3. You have discovered a coat color gene ("W" gene) in rabbits that has two alleles, W and w. Rabbits that are homozygous for W (WW) are red. white. Heterozygous rabbits (Ww) are pink. Rabbits that are homozygous for w (ww) are You have also identified a DNA sequence of about 100...
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 10kb Probe А 5kb -UI III TIT II Probe B 11kb 7kb — color pink white red pink red pink pink white pink white red pink red white pink pink red pink white pink red pink c. probes A and B: list the animals that show recombination between these markers and then calculate CM distance between them. Which...
In lizards, the gene for color and the gene for pattern are located on the same chromosome and exhibit recombination. The lined pattern allele (l) is recessive to the solid unpatterned allele (L), and the aqua color allele (b) is recessive to the blue color allele (B). Geneticists cross a homozygous lined aqua lizard with a homozygous solid blue lizard. Then, they carry out a testcross using the F1 progeny. The parental chromosomes for the testcross are given. Identify all...
The results for the F2 progeny are shown for a F1 cross using
the two X-linked markers:
w+ and m+.
As shown in the figure, the F1 cross is between the following
two parents:
- a phenotypically wild-type (red eye, full wing), heterozygous
female: w+m+/wm
- a double mutant male (white eye, miniature wing):
wm/Y
Answer the following questions:
1) In the F1 cross, can you infer the genotype for the phenotype
in the female? Why or why not?
2)...
4. You are studying three genes in corn and performed a three-point cross. The three genes are green (V) vs Viniscent (V) seedlings, blue (Pr) vs red kernels (pr) and white (B ) or brown (bm) leaf midrib color. The table below list the offspring resulting from a testcross with a female F1 heterozygote. Type of gamete (NR, SCO, DCO Phenotype Progeny Gamete from F1 heterozygote Virescent seedling red 101 kernel Red kernel 175 Red kernel, brown midrib 259 Virescent...
15. You are studying com plants. You were able to determine that the gene for kernel color (C-colored and no color) and kernel waxiness (W = waxy and not waxy) are on the same chromosome and are 30 CM apart. You are ultimately interested in determining the potential phenotypic ratio if crossing over occurs. You start by crossing a male who is homozygous dominant for both genes with a female who is homozygous recessive for both genes. The F1 heterozygotes...
PPPPPP Step 5: Determine the phenotype of cach possible allele combination shown in the offspring. In this instance, three of the offspring Chose with the genotypes EE, Ee, and Be) have free earlobes, because the free earlobe allele is dominant to the attached-carlobe allele. One offspring, ec, has attached earlobes. Therefore, the proba- bility of having offspring with free earlobes is 3/4 and with attached earlobes is 1/4. Additional Single-Factor Inheritance Problems (One Trait Followed from One Generation to the...
can
u tell me if these answers are correct please!??!!!
Choose the best answer for the following questions. Place your answer on the line. If your answer is not on the line.it does not count 1 Mender's discovery that characteristics are inherited due to the transmission of hereditary factors resulted from his (1) dissections to determine how fertilization occurs in pea plants (2analysis of the offspring produced from many pea plant crosses (3) careful microscopic examinations of genes and chromosomes...
Question 5 (14 marks) The allele cn gives fruit flies cinnabar eyes, whereas the allele cn gives red, the wild-type phenotype. The allele wx of a separate allele gives waxy wings, and wx gives non-waxy wings (wild-type). The allele b of a third gene gives fruit flies a black body, whereas the allele b gives a brown (wild-type) body colour. A geneticist knows these three genes are on the same chromosome, but does not know what order the genes are...
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