Paragraph Styles Dictate Sensitivity Font Editor I You breed two monster parents (green, hairless) GGhh and ggHH (white, hairy) to produce F1 offspring (green, hairy). Genes G and H are on the same chromosome. a. What are the dominant phenotypes? b. What are ALL of the potential gamete genotypes that could be created by an F1? Clearly indicate the recombinant genotypes. C. Explain how recombinants are created. What are the potential offspring genotypes from the cross of a F1 monster with a homozygous recessive monster. From this test cross you get 152 green, hairless; 23 green, hairy; 148 white, hairy, and 27 e. white, hairless. Calculate the chi squared to test the independent assortment hypothesis and determine if these genes assort independently. d.

Paragraph Styles Dictate Sensitivity Font Editor I You breed two monster parents (green, hairless) GGhh and ggHH (white, hairy) to produce F1 offspring (green, hairy). Genes G and H are on the same chromosome. a. What are the dominant phenotypes? b. What are ALL of the potential gamete genotypes that could be created by an F1? Clearly indicate the recombinant genotypes. C. Explain how recombinants are created. What are the potential offspring genotypes from the cross of a F1 monster with a homozygous recessive monster. From this test cross you get 152 green, hairless; 23 green, hairy; 148 white, hairy, and 27 e. white, hairless. Calculate the chi squared to test the independent assortment hypothesis and determine if these genes assort independently. d.

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Description: ParagraphStylesDictateSensitivityFontEditorIYou breed two monster parents (green, hairless) GGhh and ggHH (white, hairy) to produce F1offspring (green, hairy). Genes G and H are on the same chromosome.a.What are the dominant phenotypes?b.What are AL

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 2QP: Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To...

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. If you intercrossed F1 heterozygotes of the formA b / a B in mice, the phenotypic ratio among the F2progeny would vary with the map distance betweenthe two genes. Is there a simple way to estimate themap distance based on the frequencies of the F2phenotypes, assuming rates of recombination areequal in males and females? Could you estimatemap distances in the same way if the mouse F1heterozygotes were A B / a b?
F2 plants segregate colored : colorless. If acolored plant is picked at random and selfed, what is theprobability that both colored and colorless plants will beseen among a large number of its progeny? Please solve with step by step please
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
mapping three linked genes, the heterozygous AaBbCc was crossed to aabbcc individual the following phenotypes were obtained: ABC 4 abc 2 ABc 150 abC 141 aBc 18 AbC 20 Abc 30 aBC 35 A) What are the alleles found on each parental chromosome(genotypes of the parents)? B) How many double recombinants(cross over) were there? c) what is the linear arrangement of loci on the chromosome?
Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PCR was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PCR products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?
The drug ivacaftor has recently been developed totreat cystic fibrosis in children with the rare G551Dmutant allele of CFTR.a. Do you think that ivacaftor would be effective onlyin patients homozygous for the G551D mutation,or might it work as well in compound heterozygotes in which one copy of chromosome 7 hadG551D and the other copy a different allele ofCFTR, such as the more prevalent allele ΔF508?(The protein encoded by G551D folds up properlyand inserts into the cell membrane, but is inefficient in chloride ion transport. Ivacaftor increasesthe efficiency of G551D’s ion transport. TheΔF508 protein does not fold up properly and therefore does not get inserted into the cell membrane.)b. Why do you think ivacaftor would be more effectivein children than in older cystic fibrosis patients?c. The scientists who developed ivacaftor had a modelfor cystic fibrosis: a line of cells that grow in culture and that are homozygous for G551D. Thesecells accumulate mucus at their surfaces that…
When true-breeding mice with brown fur and short tails (BBtt)were crossed to true-breeding mice with white fur and long tails(bbTT), all of the F1 offspring had brown fur and long tails. TheF1 offspring were crossed to mice with white fur and short tails.What are the possible phenotypes of the F2 offspring? Which F2offspring are recombinant, and which are nonrecombinant? Whatare the ratios of phenotypes of the F2 offspring if independentassortment is taking place? How are the ratios affected by linkage?
2. a. A Drosophila male from a true-breeding stockwith scabrous eyes was mated with a female from atrue-breeding stock with javelin bristles. Both scabrous eyes and javelin bristles are autosomal recessive mutant traits. The F1 progeny all had normaleyes and bristles. F1 females from this cross weremated with males with both scabrous eyes andjavelin bristles. Write all the possible phenotypicclasses of the progeny that could be produced from the cross of the F1 females with the scabrous, javelin males, and indicate for each class whether it is arecombinant or parental type.b. The cross in part (a) yielded the following progeny:77 scabrous eyes and normal bristles; 76 wild type(normal eyes and bristles); 74 normal eyes andjavelin bristles; and 73 scabrous eyes and javelinbristles. Are the genes governing these traits likelyto be linked, or do they instead assort independently? Why?c. Suppose you mated the F1 females from the crossin part (a) to wild-type males. Why would thiscross fail…
Mitotic recombination can occasionally produce a twin spot.Let’s suppose an animal species is heterozygous for two genesthat govern fur color and length: one gene affects pigmentation,with dark pigmentation (A) dominant to albino (a); the other geneaffects hair length, with long hair (L) dominant to short hair (l).The two genes are linked on the same chromosome. Let’s assume ananimal of this species is AaLl; A is linked to l, and a is linked to L.Draw the chromosomes labeled with these alleles, and explainhow mitotic recombination could produce a twin spot with onespot having albino pigmentation and long fur and the other havingdark pigmentation and short fur.
What is mitotic recombination? A heterozygous individual (Bb)with brown eyes has one eye with a small patch of blue. Providetwo or more explanations for how the blue patch may have occurred?
For this particular family, what is the recombination rate between the D17S74 marker and the breast cancer gene? Lets say that at age 45 the third granddaughter III-3, is diagnosed with breast cancer. Now recalculate the recombination rate between the marker locus D17S74 and the breast cancer gene? Use only the 5 women in your analysis:the two daugthers in generation II and the three granddaughters in generation II
Two plants in a cross were each heterozygous for two gene pairs (AB/ab) whose loci are linked and 10 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part D If the two genes are 15 mu apart and the plant is (Ab/aB), what proportion of gametes from a signal plant will be ab? Part E What proportion of the offspring of two plants ( both (Ab/aB)) will be A_B_ if the genes are 15 mu apart? Part F What proportion of the offspring of two plants ( both (Ab/aB)) will be A_bb if the genes are 15 mu apart? Part G What proportion of the offspring of two plants ( both (Ab/aB)) will be aaB_ if the genes are 15 mu apart? Part H What proportion of the offspring of two plants ( both (Ab/aB)) will be aabb if the genes are 15 mu apart? How would I solve these?