When homozygous yellow mutant rats [J] are crossed with homozygous black mutant rats [N], all F1 is grey [G]. In F2, we obtain 10 [J], 28 [G], 8 [N] and 2 light individuals. Tick the plausible explanation(s): O O O O O O the allele responsible for the grey colour is dominant over the alleles responsible for the black and yellow colour no genetic interaction here, the 2 genes act independently no genetic interaction here, it is a monohybridism with one allelic series no correct answer the allele responsible for the yellow colour is epistatic on the gene responsible for the black colour the allele responsible for the black colour is epistatic on the gene responsible for the grey colour

When homozygous yellow mutant rats [J] are crossed with homozygous black mutant rats [N], all F1 is grey [G]. In F2, we obtain 10 [J], 28 [G], 8 [N] and 2 light individuals. Tick the plausible explanation(s): O O O O O O the allele responsible for the grey colour is dominant over the alleles responsible for the black and yellow colour no genetic interaction here, the 2 genes act independently no genetic interaction here, it is a monohybridism with one allelic series no correct answer the allele responsible for the yellow colour is epistatic on the gene responsible for the black colour the allele responsible for the black colour is epistatic on the gene responsible for the grey colour

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter3: Transmission Of Genes From Generation To Generation

Section: Chapter Questions

Problem 18QP: More Crosses with Pea Plants: The Principle of Independent Assortment Determine the possible...

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More Crosses with Pea Plants: The Principle of Independent Assortment Determine the possible genotypes of the following parents by analyzing the phenotypes of their children. In this case, we will assume that brown eyes (B) is dominant to blue (b) and that right-handedness (R) is dominant to left-handedness (r). a. Parents: brown eyes, right-handed brown eyes, right-handed Offspring: 3/4 brown eyes, right-handed 1/4 blue eyes, right-handed b. Parents: brown eyes, right-handed blue eyes, right-handed Offspring: 6/16 blue eyes, right-handed 2/16 blue eyes, left-handed 6/16 brown eyes, right-handed 2/16 brown eyes, left-handed c. Parents: brown eyes, right-handed blue eyes, left-handed Offspring: 1/4 brown eyes, right-handed 1/4 brown eyes, left-handed 1/4 blue eyes, right-handed 1/4 blue eyes, left-handed
Height in humans is controlled by the additive action of genes and the action of environmental factors. For the purposes of this problem, assume that height is controlled by four genesA, B, C, and Dand that there are no environmental effects. Assume further that additive alleles contribute two units of height and partially additive alleles contribute one unit of height. a. Given these assumptions, can two individuals of moderate height produce offspring that are much taller and shorter than either parent? If so, how can this happen? b. Can someone of minimum height and someone of intermediate height have children taller than the parent of intermediate height? Why or why not?
The dominant white allele (W) is a homozygous lethal and is dominant over the full-color allele (w) in horses. The Overo paint allele (O) is also lethal when homozygous, and it is dominant over the nonpaint allele (o). The dominant at the white locus, however, masks the paint phenotype. Two horses of the genotype WwOo are crossed. At what frequency is a solid white horse expected from the this cross?
Albino rabbits (lacking pigment) are homozygous forthe recessive c allele (C allows pigment formation).Rabbits homozygous for the recessive b allele makebrown pigment, while those with at least one copy ofB make black pigment. True-breeding brown rabbitswere crossed to albinos, which were also BB. F1 rabbits, which were all black, were crossed to the doublerecessive (bb cc). The progeny obtained were 34black, 66 brown, and 100 albino.a. What phenotypic proportions would have beenexpected if the b and c loci were unlinked?b. How far apart are the two loci?
. A snapdragon plant that bred true for white petals wascrossed with a plant that bred true for purple petals, andall the F1 had white petals. The F1 was selfed. Among theF2, three phenotypes were observed in the followingnumbers:white 240solid purple 61spotted purple 19Total 320a. Propose an explanation for these results, showinggenotypes of all generations (make up and explain yoursymbols).b. A white F2 plant was crossed with a solid purple F2plant, and the progeny werewhite 50%solid purple 25%spotted purple 25%What were the genotypes of the F2 plants crossed?
There is an autosomal gene in cats that controls whether or not they also have white patches of fur (W = white spots, w= no white spots). Considering this along with the Orange gene (O0 and OB) from the previous question, and how it works, how many possible genotypes (using these 4 alleles) are there for cats that are orange, black, and white (in other words, calico)?
. As we learned in this chapter, the white mutation ofDrosophila studied by Thomas Hunt Morgan isX-linked and recessive to wild type. When truebreeding white-eyed males carrying this mutationwere crossed with true-breeding purple-eyed females,all the F1 progeny had wild-type (red) eyes. When theF1 progeny were intercrossed, the F2 progeny emergedin the ratio 3/8 wild-type females: 1/4 white-eyedmales: 3/16 wild-type males: 1/8 purple-eyed females:1/16 purple-eyed males.a. Formulate a hypothesis to explain the inheritanceof these eye colors.b. Predict the F1 and F2 progeny if the parental crosswas reversed (that is, if the parental cross wasbetween true-breeding white-eyed females andtrue-breeding purple-eyed males).
. In the tiny model plant Arabidopsis, the recessive allele hyg confers seed resistance to the drug hygromycin, and her, a recessive allele of a different gene, confers seed resistance to herbicide. A plant that was homozygous hyg/hyg ⋅ her/her was crossed with wild type, and the F1 was selfed. Seeds resulting from the F1 self were placed on petri dishes containing hygromycin and herbicide.a. If the two genes are unlinked, what percentage of seeds are expected to grow?b. In fact, 13 percent of the seeds grew. Does this percentage support the hypothesis of no linkage? Explain. If not, calculate the number of map units between the loci.c. Under your hypothesis, if the F1 is testcrossed, what proportion of seeds will grow on the medium containing hygromycin and herbicide?
One of the coat color patterns in cats is determined by a pair of codominant alleles located onthe X The allele (B) produces black color, and its allele (B') produces yellow color.The heterozygote (BB') has a tortoise-shell pattern. Determine the genotypic and phenotypic frequencies expected from a cross between a black male and yellow female. Determine the genotypic and phenotypic frequencies expected from a cross between a black male and tortoise-shell female.
The dominant allele H reduces the number of bodybristles in fruit flies, giving rise to a hairless phenotype. In the homozygous condition, H is lethal. Thedominant allele S has no effect on bristle number except in the presence of H, in which case a single S allele suppresses the hairless phenotype, thus restoringthe bristles. However, S is also lethal in homozygotes.a. What ratio of flies with normal bristles to hairlessindividuals would we find in the live progeny of across between two normal flies both carrying theH allele in the suppressed condition?b. When the hairless progeny of the previous cross arecrossed with one of the parental normal flies frompart (a) (meaning a fly that carries H in the suppressed condition), what phenotypic ratio wouldyou expect to find among their live progeny?
The non-wild-type alleles are k (clipped wings), l (long tail), and m (magical powers). The parental stocks are homozygous doubly recessive flies of genotype k+/k+ · l/l · m/m and homozygous singly recessive flies of genotype k/k · l+/l+ · m+/m+. From this cross, triply heterozygous progeny of genotype k/k+ · l/l+ · m/m+ are obtained, and females of this genotype are testcrossed to triple recessives of genotype k/k · l/l · m/m. The genotypes determining the eight progeny types from this testcross (CROSS C) are shown here with their numbers, out of a total sample of 1,572 flies. What is the correct order of the genes? What number of progeny for each genotype would you predict in the case of independent assortment? What is a non-recombinant genotype? What is a genotype of a single crossover? What is a genotype of a double crossover? What is the longest distance between any of the two dragon alleles above? What is the shortest distance between any of the two dragon alleles above?…
In a cross involving independently assorting loci, the cross BBYY x bbyy (assume "B" and "Y" are dominant to "b" and "y," respectively) will yield all dihybrid offspring. When those offpspring are crossed, the result is usually a 9:3:3:1 phenotypic ratio. But what if the locus "B" and "Y" did not sort independently (i.e., they were linked, at least partially)? It would mean "BY" and "by" are now more likely. How would this specifically affect the expected 9:3:3:1 ratio? To be more clear, which one of these numbers in the ratio (9, 3, 3, and/or 1) would be expected to decrease? Question 3 options: the 9 would decrease both 3's would decrease one of the 3's would decrease the 1 would decrease