Concept explainers
In mice, a trait called snubnose is recessive to a wild-type nose, a trait called pintail is dominant to a normal tail, and a trait called jerker (a defect in motor skills) is recessive to a normal gait. Jerker mice with a snubnose and a pintail were crossed to normal mice, and then the
560 jerker, snubnose, pintail
548 normal gait, normal nose, normal tail
102 jerker, snubnose, normal tail
104 normal gait, normal nose, pintail
77 jerker, normal nose, normal tail
71 normal gait, snubnose, pintail
11 jerker, normal nose, pintail
9 normal gait, snubnose, normal tail
Construct a genetic map that shows the order of these genes and distances between them.
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Chapter 6 Solutions
Genetics: Analysis and Principles
- In snapdragons, red flower color (R) is incompletely dominant over white flower color (r), so that the heterozygote has pink flowers. A red snapdragon is crossed with a white snapdragon, and the F1 are intercrossed to produce F2. What is the genotype and phenotypes of the F1 and F2, along with their expected proportions? If the F1 are backcrossed to the white parent, what will be the expected genotype and phenotype of the offspring? If the F1 are backcrossed to the red parent and the resulting progeny are intercrossed, what is the expected proportions of genotypes and phenotypes in the progeny?arrow_forwardIn pigs, let’s assume that a straight tail (T) is dominant to a curly tail (t). Let us also assume that brown color (B) is dominant to spotted color (b). Find the phenotypic ratio of the offspring of each for each of the three crosses below. In order to receive credit, you must show your work. TTbb x Ttbb Ttbb x TtBb TtBB x ttBbarrow_forwardIn rabbits, the dominant allele B causes black fur and the recessive allele b causes brown fur; for an independently assorting gene, the dominant allele R causes long fur and the recessive allele r (for rex) causes short fur. A homozy-gous rabbit with long, black fur is crossed with a rabbit with short, brown fur, and the offspring are intercrossed. In the F2, what proportion of the rabbits with long, black fur will be homozygous for both genes? Is there a faster way to do this than to write every genotype out?arrow_forward
- Guinea pigs have coats that are either rough or smooth. Rough coats are dominant, so they are represented by (R), and smooth coats are recessive, so they are represented by (r). If we crossed a homozygous rough-coated guinea pig with a smooth-coated guinea pig, what are the phenotypic and genotypic ratios?arrow_forwardIn cats, pointed ears (E) are dominant to round ears (e), while black fur (F) is dominant to red (f). A cat with round ears and red fur is crossed with a completely heterozygous black cat with pointed ears.arrow_forwardIn rabbits, black hair is due to a dominant allele B and brown hair to its recessive allele b. Short hair (H) is dominant to long hair (h). In a cross between a homozygous black, longhaired rabbit and a brown, homozygous shorthaired one, what would the F1 generation look like? If you did not know the genotype of an F1 rabbit, you could determine its genotype by a test cross in which it is crossed with an animal with which phenotype AND genotype If you carried out this test cross, what phenotypes and in what ratio would you expect? What phenotypes in what ratio would be expected in the F2 generation?arrow_forward
- In mice, a trait called snubnose is recessive to a wild-type nose, atrait called pintail is dominant to a normal tail, and a trait calledjerker (a defect in motor skills) is recessive to a normal gait. Jerkermice with a snubnose and a pintail were crossed to normal mice,and then the F1 mice were crossed to jerker mice that have a snubnoseand a normal tail. The outcome of this cross was as follows: 560 jerker, snubnose, pintail548 normal gait, normal nose, normal tail102 jerker, snubnose, normal tail104 normal gait, normal nose, pintail77 jerker, normal nose, normal tail71 normal gait, snubnose, pintail11 jerker, normal nose, pintail9 normal gait, snubnose, normal tail Construct a genetic map that shows the order of these genes anddistances between them.arrow_forwardCoat color in mice is influenced by two genes, one for color (A) and one for the amount of pigment production (C). Mice with the wild type agouti coat color have a yellowish to brownish color. Mice also have a gene that determines the amount of pigment the hair produces. Mouse Coat Phenotype and Genotype Correlation Agouti coat AA, Aa Solid gray/black coat aa Pigment production CC, Cc Albinism cc Multiple crosses were made between male and female mice that were each heterozygous for both traits (AaCc). The data table shows the number of mice of each coat type. Calculate the average F1 generation coat color to answer the question. Coat Color Number of Mice Born in Each Trial Cross Mean Number of Mice 1 2 3 4 5 6 7 8 9 10 Agouti coat 11 8 9 9 10 8 10 7 10 9 Solid gray/black coat 5 2 1 3 4 3 4 3 3 2 Albinism 3 3 6 4 4 3 3 7 4 3 Which…arrow_forwardWhen 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?arrow_forward
- In the case of carnation flower, what do you think a flower carnation would look like if the red and white phenotypes were codominant instead?arrow_forwardIn watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. a. What are the phenotypic ratios in the F2? b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? c. If an F1 plant is backcrossed with the sweet, unspotted parent, what phenotypes and proportions are expected in the offspring?arrow_forwardIn Drosophila, white eyes (w) are recessive to red eyes (w+) at one locus and black body (b) is recessive to gray body (b+). A homozygous white eyes, gray bodied female is crossed with a homozygous red eyes, black bodied male to produce the F1 progeny. The F1 progeny are testcrossed and produce the following progeny: White eyes, black body: 212 White eyes, gray body: 288 Red eyes, black body: 308 Red eyes, gray body: 192 Does the evidence indicate that w and b loci are linked? Explain why or why not? If they are linked, what is the map distance between the two loci? If they are not linked, what is the map distance between the two loci? If they are linked, are the allels in the F1 in coupling or repulsion? How do you know? Draw the genotypes of all individuals described in the problem (original parents, F1, testcross, and F2 progeny) using the appropriate notation.arrow_forward
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