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)

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)

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Description: 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 p

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter6: Cytogenetics: Karyotypes And Chromosome Aberrations

Section: Chapter Questions

Problem 2QP: Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the...

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In cats, black fur (B) is dominant over brown fur (b). Also, striped fur (A) is dominant over solid colored fur (a). What is the phenotype of a cat with the genotype Bbaa? What combinations of gametes could the Bbaa cat make? If the Bbaa cat mated with a cat that was heterozygous for both the color and pattern genes, what fraction of their kittens would be likely to have solid brown fur? Hint: draw your square on a separate paper to figure it out.
In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?
In rabbits, white coat color (CW) and black coat color (CB) are codominant, and both of these alleles are dominant over albino (c); heterozygotes (CWCB) are spotted. Consider a cross between a heterozygous black-coated rabbit and a homozygous white-coated rabbit.
Another recessive mutation in Drosophila, ebony (e), is on anautosome (chromosome 3) and causes darkening of the bodycompared with wild-type flies. What phenotypic F1 and F2 maleand female ratios will result if a scalloped-winged female withnormal body color is crossed with a normal-winged ebony male?Work this problem by both the Punnett square method and theforked-line method.
IN DROSOPHILA, AN X-LINKED RECESSIVE MUTATION, Xm CAUSES MINIATURE WINGS. LIST THE F₂ PHENOTYPIC RATIOS IF: A MINIATURE-WINGED FEMALE IS CROSSED WITH A NORMAL MALE AND A MINIATURE-WINGED MALE IS ● ● CROSSED WITH A NORMAL FEMALE. WHAT WOULD THE PHENOTYPIC RATIO FROM (A) BE IF THE MINIATURE- WINGED GENE WERE AUTOSOMAL? ASSUME IN ALL CASES THAT THE P1 INDIVIDUALS ARE TRUE-BREEDING.
In sheep, lustrous fleece results from an allele (L) that is dominant over an allele (l) for normal fleece. A ewe (adult female) with lustrous fleece is mated with a ram (adult male) with normal fleece. The ewe then gives birth to a single lamb with normal fleece. From this single offspring, is it possible to determine the genotypes of the two parents? If so, what are their genotypes? If not, why not?
Suppose a man is heterozygous for heterochromia, an autosomal dominant disorder which causes two different‑colored eyes in an individual, produced 25‑offspring with his normal‑eyed wife. Of their children, 16 were heterochromatic and 9 were‑normal. Calculate the chi‑square value for this observation.
In cattle, the gene for hornless (H) is dominant to the gene for horned (h), the gene for black (B) is dominant to that of red (b), and the gene for white face, or Hereford spotting, (S) is dominant to that for solid color (s). A cow (female) that is heterozygous for all traits is inseminated by a bull (male) of the genotype bbhhSs. What is the probability of obtaining a calf that is a black, hornless bull with Hereford spotting?
In mice, the wild-type coat color, agouti (AA) [left mouse], is dominant to solid-colored, black fur (aa) [middle mouse]. However, a separate gene (C) is necessary for pigment production. A mouse that is homozygous for a recessive c allele at this locus is unable to produce pigment and is albino [right mouse] regardless of the allele present at locus A. Thus, the following genotypes are listed with their associated phenotypes: ~ A/A; C/C or A/A; C/c or A/a; C/C or A/a; C/c ==> "agouti" ~ a/a; C/C or a/a; C/c ==> "black" ~ A/A; c/c or A/a; c/c or a/a; c/c or a/a; c/c ==> "albino" This is an example of recessive epistasis, in which the recessive c allele "stands upon" the possible genotypes for the A gene (A/A, A/a, or a/a). If two agouti mice with the A/a; C/c genotype are mated, what is the expected phenotypic ratio in their offspring? A.9 albino, 4 agouti, 3 black B. 9 agouti, 4 albino, 3 black C.9 black, 4 albino, 3 agouti D.9 agouti, 4 black, 3 albino
In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). a. What will the results of the testcross be if the loci that control banding and color are linked with no crossing over? b. What will the results of the testcross be if the loci assort independently? c. What will the results of the testcross be if the loci are linked and 20 m.u. apart?
In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). Q.What will the results of the testcross be if the loci are linked and 20 m.u. apart?
In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). Q.What will the results of the testcross be if the loci assort independently?