can you do question 8 The next two questions go together.Suppose the tail length of two populations of lizards is controlled by one gene. To determine themode of inheritance, a true-breeding short-tailed female is crossed with a true-breeding long-tailedmale. Then, siblings from the F1 are crossed and the number of short- and long-tailed animals arecounted to determine the phenotype ratio.1. What phenotypic ratio do you expect in the F2 generation if tail length is controlled by anautosomal gene, and long tails are dominant to short tails?1 long: 1 shortmales 1 long: 1 short, females all short3 long: 1 shortO 1 long: 3 shortQuestion 82. What phenotypic ratio do you expect in the F2 generation if tail length is controlled by a sex-linked gene, and long tails are dominant to short tails? Assume a chromosomal sex determinationsystem as in mammals, with XX females and XY males.1 long: 3 shortmales 1 long: 1 short, females all short1 long: 1 short3 long: 1 short

The alleles are the alternative forms of a gene that are located on the same locus of a homologous…

Question 8 2. What phenotypic ratio do you expect in the F2 generation if tail length is controlled by a sex- linked gene, and long tails are dominant to short tails? Assume a chromosomal sex determination system as in mammals, with XX females and XY males. O 1long: 3 short males 1 long: 1 short, females all short O 1 long: 1 short 3 long: 1 short

Question 8 2. What phenotypic ratio do you expect in the F2 generation if tail length is controlled by a sex- linked gene, and long tails are dominant to short tails? Assume a chromosomal sex determination system as in mammals, with XX females and XY males. O 1long: 3 short males 1 long: 1 short, females all short O 1 long: 1 short 3 long: 1 short

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 11QP: Crossing Pea Plants: Mendels Study of Single Traits Stem length in pea plants is controlled by a...

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The next two questions go together. Suppose the tail length of two populations of lizards is controlled by one gene. To determine the mode of inheritance, a true-breeding short-tailed female is crossed with a true-breeding long-tailed male. Then, siblings from the F1 are crossed and the number of short- and long-tailed animals are counted to determine the phenotype ratio. 1. What phenotypic ratio do you expect in the F2 generation if tail length is controlled by an autosomal gene, and long tails are dominant to short tails? 1 long: 1 short males 1 long: 1 short, females all short O 3 long: 1 short O 1 long: 3 short
The next two questions go together. Suppose the tail length of two populations of lizards is controlled by one gene. To determine the mode of inheritance, a true-breeding short-tailed female is crossed with a true-breeding long-tailed male. Then, siblings from the F1 are crossed and the number of short- and long-tailed animals are counted to determine the phenotype ratio. 1. What phenotypic ratio do you expect in the F2 generation if tail length is controlled by an autosomal gene, and long tails are dominant to short tails? 1 long: 1 short males 1 long: 1 short, females all short O 3 long: 1 short O 1 long: 3 short 2. What phenotypic ratio do you expect in the F2 generation if tail length is controlled by a sex- linked gene, and long tails are dominant to short tails? Assume a chromosomal sex determination system as in mammals, with XX females and XY males. 1 long: 3 short males 1 long: 1 short, females all short O 1 long: 1 short 3 long: 1 short
. Many kinds of wild animals have the agouti coloring pattern, in which each hair has a yellow band around it. a. Black mice and other black animals do not have the yellow band; each of their hairs is all black. This absence of wild agouti pattern is called nonagouti. When mice of a true-breeding agouti line are crossed with nonagoutis, the F1 is all agouti and the F2 has a 3 :1 ratio of agoutis to nonagoutis. Diagram this cross, letting A represent the allele responsible for the agouti phenotype and a, nonagouti. Show the phenotypes and genotypes of the parents, their gametes, the F1, their gametes, and the F2. b. Another inherited color deviation in mice substitutes brown for the black color in the wild-type hair. Such brown-agouti mice are called cinnamons. When wildtype mice are crossed with cinnamons, all of the F1 are wild type and the F2 has a 3 :1 ratio of wild type to cinnamon. Diagram this cross as in part a, letting B stand for the wild-type black allele and b stand for the…
O e. Parent 2: Parent 1: Parent 2: ¡Ai QUESTION 9 Bi QUESTION 8 Let's assume that, in dragons, red scales (B) are dominant to green scales (b), and long tongues (S) are dominant to short tongues (s). The genes that determine these characteristics assort independently. A homozygous red, long-tongued dragon is crossed with a homozygous green, short-tongued dragon. If an F1 dragon is crossed to a homozygous green and homozygous short-tongued dragon, what phenotypes and proportions are expected in the offspring? O a. 100% green and long-tongued O b. ½ green and short-tongued & ½ red and long-tongued O c. ½ red and short-tongued & ½ green and long-tongued O d.% red and long-tongued, % red and short-tongued, ½ green and long-tongued, % green and short-tongued O e. 9/16 red, long-tongued, 3/16 green, long-tongued, 3/16 red, short-tongued, 1/16 green, short-tongued Save and Submit to save and submit. Click Save All Answers to save all answers. 000 MacBook Air
Inheritance of multiple genes: 8. A pure-bred grey cat mates with a pure-bred chocolate cat: a. What are the possible phenotypes of their kittens? b. For each of these phenotypes, what is their probability? black [Choose ] [Choose ] 0.75 gray 0.5 0 1.0 0.25 chocolate [Choose ] lilac [Choose ] cinnamon [Choose ] fawn [Choose ] orange [Choose ] cream [Choose ]
One Trait (Monohybrid) Crosses-Complete Dominance 1. In peas, inflated pod is dominant over wrinkled. If a homozygous inflated pod plant is crossed with a homozygous wrinkled pod plant, what will be the F₂ phenotypic ratio? 2. A tall pea plant crossed with a dwarf pea plant produces offspring of which about half are tall and half are dwarf. What are the genotypes of: a. the tall parent b. the dwarf parent 3. What does "test cross" or "backcross" mean?
Paragraph Styles DIHYBRID GENETICS PROBLEMS Name In garden peas, tall (T) vine is dominant over short (t) vine, and round pea pod (R) is dominant over wrinkled pea pod (r). Cross a homozygous tall, round pea plant with a homozygous short, wrinkled pea plant. Show the genotypes and phenotypes of the F1 offspring, and the probability of each. 1. Genotype of the homozygous tall, round pea plant Genotype of the homozygous short, wrinkled pea plant. Genotypes Phenotypes acer
You breed two black dogs. It results in 3 offspring, the first which is black, the second which is brown, and the third which is white. Give two genetic explanations as to what could cause their fur color. Make sure each explanation includes two genes and follows along with the provided data. Also, explain the genotype of each color of dog and a ratio of the F2 phenotypes you would expect to see from a dihybrid cross. After you're finished, pick one of the explanations and create a hypothesis for the function of each gene which would explain the molecular basis of the fur color Finally, for each of the two genes, explain how a specific functional difference between the two alleles leads to a difference in their functions
For number 15-24. In man, assume that spotted skin (S) is dominant over non-spotted skin (s) and that wooly hair (W) is dominant over non-wooly hair (w). Cross a heterozygous parent (SsWw) with a heterozygous parent (Ssww) Give genotypic and phenotypic ratios of offspring. 15-18. What is the probability of producing spotted skin with non-wooly hair? 8/16 4/16 3/16 6/16 8/16 A. B. C. D. E. 19-21. What is the probability of producing non-spotted skin with non-wooly hair? A 2/16 B. 4/16 C. 1/16 D. 6/16 22-24. What is the probability of producing non-spotted skin with wooly hair? A 3/16 В. 4/16 с. 1/16 D. 9/16 E. 8/16
for abor Monohybrid crossing the monohybrid and dihybrid crosses. 1. A baby with type Ab blood is born into a family where the mother has type B blood and father type O blood. (i) Determine whether the husband is the biological father of the baby. Explain your answer in (i) above. (ii) 2. Silky feathers in fowls are caused by an allele which is recessive to the allele for normal feathers. 200 birds were raised from a cross between two heterozygous individuals. Predict the number of the birds that would expected to be silky and those expected to be normal. If about half of the offspring turned out to be silky, comment on the nature of the normal. If you had normal feather birds, propose the easiest way to determine whether its homozygous or heterozygous. 3. In dogs, black fur is dominant to white fur. Determine the possible offspring from crosses between: Homozygous black and white. (1) Heterozygous black and heterozygous black. (iii) Detach this page from your manual and att ch it to…
3. Short-tailed pup distribution. In mice, the dominant T allele results in a short tail. Homozygous T/T genotype is lethal, which means mouse embryos with this genotype die before they are born. Homozygous t/t is normal, with normal tail. A cross between two short-tailed mice produces a litter of 5 pups. a). What are the genotype(s) of the two short-tailed mice used in the cross? b). What are the possible genotypes and phenotypes of the 5 pups? What are their perspective ratio?
Kernel color in wheat Red kernel color in wheat results from the interaction between two dominant alleles. With only one dominant allele, the phenotype is brown while in the absence of any dominant allele, the phenotype is white. Suppose that plants of a variety that is true breeding for red kernels are crossed with plants true breeding for white kernels. What specific type of interaction can be observed? A. duplicate dominant genes B. duplicate genes with cumulative effects C. complementary genes D. dominant epistasis (case 2)