normal levels of dystrophin in muscle. Also, the dominant allele F at an autosomal locus is responsible for polydactyly (the presence of extra digits on each foot) and the expression of its recessive counterpart f results in the absence of polydactyly (having the 5 digits per front paw, and 4 digits per hind paw). A healthy female cat that has polydactyly gives birth to a male kitten with the normal dition, this male kitten has muscular dystrophy. What can you conclude : Unknown alleles should be indicated with a "?" r?

normal levels of dystrophin in muscle. Also, the dominant allele F at an autosomal locus is responsible for polydactyly (the presence of extra digits on each foot) and the expression of its recessive counterpart f results in the absence of polydactyly (having the 5 digits per front paw, and 4 digits per hind paw). A healthy female cat that has polydactyly gives birth to a male kitten with the normal dition, this male kitten has muscular dystrophy. What can you conclude : Unknown alleles should be indicated with a "?" r?

Human Biology (MindTap Course List)

11th Edition

ISBN:9781305112100

Author:Cecie Starr, Beverly McMillan

Publisher:Cecie Starr, Beverly McMillan

Chapter19: Introduction To Genetics

Section: Chapter Questions

Problem 9CT

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In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant W form (WW or Ww), this gene blocks the formation of fur color and all the offspring are white; ww individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross AaWw Aa Ww?
BIU A- == 三E 12 - Match each of the following examples to the appropriate type of non-Mendelian inheritance. 1. A homozygous recessive genotype for the gene that encodes phenylalanine hydroxylase (which breaks down the amino acid phenylalanine) causes lighter skin color, a musty odor, differences in intellectual development, and seizures. 2. In pea plants, alleles of Gene W control flower color, with the dominant allele (W) leading to purple flower.color, and the recessive allele (w) leading to white flower color. Usually, a genotype of WW or Ww leads to purple flowers. However, when Gene C is homozygous recessive, WW or Ww plants always have white flowers. 3. In mallard ducks, feather coloring is controlled by Gene F. A dominant allele (F) leads to green head feathers, while a recessive allele (f) leads to brown head feathers. In male mallards, inheritance of one or more F alleles always leads to the green head feather trait. But female mallards always have brown head feathers,…
4. In humans, cystic fibrosis and albinism are both inherited as independently assorting autosomal recessive traits. Use fto represent the recessive allele for cystic fibrosis and Ffor the allele for the unaffected condition. Use A to represent the dominant allele for normal melanin man is albino and heterozygous for cystic fibrosis. His wife has the unaffected phenotype for both traits, although her father had cystic fibrosis and a for the recessive allele for albinism. and her mother is albino. 4a. What is the genotype of the wife? 4b. What is the probability that a child of this couple is albino and has cystic fibrosis?
3) In the snail Capaea nemoralis, an autosomal allele causing a banded shell (B®) is recessive to the allele for an unbanded shell (B°). Genes at a different locus determine the background color of the shell; here yellow (C') 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 test cross). a. What will the results of the testcross be in the loci that control banding and color ARE linked with NO crossing over.
12. please match the following terms
1 pts The pedigree below shows the expression of Huntington's disease in a family. Huntington's disease is a fatal genetic disorder that causes the progressive breakdown of nerve cells in the brain. It deteriorates a person's physical and mental abilities usually during their prime working years and has no cure. Huntington's is inherited as a dominant allele (H). Examine the pedigree below and determine the genotype for individual "2" DODO0O 2. O HH O Hh О h O either HH or Hh
6. A person is simultaneously heterozygous for two autosomal genetic traits. One is a recessive condition foralbinism (alleles A and a); this albinism gene is foundnear the centromere on the long arm of an acrocentricautosome. The other trait is the dominantly inheritedHuntington disease (alleles HD and HD+). TheHuntington gene is located near the telomere of oneof the arms of a metacentric autosome. Draw all copies of the two relevant chromosomes in this person asthey would appear during metaphase of (a) mitosis,(b) meiosis I, and (c) meiosis II. In each figure, labelthe location on every chromatid of the alleles forthese two genes, assuming that no recombinationtakes place.
3) In the snail Capaea nemoralis, an autosomal allele causing a banded shell (B®) is recessive to the allele for an unbanded shell (B°). Genes at a different locus determine the background color of the shell; here yellow (C") 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 test cross). b. What will the results be of the testcross be if the loci assort independently?
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…
. In mice, the dominant allele Gs of the X-linked geneGreasy produces shiny fur, while the recessive wildtype Gs+ allele determines normal fur. The dominantallele Bhd of the X-linked Broadhead gene causesskeletal abnormalities including broad heads andsnouts, while the recessive wild-type Bhd+ alleleyields normal skeletons. Female mice heterozygousfor the two alleles of both genes were mated withwild-type males. Among 100 male progeny of thiscross, 49 had shiny fur, 48 had skeletal abnormalities,2 had shiny fur and skeletal abnormalities, and 1 waswild type.a. Diagram the cross described and calculate thedistance between the two genes.b. What would have been the results if you hadcounted 100 female progeny of the cross?
A recessive epistasis cross gives in the F2 a 9:3:4 phenotypic ratio. What is the expected phenotypic ratio when the F1 (BbEe) is crossed with the double recessive homozygote (bbee)? Keeping in mind ee masks the expression of B.
5A. In shorthorn cattle, the polled condition is dominant over horned. Also, the heterozygous condition (RW) of red coat (R) and white coat (W) is roan, an example of codominance. (The roan coat contains both red and white fur because neither allele is dominant over the other). If a homozygous polled, white male is bred to a horned, red female, what will be the appearance of the F1 generation? a. Possible genotypes for Polled condition: Horned: b. Possible genotypes for Red coat:White Coat: Roan Coat: c. Male genotype Male possible gametes: Female genotypeFemale possible gametes: e. f. Genotypic Ratio: g. Phenotypic Ratio: