4. In Drosophila, three autosomal, recessive mutations are ebony body (e), sepia eye color (s),and hairless bristles (h). A female fly heterozygous at all loci was testcrossed with an ebony,sepia, hairless male; the progeny are shown below. Using the gene symbols stated åbove,answer the following questions.PhenotypeObserved # of offspringEbonysepia, hairlessebony, sepia5753344346 ralHairlessebony, hairlessSepiawild typeebony, sepia, hairlessTotal offspring119.89911000A. Give the genotypes of the flies in the testcross. Write the answer using the gene symbolsstated above and using proper notation.B. Deduce the linkage map showing all distances among the three genes.C. Calculate the interference. Show all work.

In genetics, crossing over occurs in a chromosome, during the prophase stage of meiosis. During this…

4. In Drosophila, three autosomal, recessive mutations are ebony body (e), sepia eye color (s), and hairless bristles (h). A female fly heterozygous at all loci was testcrossed with an ebony, sepia, hairless male; the progeny are shown below. Using the gene symbols stated åbove, answer the following questions. Phenotype Observed # of offspring Ebony sepia, hairless ebony, sepia Hairless ebony, hairless Sepia wild type ebony, sepia, hairless | Total offspring 57 53 344 346 11 6. 89 91 1000 A. Give the genotypes of the flies in the testcross. Write the answer using the gene symbols stated above and using proper notation.

4. In Drosophila, three autosomal, recessive mutations are ebony body (e), sepia eye color (s), and hairless bristles (h). A female fly heterozygous at all loci was testcrossed with an ebony, sepia, hairless male; the progeny are shown below. Using the gene symbols stated åbove, answer the following questions. Phenotype Observed # of offspring Ebony sepia, hairless ebony, sepia Hairless ebony, hairless Sepia wild type ebony, sepia, hairless | Total offspring 57 53 344 346 11 6. 89 91 1000 A. Give the genotypes of the flies in the testcross. Write the answer using the gene symbols stated above and using proper notation.

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 6TYK

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In addition to the two genes in problem 4, assume you now study a third independently assorting gene that has the alleles C and c. For each of the following genotypes, indicate what types of gametes will be produced: a. AA BB CC b. Aa BB Cc c. Aa BB cc d. Aa Bb Cc
Refer to problem 2. Assume a third gene has alleles C and c. For each genotype listed, what allele combinations will occur in gametes, assuming independent assortment? a. AABBCC b. AaBBcc c. AaBBCc d. AaBbCc
1.In Drosophila melanogaster body color is controlled by one gene while wing shape is controlled by a second gene. Gray body color is dominant to black body color, and normal wings are dominant to vestigial wings. Flies homozygous for gray body color and vestigial wings are crossed with flies homozygous for black body color and normal wings. A.compare the possible f2 generation genotypes and phenotypes and proportions if these two traits. B.How does your answer change if one of the original parents is homozygous for gray body color and normal wings while the other has black body color and vestigial wings?
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…
3) A. How are the following proteins involved in cell division: condensing, cohesin, kinetochore complex, tubulin? B. Why does phenotype not always reveal genotype? C. How can Mendel's postulates of segregation and independent assortment be explained by meiosis? D. Explain the genotypic and phenotypic ratios predicted when performing a monohybrid and a dihybrid cross of heterozygotes? E. How can you use the product rule and sum rule to calculate the probability of a particular genotype resulting from crossing parents of specific genotypes?
7. In rabbits, a locus involved in the control of coat colour may be occupied by any of fouralleles: Full colour (C), Sepia (ck), Cream (cd), or Albino (ca). A geneticist counted the numberof Full colour (C), Sepia (ck), Cream (cd) and Albino coat offspring resulting after crossesbetween Sepia (ck) and Cream (cd) coloured coated parents. The results were as follows.• Full colour (C): 152• Sepia (ck): 53• Cream (cd): 39• Albino (ca): 6Mendelian inheritance of this trait predicts that the ratio of Full colour (C) to Sepia (ck) to Cream(cd) to Albino (ca) should be 9:3:3:1. Do the experimental results support this mode ofinheritance?
1. A sex-linked recessive allele rg causes a red-green colorblindness in humans. A normal woman whose father was colorblind marries a colorblind man. A. Of all the children of these parents (sex unspecified), what proportion can be expected to have normal color vision?
7. In guinea pigs, black hair colour (B) is dominant and brown hair colour (b) is recessive. Long hair (L) is dominant and short hair (l) is recessive. Answer the following questions: (a) Diagram the cross: BbLl x BbLL (b) What are the phenotypes of the parent generation? (c) What are the genotypes and phenotypes of the F1 generation?
0. In Drosophila, the gene for cinnabar eye color is onchromosome 2, and the gene for scarlet eye color is onchromosome 3. A fly homozygous for both recessivecinnabar and scarlet alleles (cn/cn; st/st) is white-eyed.a. If male flies (containing chromosomes with thenormal gene order) heterozygous for cn and st allelesare crossed to white-eyed females homozygous forthe cn and st alleles, what are the expected phenotypes and their frequencies in the progeny?b. One unusual male heterozygous for cn and st alleles,when crossed to a white-eyed female, produced onlywild-type and white-eyed progeny. Explain the likelychromosomal constitution of this male.c. When the wild-type F1 females from the cross withthe unusual male were backcrossed to normal cn/cn;st/st males, the following results were obtained:wild type 45%cinnabar 5%scarlet 5%white 45%Diagram a genetic event at metaphase I that couldproduce the rare cinnabar or scarlet flies among theprogeny of the wild-type F1 females.
4. An albino corn snake is crossed with a normal-coloredcorn snake. The offspring are all normal-colored.When these first-generation progeny snakes arecrossed among themselves, they produce 32 normalcolored snakes and 10 albino snakes.a. How do you know that only a single gene is responsible for the color differences between these snakes? b. Which of these phenotypes is controlled by thedominant allele?c. A normal-colored female snake is involved in atestcross. This cross produces 10 normal-coloredand 11 albino offspring. What are the genotypes ofthe parents and the offspring?
2. Uniparental disomy is a rare phenomenon in whichonly one of the parents of a child with a recessivedisorder is a carrier for that trait; the other parent ishomozygous normal. By analyzing DNA polymorphisms, it is clear that the child received both mutantalleles from the carrier parent but did not receive anycopy of the gene from the other parent.a. Diagram at least two ways in which uniparentaldisomy could arise. (Hint: These mechanismsall require more than one error in cell division,explaining why uniparental disomy is so rare.)Is there any way to distinguish between thesemechanisms to explain any particular case ofuniparental disomy?
2. Uniparental disomy is a rare phenomenon in whichonly one of the parents of a child with a recessivedisorder is a carrier for that trait; the other parent ishomozygous normal. By analyzing DNA polymorphisms, it is clear that the child received both mutantalleles from the carrier parent but did not receive anycopy of the gene from the other parent.a. Diagram at least two ways in which uniparentaldisomy could arise. (Hint: These mechanismsall require more than one error in cell division,explaining why uniparental disomy is so rare.)Is there any way to distinguish between thesemechanisms to explain any particular case ofuniparental disomy?b. How might the phenomenon of uniparental disomyexplain rare cases in which girls are affected withrare X-linked recessive disorders but have unaffectedfathers, or other cases in which an X-linked recessive disorder is transmitted from father to son?c. If you were a human geneticist and believed oneof your patients had a disease syndrome caused…