stion 6 of 18Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver.She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to determinewhether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous forquiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F, females in atestcross. She obtains the flies from this testcross.Phenotype Number of fliesvg* qu+230vg qu224vg qutvg* qu9799Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X², for thishypothesis. Provide the X2 to one decimal place.X2 =Does the X value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not?the partial table of critical values for X2 calculations to test this hypothesis. vg* qut230vg qu224vg qu+97vg+nb99Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X², for thishypothesis. Provide the X2 to one decimal place.Does the X value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not? Usethe partial table of critical values for X calculations to test this hypothesis.No, the X4 value indicates that there are too many phenotypes for independent assortment.Yes, the X2 value indicates that the observed and expected number of progeny are equal in number.No, the X2 value indicates that the observed progeny are significantly different from what would be expected withindependent assortment of the two genes.Yes, the X2 value indicates that the genes vestigial and quiver assort independently.DELL

Name: stion 6 of 18Suppose that a geneticist discovers a new mutation in Dr
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Description: stion 6 of 18Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver.She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to deter

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vg* qu* 230 vg qu 224 vg qu+ 97 vg* qu 99 Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X, for this hypothesis. Provide the X to one decimal place. x2 = Does the X value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not? Use the partial table of critical values for X2 calculations to test this hypothesis. No, the X value indicates that there are too many phenotypes for independent assortment. Yes, the X2 value indicates that the observed and expected number of progeny are equal in number. independent assortment of the two genes. Yes, the X2 value indicates that the genes vestigial and quiver assort independently. No, the X2 value indicates that the observed progeny are significantly different from what would be expected with

vg* qu* 230 vg qu 224 vg qu+ 97 vg* qu 99 Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X, for this hypothesis. Provide the X to one decimal place. x2 = Does the X value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not? Use the partial table of critical values for X2 calculations to test this hypothesis. No, the X value indicates that there are too many phenotypes for independent assortment. Yes, the X2 value indicates that the observed and expected number of progeny are equal in number. independent assortment of the two genes. Yes, the X2 value indicates that the genes vestigial and quiver assort independently. No, the X2 value indicates that the observed progeny are significantly different from what would be expected with

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

Chapter13: Genes, Chromosomes, And Human Genetics

Section: Chapter Questions

Problem 8TYK: Another gene in Drosophila determines wing length. The dominant wild-type allele of this gene...

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A Drosophila male is heterozygous for a reciprocaltranslocation between an autosome and the Y chromosome. The part of the autosome now present onthe Y chromosome contains the dominant mutationLyra (shortened wings); the other (normal) copyof the same autosome is Lyra+. This male is nowmated with a true-breeding, wild-type female. Whatkinds of progeny would be obtained, and in whatproportions?
The Drosophila gene Sex lethal (Sxl) is deserving of itsname. Certain alleles have no effect on XY animals butcause XX animals to die early in development. Other alleles have no effect on XX animals but cause XY animals to die early in development. Thus, some Sxl allelesare lethal to females, while others are lethal to males.a. Would you expect a null mutation in Sxl to causelethality in males or in females? b. Why do Sxl alleles of either type cause lethality ina specific sex?The gene transformer (tra) gets its name from sexualtransformation, as some tra alleles can change XXanimals into morphological males, while other traalleles can change XY animals into morphologicalfemales.c. Which of these sex transformations would becaused by null alleles of tra and which would becaused by constitutively active alleles of tra?d. In contrast with Sxl, null tra mutations do notcause lethality either in XX or in XY animals.However, the Sxl protein regulates the productionof the Tra protein. Why…
A Drosophila embryo dies during early embryogenesis due to arecessive maternal effect allele called bicoid−. The wild-type alleleis designated bicoid+. What are the genotypes and phenotypes ofthe embryo’s mother and maternal grandparents?
In Drosophila, the autosomal recessive brown eyecolor mutation displays interactions with both theX-linked recessive vermilion mutation and the autosomal recessive scarlet mutation. Flies homozygous for brown and simultaneously hemizygous orhomozygous for vermilion have white eyes. Fliessimultaneously homozygous for both the brown andscarlet mutations also have white eyes. Predict theF1 and F2 progeny of crossing the following truebreeding parents:a. vermilion females × brown malesb. brown females × vermilion malesc. scarlet females × brown malesd. brown females × scarlet males
A new recessive mutant allele doesn’t show pseudodominance with any of the deletions that span Drosophilachromosome 2. What might be the explanation?
The maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). How is it possible for a male homozygous for the mutationto exist?
The Drosophila chromosome 4 is extremely small;virtually no recombination occurs between genes onthis chromosome. You have available three differentlymarked chromosome 4s: one has a recessive allele ofthe gene eyeless (ey), causing very small eyes; one hasa recessive allele of the cubitus interruptus (ci) gene,which causes disruptions in the veins on the wings;and the third carries recessive alleles of both genes.Drosophila adults can survive with two or three, butnot with one or four, copies of chromosome 4.a. How could you use these three chromosomes tofind Drosophila mutants with defective meiosescausing an elevated rate of nondisjunction?b. Would your technique allow you to discriminatenondisjunction occurring during the first meioticdivision from nondisjunction occurring during thesecond meiotic division?
The Drosophila chromosome 4 is extremely small;virtually no recombination occurs between genes onthis chromosome. You have available three differentlymarked chromosome 4s: one has a recessive allele ofthe gene eyeless (ey), causing very small eyes; one hasa recessive allele of the cubitus interruptus (ci) gene,which causes disruptions in the veins on the wings;and the third carries recessive alleles of both genes.Drosophila adults can survive with two or three, butnot with one or four, copies of chromosome 4.a. How could you use these three chromosomes tofind Drosophila mutants with defective meiosescausing an elevated rate of nondisjunction?b. Would your technique allow you to discriminatenondisjunction occurring during the first meioticdivision from nondisjunction occurring during thesecond meiotic division?c. What progeny types would you expect if a flyrecognizably formed from a gamete produced bynondisjunction were testcrossed to a fly homozygous for a chromosome 4 carrying both ey…
A maternal effect gene in Drosophila, called torso, is found as afunctional allele (torso+) and a nonfunctional, recessive allele(torso−) that prevents the correct development of anterior- andposterior-most structures. A wild-type male (torso+ torso+) is crossedto a female of unknown genotype. This mating produces 100% larvathat are missing their anterior- and posterior-most structures andtherefore die during early development. What is the genotype andphenotype of the female fly in this cross? What are the genotypesand phenotypes of the female fly’s parents?
The recessive, X-linked z1mutation of the Drosophilagene zeste (z) can produce a yellow (zeste) eye coloronly in flies that have two or more copies of the wildtype white (w) gene. Using this property, tandem duplications of the w+ gene called w+Rwere identified.Males with the genotype y+ z1w+R spl+ / Y thus havezeste eyes. These males were crossed to females withthe genotype y z1 w+R spl / y+ z1 w+R spl+. (These fourgenes are closely linked on the X chromosome, in theorder given in the genotype, with the centromere tothe right of all these genes: y = yellow bodies; y+ =tan bodies; spl = split bristles; spl+ = normal bristles.) Out of 81,540 male progeny of these females,the following exceptions were found:Class A 2430 yellow bodies, zeste eyes, wild-type bristlesClass B 2394 tan bodies, zeste eyes, split bristlesClass C 23 yellow bodies, wild-type eyes, wild-type bristlesClass D 22 tan bodies, wild-type eyes, split bristlesa. What were the phenotypes of the remainder of the81,540 males…
Male Drosophila expressing the autosomal recessivemutations sc (scute), ec (echinus), cv (crossveinless),and b (black) were crossed to phenotypically wildtype females, and the 3288 progeny listed wereobtained. (Only mutant traits are noted.)653 black, scute, echinus, crossveinless670 scute, echinus, crossveinless675 wild type655 black71 black, scute73 scute73 black, echinus, crossveinless74 echinus, crossveinless87 black, scute, echinus84 scute, echinus86 black, crossveinless83 crossveinless1 black, scute, crossveinless1 scute, crossveinless1 black, echinus1 echinusa. Diagram the genotype of the female parent.b. Map these loci.c. Do the data provide evidence of interference?Justify your answer with numbers.
The following list of four Drosophila mutations indicates the symbol for the mutation, the name of thegene, and the mutant phenotype:Allele symbol Gene name Mutant phenotypedwp dwarp small body, warped wingsrmp rumpled deranged bristlespld pallid pale wingsrv raven dark eyes and bodiesYou perform the following crosses with the indicatedresults:Cross #1: dwarp, rumpled females × pallid, raven males→ dwarp, rumpled males and wild-type femalesCross #2: pallid, raven females × dwarp, rumpled males→ pallid, raven males and wild-type femalesF1 females from cross #1 were crossed to males froma true-breeding dwarp rumpled pallid raven stock.The 1000 progeny obtained were as follows:pallid 3pallid, raven 428pallid, raven, rumpled 48pallid, rumpled 23dwarp, raven 22dwarp, raven, rumpled 2dwarp, rumpled 427dwarp 47Indicate the best map for these four genes, includingall relevant data. Calculate interference values whereappropriate.