Questions 1. List the phenotypes and expected proportions of the F, flies from Exercise II.Scenario 1: vestigial gene is on chromosome 2NonrecombinantDnenotyPeRecombinantDhenotyRe[Cht vgte] wild type[ cntvgte] E bonyCcn Vg e] Cinnabar, vegista lCentvget] vegiętealCinnabor, vegistal, ebonyScenario 2: vestigial gene is on chromosome 32. What are the main differences between the two scenarios regarding the F, flies?3. Which chromosome is the vestigial gene on? Then you would backcross the F, flies to the recessive flies, and observe the F, progeny numbers. Inorder to predict the progeny ratios, we need to consider two different scenarios: 1) vestigial gene ison chromosome 2 with cinnabar and 2) vestigial gene is on chromosome 3 with ebony. In order tohelp you get started, Figure 3.6 illustrates the P and F, generations for each scenario.If the vestigial gene is on chromosome 2 with cinnabar:a.P generationcn* vg*ete*cn vg*cn vgcn vgееbackcross the female F,flies with recessive parentF, generationcn* vg*e*ecn vgcn vgcn vgееFa?b. If the vestigial gene is on chromosome 3 with ebony:generationet vg*et vg*e vge vgcn*cn+ xcn cnbackcross the female F,flies with recessive parentF, generationet vg*e vge vge vgcn*cncn cnFigure 3.61. Using your whiteboard complete Figure 3.6 by listing the phenotypes and the expectedproportions of the F, flies for each scenario (you should have 8 phenotypic classes). Then transferthe information to #1 in the Data Sheet. Note: For the linked genes, you will not be able topredict exact progeny ratios, because recombination is a stochastic process. Rather you canonly state whether you would expect a lower or higher number of progeny, because we expectrecombination to happen rarely when genes are closely linked.2. Looking at your F, flies for each scenario, what are the main differences? Answer this question in#2 of your Data Sheet.3. Now let's work with some data. Suppose you obtained the following flies in your F, generation.Based on this data, which chromosome is the vestigial gene on? Answer this in #3 on your DataSheet.Phenotype# of fliescinnabar42120cinnabar, ebony, vestigial16ebony, vestigialvestigial33cinnabar, ebony8wild-type114ebony339cinnabar, vestigial328

A dihybrid cross is a cross between two individuals who differ in two features controlled by two…

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Biology

n differences betweer

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 2QP: Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To...

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You design Drosophila crosses to provide recombinationdata for gene a, which is located on the chromosome shownin Figure 15.12. Gene a has recombination frequencies of14% with the vestigial wing locus and 26% with the browneye locus. Approximately where is a located along thechromosome?
three recessive genes a, b, and c in the model plant Arabidopsis are found to be linked on chromosome 4. A three point test cross is done with a homozygous recessive plant with a heterozygous for all three genes. Following is the number of progenies a b C 65 A B c 56 A B C 1267 a b c 1310 A b C 550 a B c 515 a B C 470 A b c 489 Total = 4,722 Determine the middle locus by your choice of method and after that calculate the map distance between the genes in map unit (m.u.).
. Chromosome 3 of corn carries three loci (b for plant-color booster, v for virescent, and lg for liguleless). A testcross of triple recessives with F1 plants heterozygous forthe three genes yields progeny having the followinggenotypes: 305 + v lg, 275 b + +, 128 b + lg, 112 + v +,74 + + lg, 66 b v +, 22 + + +, and 18 b v lg. Give the genesequence on the chromosome, the map distances between genes, and the coefficient of coincidence.
Mitotic recombination can occasionally produce a twin spot.Let’s suppose an animal species is heterozygous for two genesthat govern fur color and length: one gene affects pigmentation,with dark pigmentation (A) dominant to albino (a); the other geneaffects hair length, with long hair (L) dominant to short hair (l).The two genes are linked on the same chromosome. Let’s assume ananimal of this species is AaLl; A is linked to l, and a is linked to L.Draw the chromosomes labeled with these alleles, and explainhow mitotic recombination could produce a twin spot with onespot having albino pigmentation and long fur and the other havingdark pigmentation and short fur.
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome . To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness , their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters’ sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia 24 have neither of the traits , 1 has color blindness only and 1 has hemophilia only . how many centimorgans seperate the hemophilia locus from the locus of the color blindness.
In the experiment of Figure shown, Stern followed the inheritancepattern in which females carried two abnormal X chromosomesto correlate genetic recombination with the physical exchange ofchromosome pieces. Is it necessary to use a strain carrying twoabnormal chromosomes, or could he have used a strain in whichfemales carried one normal X chromosome and one abnormal Xchromosome with a deletion at one end and an extra piece of the Ychromosome at the other end?
For this particular family, what is the recombination rate between the D17S74 marker and the breast cancer gene? Lets say that at age 45 the third granddaughter III-3, is diagnosed with breast cancer. Now recalculate the recombination rate between the marker locus D17S74 and the breast cancer gene? Use only the 5 women in your analysis:the two daugthers in generation II and the three granddaughters in generation II
The a, b, and c loci are all on different chromosomesin yeast. When a b+ yeast were crossed to a+ b yeastand the resultant tetrads analyzed, it was found thatthe number of nonparental ditype tetrads was equal tothe number of parental ditypes, but there were no tetratype asci at all. On the other hand, many tetratypeasci were seen in the tetrads formed after a c+ wascrossed with a+ c, and after b c+ was crossed withb+ c. Explain these results.
DRAW IT Suppose, as in the questionat the bottom of Figure 15.9, the parental(P generation) flies were true-breedingfor gray body with vestigial wings andblack body with normal wings. Draw thechromosomes in each of the four possiblekinds of eggs from an F1 female, andlabel each chromosome as “parental”or “recombinant.”
In a species of rat, the map unit distance between genes A, B, C and D were asfollows: (in the picture) (a) What is the order of these genes on the chromosome? (b) What is the map unit distance between B and D?
Consider the following "pericentric" ("around the center") inversion. In one an individual, a simplified 8-gene sequence along one chromosome is 1234 5678 while a pericentric inversion occurred on their other homologous chromosome, resulting in the sequence 1265 4378 (the dot represents the centromere). Draw these homologous chromosomes lined up (as during crossing over, in Prophase l). If a crossover event occurred between gene locus 3 and 4 (slicing between them) on the original chromosome, what would the crossover products look like? What would be the "problem" with each of the chromosomes that result?
A crossover has occurred in the bivalent shown in figure.What is the outcome of this single crossover event? If a secondcrossover occurs somewhere between A and C, explain which twochromatids it would involve and where it would occur (i.e., betweenwhich two genes) to produce the chromosomes shown here:A. ABC, AbC, aBc, and abcB. Abc, Abc, aBC, and aBCC. ABc, Abc, aBC, and abCD. ABC, ABC, abc, and abc