You have carried out the test cross above and get four phenotypes of cats with the following numbers: 95 105 23 27 Somehow, you forgot to note which phenotypes went with which numbers. But, given the information above, you can figure it out. Assuming the gene that controls paw color is gene P and the gene that controls fur length is gene L (with P and L being white paws and short fur), which of the following numbers could be associated with the PL gamete (cats with white paws and short fur)?
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- stion 6 of 18 Suppose 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 determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F, females in a testcross. She obtains the flies from this testcross. Phenotype Number of flies vg* qu+ 230 vg qu 224 vg qut vg* qu 97 99 Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X², for this hypothesis. 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.In the Human Genome Project, researchers have collectedlinkage data from many crosses in which the male washeterozygous for molecular markers and many crosses wherethe female was heterozygous for the markers. The distancebetween the same two markers, computed in map units, isdifferent between males and females. In other words, thelinkage maps for human males and females are not the same.Propose an explanation for this discrepancy. Do you think thesizes of chromosomes (excluding the Y chromosome) in humanmales and females are different? How could physical mappingresolve this discrepancy?. In 1919, Calvin Bridges began studying an X-linkedrecessive mutation causing eosin-colored eyes inDrosophila. Within an otherwise true-breedingculture of eosin-eyed flies, he noticed rare variantsthat had much lighter cream-colored eyes. By intercrossing these variants, he was able to make a truebreeding cream-eyed stock. Bridges now crossedmales from this cream-eyed stock with true-breedingwild-type females. All the F1 progeny had red (wildtype) eyes. When F1 flies were intercrossed, the F2progeny were 104 females with red eyes, 52 maleswith red eyes, 44 males with eosin eyes, and14 males with cream eyes. Assume that thesenumbers represent an 8:4:3:1 ratio.a. Formulate a hypothesis to explain the F1 and F2results, assigning phenotypes to all possiblegenotypes.b. What do you predict in the F1 and F2 generations if the parental cross is between truebreeding eosin-eyed males and true-breedingcream-eyed females?c. What do you predict in the F1 and F2 generationsif the parental cross is…
- A friend tells you that her biological father has an inherited disorder determined by a dominant allele but neither your friend nor her three siblings are affected. Is this possible? If not, why not? If so, explain how it is possible. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph Arial 14px A V x² X, ABC 田 田 | 国田用田 Ť {;} III !!! +)Consider two maize plants:a. Genotype C/cm ; Ac/Ac+, where cm is an unstableallele caused by a Ds insertionb. Genotype C/cm, where cm is an unstable allele causedby Ac insertionWhat phenotypes would be produced and in whatproportions when (1) each plant is crossed with a basepair-substitution mutant c/c and (2) the plant in part a iscrossed with the plant in part b? Assume that Ac and care unlinked, that the chromosome-breakage frequencyis negligible, and that mutant c /C is Ac+.What would justify the following ratio appearing after phenotyping the outcome of a crossing trial: 8.9: 2.9: 3.2:1? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b C d Obviously this represents independent assortment based on crossing dihybrid heterozygotes. Obviously this represents gene linkage based on test crossing a dihybrid heterozygote. Obviously this represents the results of a trihybrid test cross. Obviously this represents independent assortment based on crossing monohybrid heterozygotes.
- An organism of the genotype AaBbCc was testcrossed to a triplyrecessive organism (aabbcc). The genotypes of the progeny are inthe following table.AaBbCc 20 AaBbcc 20aabbCc 20 aabbcc 20AabbCc 5 Aabbcc 5aaBbCc 5 aaBbcc 5 a.) Assuming simple dominance and recessiveness in each genepair, if these three genes were all assorting independently,how many genotypic and phenotypic classes would result inthe offspring, and in what proportion? b.) Answer part (a) again, assuming the three genes are sotightly linked on a single chromosome that no crossovergametes were recovered in the sample of offspring. c.) What can you conclude from the actual data about thelocation of the three genes in relation to one another?E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings . Q. Give the genotypes for the green, virescent-white, and yellow progeny.A wild-type fruit fly (heterozygous for gray body color andnormal wings) is mated with a black fly with vestigial wings.The offspring have the following phenotypic distribution: wildtype, 778; black vestigial, 785; black normal, 158; gray vestigial,162. What is the recombination frequency between these genesfor body color and wing size? Is this consistent with the resultsof the experiment in Figure 15.9?
- GsnKivd010j2gIRWLIZOMZZ-VibKYvBbo61ylATAQ/viewform RECOMBINATION". For numbers 7-35, reler to the given data below. Glven the following testcross data for com In whlch the genes for fine stripe (f), bronze gleurone (bz) and knotted leaf (Kn) are involved: + = wild type f fine stripe +=wild type bz = bronze gleurone +=wild type Kn knotted leaf %3D Genotype Ko f Number 451 Ko 134 97 436 bz bz bz Ko 18 119 f 24 Kn f bz 86 Total: Your answer 7-8. What would be the recombination frequency or the frequency of the recombinant type between +/Kn and +/f genes? Oa. 16% O b. 16 map units + + + + +A corn geneticist wants to obtain a corn plant that hasthe three dominant phenotypes: anthocyanin (A), longtassels (L), and dwarf plant (D). In her collection ofpure lines, the only lines that bear these alleles are AALL dd and aa ll DD. She also has the fully recessive lineaa ll dd. She decides to intercross the first two and testcross the resulting hybrid to obtain in the progeny aplant of the desired phenotype (which would have to beAa Ll Dd in this case). She knows that the three genesare linked in the order written, that the distance between the A/a and the L/l loci is 16 m.u., and that thedistance between the L/l and the D/d loci is 24 m.u.a. Draw a diagram of the chromosomes of the parents,the hybrid, and the tester.b. Draw a diagram of the crossover(s) necessary toproduce the desired genotype.c. What percentage of the testcross progeny will be ofthe phenotype that she needs?d. What assumptions did you make (if any)?In Figure 1-6, the students have 1 of 15 different heights,plus there are two height classes (4′11″ and 5′ 0″) forwhich there are no observed students. That is a total of17 height classes. If a single Mendelian gene can account for only two classes of a trait (such as purple orwhite flowers), how many Mendelian genes would beminimally required to explain the observation of 17height classes?