Mouse tail length is controlled by 3 gene pairs. The longest possible length is 93 mm (B₁B₁B₂B₂B3B3), while the shortest possible length is 57 mm (b₁b₁b₂b₂b3b3). Assume that the alleles have equal contribution and have cumulative effects. The F₁ and F₂ of the two parents below were studied. Parent 1 Parent 2 b₁b₁B₂B₂b3b3 B₁B₁b₂b₂B3B3 X Q: What is the tail length of the F1? A. 63 mm B. 66 mm C. 69 mm D. 75 mm/allele
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- Kernel color in wheat is controlled by 2 pairs of genes (AABB). Determine the color of the offspring with the following genotypes: (Note: 4 contributing alleles – red; 3 contributing alleles – medium red; 2 contributing alleles – intermediate red; 1 contributing allele – red; and 0 – white (AAbb x AaBb, AABb x Aabb, aaBb x Aabb, AABb x aabb, AABb x AaBb) Refer to the problem above, a wheat plant producing medium red seed is crossed with another plant producing intermediate red seed. How many individuals will be? Red Medium red Intermediate red Light red WhiteIn a certain plant, leaf size is determined by fourgenes whose alleles assort independently and actadditively. Thus, alleles A, B, C, and D each add4 cm to leaf length and alleles A′, B′, C′, andD′ each add 2 cm to leaf length. Therefore,an AA BB CC DD plant has leaves 32 cm longand an A′A′ B′B′ C′C′ D′D′ plant has leaves16 cm long.a. If true-breeding plants with leaves 32 cm longare crossed to true-breeding plants with leaves16 cm long, the F1 will have leaves 24 cm longand the genotype AA′ BB′ CC′ DD′. Listall possible leaf lengths and their expectedfrequencies in the F2 generation produced fromthese F1 plants.Kernel color in wheat is controlled by 2 pairs of genes (AABB). Determine the color of the offspring with the following genotypes: (Note: 4 contributing alleles – red; 3 contributing alleles – medium red; 2 contributing alleles – intermediate red; 1 contributing allele – red; and 0 – white (AAbb x AaBb, AABb x Aabb, aaBb x Aabb, AABb x aabb, AABb x AaBb) Show the solution through punnett square. Refer to the problem above, a wheat plant producing medium red seed is crossed with another plant producing intermediate red seed. How many individuals will be?Show the solution through punnett square. Red Medium red Intermediate red Light red White
- 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?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.Given the following testcross data for corn in which the genes for fine stripe (f), bronze aleurone (bz) and knotted leaf (Kn) are involved: Phenotype Number Kn + + 451 Kn f + 134 + + + 97 + f bz 436 Kn + bz 18 + + bz 119 + f + 24 Kn f bz 86 Total: 1,365 Determine the sequence (order) of the three genes and the calculate the distance between them.
- In a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.What is the genetic distance between the eye colour locus (w) and the bristle locus (sn)? answer in Mu (Map units) and to 2 decimal places? The following table summarises the results of the 2022-2023 Drosophila three-point cross involving the loci white eyes (w), miniature wings (m) and singed bristles (sn). The data is also available in the Excel file 'Drosophila Counts 2022-2023'. We strongly recommend working in Excel during this exercise.Phenotype Count+ + + 584w m sn 324w + + 227+ m sn 150+ m + 134w + sn 196 + + sn 134 w m + 92A presumed dihybrid in Drosophila, B/b ; F/f, is testcrossed with b/b ; f/f. (B = black body;b = brown body;F = forked bristles; f = unforked bristles.) The results areblack, forked 230 brown, forked 240black, unforked 210 brown, unforked 250Use the χ2 test to determine if these results fit the resultsexpected from testcrossing the hypothesized dihybrid
- A student crossed a female worm homozygous for the dpy-17 e164 allele with a wild-type male worm heterozygous for dpy-17 e164. He then scored all of the developing eggs/larva for the Dpy phenotype. He found 88 with the Dpy phenotype and 98 wild type. Do a Chi square test on phenotypic data to determine if your data supports the hypothesis that the mutation segregates as a single-gene mutation. Show your work in a table and include Null hypothesis Χ2 value dof p value ConclusionConsider the results of a three point linkage mapping testcross between a female (AaBbCc) and a male (aabbcc). Phenotype | no. Of progeny ABc 849 abC 820 Abc 58 aBC 60 AbC 102 aBc 98 ABC 6 abc 7 a) what is the correct order of the gene loci? ( which gene is in the middle) B) what was the configuration of recessive alleles in the triple heterozygous mother? (Use a cis-trans statement) C) calculate the map distances for each segment of the map.You are mapping three linked loci in Drosophila melanogaster (the common laboratory fruit fly). You cross flies that are triply mutant for apricot (pale eyes), bristle (extra bristles) and clipped (notched wings) to wild-type flies. The F+ flies are wild-type in appearance. You then backcross the F+ females to pure-breeding (apricot, bristle, clipped) males and score the phenotypes of 1000 F progeny for all three loci. Here are the results: 359 wild-type 361 apricot, bristle, clipped 89 bristle, clipped 91 apricot 42 apricot, bristle 38 clipped 9 apricot, clipped 11 bristle Using these data, first determine what gametes from the F; trihybrid produced each of the eight F2 categories. Note that apricot = aa (recessive to wild-type A); bristle = bb (recessive to wild-type B); and clipped = cc (recessive to wild-type C). Then determine if each gamete is recombinant (R) or nonrecombinant (R) for each pair of alleles (that is, for each genetic interval). Complete the table by dragging the…