GsnKivd010j2gIRWLIZOMZZ-VibKYvBbo61ylATAQ/viewformRECOMBINATION".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 typef fine stripe+=wild typebz = bronze gleurone+=wild typeKn knotted leaf%3DGenotypeKofNumber451Ko13497436bzbzbzKo18119f24Knfbz86Total:Your answer7-8. What would be the recombination frequency or the frequency of therecombinant type between +/Kn and +/f genes?Oa. 16%O b. 16 map units+ + ++ +

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Description: GsnKivd010j2gIRWLIZOMZZ-VibKYvBbo61ylATAQ/viewformRECOMBINATION".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 involve

Introduction "Genotype" refers to an organism's complete genetic information. The observed…

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RECOMBINATION". For numbers 7-35, reler to the given data below. Glven the following testcross data for com in which the genes for fine stripe (f), bronze aleurone (bz) and knotted leaf (Kn) are involved: + = wild type f = fine stripe +=wild type bz = bronze aleurone +=wild type Kn knotted leaf %3D %3D Genotype Ka f Number 451 134 97 Kn 436 bz bz bz Kn 18 119 f 24 Kn f bz 86 Total: Your answer + + +

RECOMBINATION". For numbers 7-35, reler to the given data below. Glven the following testcross data for com in which the genes for fine stripe (f), bronze aleurone (bz) and knotted leaf (Kn) are involved: + = wild type f = fine stripe +=wild type bz = bronze aleurone +=wild type Kn knotted leaf %3D %3D Genotype Ka f Number 451 134 97 Kn 436 bz bz bz Kn 18 119 f 24 Kn f bz 86 Total: Your answer + + +

Concepts of Biology

1st Edition

ISBN:9781938168116

Author:Samantha Fowler, Rebecca Roush, James Wise

Publisher:Samantha Fowler, Rebecca Roush, James Wise

Chapter8: Patterns Of Inheritance

Section: Chapter Questions

Problem 2ACQ: Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are...

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F1 hybrids between two species of cotton, Gossypium barbadenseand Gossypium hirsutum, are very vigorous plants. However, F1crosses produce many seeds that do not germinate and a high percentageof very weak F2 offspring. Suggest two reasons for theseobservations.
You are given a Drosophila female that looks wild-type but is heterozygous for mutations intan body (t), miniature wings (m), and white eyes (w). You test cross this female with a tanbodied, miniature winged, and white-eyed homozygous mutant male, and you obtain thefollowing 1400 progeny: Phenotype : number+ + + : 608t m w : 516+ m w : 2t + + : 6+ m + : 39t + w : 46+ + w : 81t m + : 102 Calculate the distance between each pair t-m, m-w, and t-w only using the number ofrecombinants between them (i.e. ignoring the gene in the middle). Draw a linear map with thedistances between genes.
Purple flowers are dominant to white flowers. Identify the phenotypefor the following genotype Ff, FF, ff and determine if the genotype is heterozygous or homozygous. * For each row, you should select two columns. Purple flowers White flowers Heterozygous Homozygous Ff FF ff Brown eyes are dominant to blue eyes. Identify the phenotype for the following genotype BB, Bb, bb and determine if the genotype is heterozygous or homozygous. * 口ロ口
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.
Here are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…
Homozyogous wild type male Drosophila PPQQRRSSTTUUVV (all linked) are irradiated to induce the formation of chromosomal deletions and then mated with homozygous recessive females. In several of these matings a unique pattern of pseudodominance could be correlated with the loss of specific polytene chromosome bands, as shown in the table... Cross Chromosomal Bands Deleted Pattern of Pseudodominance PqRs TuV 1 2 3 4 5 5-6 1-2 3-5 6-8 1-3 PQRSTUV PqrSTUV PQRSTUV pQrSTUV In which chromosomal band is gene R located? A. 3 OB. 4 OC. 2 D.8 O E. none of these answers are correct
Give typed full explanation You are studying three linked genes in snapdragons. The flower color locus is in the center. There are 13.8 cM between the flower color locus and the plant height locus. There are 14.5 cM between the flower color locus and the leaf type locus. The coefficient of coincidence is 0.8. Pure-breeding tall, red-flowered plants with fuzzy leaves were crossed to pure-breeding dwarf, blue-flowered plants with smooth leaves. The F1 were testcrossed. Calculate the proportion of the testcross progeny that are expected to be dwarf with red flowers. Round properly to 4 decimal digits.
Drosophila females heterozygous for 3 recessive mutations a, b, and c werecrossed to males homozygous for all 3 mutations. The cross had the followingresults:wild 75 c 348 ab 306b 96 ac 110 abc 85Based on their recombination frequencies, the map distances in cM among the 3alleles are? a-b ; b-c ; a-c?
The following corn loci are on one arm of chromosome9 in the order indicated (the distances between themare shown in map units):c-bz-wx-sh-d-centromere12 8 10 20 10C gives colored aleurone; c, white aleurone.Bz gives green leaves; bz, bronze leaves.Wx gives starchy seeds; wx, waxy seeds.Sh gives smooth seeds; sh, shrunken seeds.D gives tall plants; d, dwarf.A plant from a standard stock that is homozygous for allfive recessive alleles is crossed with a wild-type plantfrom Mexico that is homozygous for all five dominantalleles. The F1 plants express all the dominant allelesand, when backcrossed to the recessive parent, give thefollowing progeny phenotypes:colored, green, starchy, smooth, tall 360white, bronze, waxy, shrunk, dwarf 355colored, bronze, waxy, shrunk, dwarf 40white, green, starchy, smooth, tall 46colored, green, starchy, smooth, dwarf 85white, bronze, waxy, shrunk, tall 84colored, bronze, waxy, shrunk, tall 8white, green, starchy, smooth, dwarf 9colored, green, waxy, smooth,…
Gene mapping using the Three-point Testcross a) Given the following alleles that control seed traits:W = wrinkled G = green R = roundw = smooth g = yellow r = oval b) Results of a cross with a triple heterozygote revealed the following phenotypes:30 smooth yellow round4 smooth green round958 wrinkled green round2 wrinkled yellow oval18 wrinkled yellow round946 smooth yellow oval16 smooth green oval26 wrinkled green oval c) Determine the order of the genes and the distance between them in centiMorgan (cM).Construct a gene map to show your results. TIP: Based on the phenotypes, determine the alleles in the gametes
Draw a preliminary linkage map of the three loci. What is their order? (E.g. is the forked locus between yellow and cross-veinless, or outside this region?) As this is a preliminary linkage map - do not account for double crossovers. You should base the distances on this map off your map distances calculated: yellow and forked: 39.3 mu yellow and cross-veinless: 11.9 mu cross-veinless and forked: 33.6 mu
The R and S loci are 35 m.u. apart. If a plant of genotype of RS/rs is selfed, what progeny phenotype will be seen and in what proportions?