Consider the following statements about linkage: 1. One map unit of distance is equivalent to a 10% recombination frequency. 2. If a two-factor testcross produces very few recombinants, the two genes are far apart.
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- If two genes are unlinked the recombination frequency will be? 25% 50% 75% 100%Use the data provided in the figure to calculate the recombination frequency between the two genes. Parents Wild Type (gray with normal wings) bb vg* vg Black with vestigial wings bb vg vg recombination frequency: % + TOOLS x10 Wild Type Black nogmal Black vestigial Gray vestigial b'bvg vg bb vg vg bb vg vg bb vg vg The number of each type of offspring is • wild type: 865 • black vestigial: 875 • gray vestigial: 160 • black normal: 100The table below shows the progeny of a test cross of a heterozygote (HhFfGg). Each row shows the number of progeny that inherited each combination of alleles from the heterozygous parent. You've already determined the gene order (G is in the middle) and labeled each type of progeny. Using this information, what is the recombination frequency between H and G? Alleles: Type: HGF hgf HgF h Gf Hgf h GF HGf hg F O 0.543 0.302 O 0.241 O 0.23 0.139 Parental Parental DCO DCO SCO (H-G) SCO (H-G) SCO (G-F) SCO (G-F) Count: 124 119 15 18 51 55 38 40 O The correct answer is not available.
- Explain the following pairs of concepts and briefly describe their major difference(s) 1. Recombination fraction & genetic distance.What would be the gene map for the following genes with known recombination frequencies? D and A is 5.1 cM, A and C is 15.8 cM, A and B is 7.8 cM, C and D is 10.7 cM, and C and B is 8.0 cM.Figure 13.4 Which of the following statements is true? Recombination of the body color and red/ cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles. Recombination of the gray/black body color and long/short aristae alleles will not occur. Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color.
- Wild-type mice have brown fur and short tails. Loss of function of a particular gene produces white fur, while loss of function of another gene produces long tails, and loss of function at a third locus produces agitated behavior. Each of these loss of function alleles is recessive. If a wild-type mouse is crossed with a triple mutant, and their F1 progeny is test-crossed, the following recombination frequencies are observed among their progeny. Produce a genetic map for these loci. Brown, short tailed, normal: 955 White, short tailed, normal: 16 Brown, short tailed, agitated: 0 White, short tailed, agitated: 36 Brown, long tailed, normal: White, long tailed, normal: Brown, long tailed, agitated: 46 0 14 White, long tailed, agitated: 933As per our guidelines, we are able to only answer three subparts in a question that is posted. Please post the remaining subparts from Part B as separate questions. this is the remaining questions 4. What are the two parental classes and the two recombinant classes of offspring? 5. What is the percentage of recombination between these two loci? Show your calculations. 6. How many map units apart are they?In your own words, explain why a testcross cannot produce more than 50% recombinant offspring. When a testcross does produce 50% recombinant offspring, what does this result mean?
- You cross two plants and get the following results: Recombinant progeny type A: 43 Recombinant progeny type B: 51 Nonrecombinant progeny type C: 129 Nonrecombinant progeny type D: 135 What is the recombination frequency? (Enter your answer as a % without the sign. For example, if your answer is 0.534, you would enter 53.4)When true-breeding mice with brown fur and short tails (BBtt) were crossed to true-breeding mice with white fur and long tails (bbTT), all the F1 offspring had brown fur and long tails. The F1 offspring were then crossed to mice with white fur and short tails (bbtt). The fur gene and the tail length gene are 7 cM apart. Part a Which of the following F2 offspring are recombinant? (Select all that apply.) a) White fur, short tail b) Brown fur, long tail c) You need more information to answer this question. d) White fur, long tail e) Brown fur, short tail Part b Using the information given in Part a, if there were 400 mice in the F2 generation, how many do you expect would have white fur and long tails? a) 25 b) 186 c) 14 d) 372 e) 28 f) 75Three linked loci are evaluated to determine the recombination frequency of the loci. To set up the test cross, an individual that is true breeding dominant for all three phenotypes is crossed with an individual that is true breeding recessive for all three phenotypes. F1 offspring are crossed with individuals that are recessive for all phenotypes. The phenotypic results are shown in the table below. Phenotype Number of Offspring Dominant 1, Dominant 2, Dominant 3 322 Dominant 1, Dominant 2, Recessive 3 82 Recessive 1, Dominant 2, Recessive 3 5 Dominant 1, Recessive 2, Dominant 3 2 Recessive 1, Dominant 2, Dominant 3 28 Recessive 1, Recessive 2, Dominant 3 89 Recessive 1, Recessive 2, Recessive 3 312 Dominant 1, Recessive 2, Recessive 3 25 1. Which two phenotypes are parental? 2. What is the map distance between locus 1 and locus 2? 3. What is the map distance between locus 1 and locus 3? 4. What is the map distance between…