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You perform a cross between two plants, one MmPp and one mmpp, and obtain offspring with the following
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- Imagine that the genes for seed color and seed shape are located on the same chromosome. A cross is made between two true-breeding plants. One plant produces green wrinkled seed (rryy) and the second parent produced round yellow seeds (RRYY). A test cross is made between the F1 generation with the following results: green, wrinkled 645 green round 36 yellow wrinkled 29 yellow round 590 a) Draw the chromosomes of parents and four types of offsprings and show the recombinant and non-recombinants in the offsprings. b) What is the recombination frequency between two loci?Part A: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be AB? Enter your answer as a decimal to three places (for example: 0.120). Part B: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be Ab? Part C: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a single plant will be aB? Part D: If the two genes are 30 mu apart and the plant is (AB/ab), what proportion of gametes from a signal plant will be ab?You have already localized the genes to the same chromosome by deletion mapping, and now decide that the best way to accomplish the mapping is to conduct two simultaneous three-point testcross experiments. The genes you are investigating are as follows: N = round leaves, n = notched leaves; H = smooth stems, h = hairy stems; R = purple flowers, r = red flowers; B = grey seeds, b = black seeds; and Y = green pods, y = yellow pods. Earlier experiments you have done already established that gene B is in the middle of this gene cluster, so you design both three-point test crosses to include that gene. Cross #1 is designed as RrHhBb x rrhhbb while cross #2 is NnBbYy x nnbbyy. The results of both crosses are given in the table below. Based on the information given, determine the arrangement of these five genes including the position of each allele in the heterozygous fly and the distances between each pair of genes. (Hint: treat each experiment separately, knowing that gene B is in the…
- You are doing a genetics experiment with the fruit fly. In the “P” generation, you cross two true-breeding flies. The female parent is brown and wingless and the male parent is black with normal wings. All of the flies in the F1 generation are brown and have normal wings. Answer all the questions below. 3a. What is the genotypes of P generation? 3b. What is the genotype of the F1 generation? 3c. You now take an F1 female and cross her to a true-breeding black, wingless male so what is the male’s genotype? 3d. You count 1600 offspring in the F2 generation. If the wing and the color traits were linked and no recombination occurred, what would you expect the numbers of: 1. brown winged flies (BbNn) 2. black winged flies (Bbnn) 3. brown wingless flies (bbNn) 4. black wingless flies (bbnn). 3e. What is the genetic distance between the color and wing genes when you count the F2 generation as 85 brown winged flies, 728 black winged flies, 712 brown wingless flies, and 75 black wingless flies.In corn, a triple heterozygote was obtained carrying themutant alleles s (shrunken), w (white aleurone), andy (waxy endosperm), all paired with their normal wildtype alleles. This triple heterozygote was testcrossed, andthe progeny contained 116 shrunken, white; 4 fully wildtype; 2538 shrunken; 601 shrunken, waxy; 626 white;2708 white, waxy; 2 shrunken, white, waxy; and 113 waxy.a. Determine if any of these three loci are linked and,if so, show map distances.b. Show the allele arrangement on the chromosomesof the triple heterozygote used in the testcross.c. Calculate interference, if appropriate.You are doing a genetics experiment with the fruit fly. In the “P” generation, you cross two true-breeding flies. The female parent is brown and wingless and the male parent is black with normal wings. All of the flies in the F1 generation are brown and have normal wings. Answer all the questions below. 3d. You count 1600 offspring in the F2 generation. If the wing and the color traits were linked and no recombination occurred, what would you expect the numbers of: 1. brown winged flies (BbNn) 2. black winged flies (Bbnn) 3. brown wingless flies (bbNn) 4. black wingless flies (bbnn). 3e. What is the genetic distance between the color and wing genes when you count the F2 generation as 85 brown winged flies, 728 black winged flies, 712 brown wingless flies, and 75 black wingless flies.
- You conduct a series of two-point mapping crosses involving five genes located on chromosome III and obtain the following percentage of recombinant offspring: h-cu eyg-h eyg-cu rai-eyg rai-h rai-cu th-cu th-h th-eyg th-rai 23.5% 9.0% 14.5% 18.5% 9.5% 33.0% 6.8% 16.7% 7.7% 26.2% The rai gene is at the near end of the chromosome. Gene th is between which two genes? A. eyg and cu B. rai and cu C. h and cu D. rai and hYou have determined that the gene order for three linked genes being studied is A B C. The number of recombinants resulting from crossover between genes A and BC are 44 and 48, respectively, while the double-crossover progeny total 4 and 6. What is the recombination frequency between genes A and B if the total progeny from the cross is 1000? Show all of your work.A genetic cross was made between two types of pea plants. One type produces round and green seeds, while the other type produces wrinkled and yellow seeds. A plant breeder crossed these two types and then allowed the F1 generation to self-fertilize. The following data were obtained: F1 generation: all produce wrinkled and yellow seeds. F2 generation: 157 produce wrinkled and yellow seeds 57 produce wrinkled and green seeds 54 produce round and yellow seeds 20 produce round and green seeds a) Propose a hypothesis that you think is consistent with the observed data. b) Test the goodness of fit between the data and your hypothesis using a chi square test. c) Explain what your calculated chi square results mean.
- Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?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 (map units) separate the hemophilia locus from the locus for color blindness?Let’s suppose that you have made a karyotype of a female fruit flywith red eyes and found that it has three X chromosomes insteadof the normal two. Although you do not know its parents, you doknow that this fly came from a mixed culture of flies in whichsome had red eyes, some had white eyes, and some had eosin eyes.Eosin is an allele of the same gene that has white and red alleles.Eosin is a pale orange color. The red allele is dominant and thewhite allele is recessive. The expression of the eosin allele, however, depends on the number of copies of the allele. When femaleshave two copies of this allele, they have eosin eyes. When femalesare heterozygous for the eosin allele and the white allele, they havelight-eosin eyes. When females are heterozygous for the red alleleand the eosin allele, they have red eyes. Males that have a singlecopy of the eosin allele have eosin eyes.You cross the XXX red-eyed female with a white-eyed male andcount the numbers of offspring. You may assume that…