A plant of phenotype 1 was selfed, and, in the progeny,there were 100 plants of phenotype 1 and 60 plants ofan alternative phenotype 2. Are these numbers compatible with expected ratios of 9:7, 13:3, and 3:1?Formulate a genetic hypothesis on the basis of yourcalculations
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A plant of
there were 100 plants of phenotype 1 and 60 plants of
an alternative phenotype 2. Are these numbers compatible with expected ratios of 9:7, 13:3, and 3:1?
Formulate a genetic hypothesis on the basis of your
calculations
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- A plant of phenotype 1 was selfed and in the progeny there were 100 individuals of phenotype 1 and 60 of an alternative phenotype 2. Are these numbers compatible with expected ratios of 9:7, 13:3 and 3:1. Formulate a genetic hypothesis based on your calculations.A plant of phenotype 1 was selfed and in the progeny there were 100 individuals of phenotype 1 and 60 of an alternative phenotype 2. Are these numbers compatible with expected ratios of 9:7, 13:3 and 3:1. Formulate a genetic hypothesis based on your calculations. DO 6:55 PM /A plant believed to be heterozygous for a pair of allelesB/b (where B encodes yellow and b encodes bronze) wasselfed, and, in the progeny, there were 280 yellow and120 bronze plants. Do these results support the hypothesis that the plant is B/b?
- In pea plants, a dihybrid for the recessive a and b is testerossed. The distribution of the phenotypes is as follows: Phenotype AB Number 138 a b 132 a B 69 A b 61 (i) Are the genes assorting independently? Test the hypothesis with a chi-square test. State your hypothesis, estimate the p value using the x table below and clearly state your conclusion. Table 1: Chi-square table Chi Square Values and Probability Degrees of Freedom P = 0.99 0.95 0.80 0.50 0.20 0.05 0.01 0.000157 0.00393 0.0642 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3. 0.115 0.352 1.005 2.366 4.642 7.815 11.345 0.297 0.711 1.649 3.357 5.989 9.488 13.277 (ii) If the genes are linked, calculate the map distance between the genes in cM. '.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.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?
- In certain plant species such as tomatoes and petunias, a highly polymorphic incompatibility gene S with more than 100 known alleles prevents self-fertilization and promotes outbreeding. In this form of incompatibility, a plant cannot accept sperm carrying an allele identical to either of its own incompatibility alleles. If, for example, pollen carrying sperm with allele sI of the incompatibility gene lands onto the stigma (a female organ) of a plant that also carries the st allele, the sperm cannot fertilize any eggs in that plant. (This phenomenon occurs because the pollen grain on the stigma cannot grow a pollen tube to allow the sperm to unite with the egg.) For the following crosses, indicate whether any progeny would be produced, and if so, list all possible genotypes of these progeny. a. 8 s'5x 9 s's? b. ¿s'sx95353 d. Explain how this mechanism of incompatibility would prevent plant self- fertilization. e. How does this incompatibility system ensure F. How do you know that peas…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.Assume the height in a particular plant is determined by two pairs of unlinked polygenes,each effective allele contributes 5cm to a base height of 10cm. What are the heights of each parent and what height is to be expected in the F1 if there are no environmental effects?
- A test cross between a plant of genotype PpSs and the tester white plant with wrinkled seed coat (ppss) gives the following numbers of progeny in four phenotypic types. 14:87:83:16 (purple flower + smooth seed coat: purple flower + wrinkled seed coat: white flower + smooth seed coat: white flower + wrinkled seed coat). a. What is the expected ratio of progeny phenotypes assuming independent assortment of alleles? b. Explain how ratios of progeny show that the two genes are linked. c. How many map units separate the purple and smooth genes? Show your calculations. d. What is the “parental” genotype of the heterozygous parent? (i.e. Which alleles of the P and S loci are present on each of the two chromosomes of the doubly heterozygous parent of this test cross?)A pure-breeding strain of squash that produced diskshaped fruits (see the accompanying illustration) wascrossed with a pure-breeding strain having long fruits.The F1 had disk fruits, but the F2 showed a new phenotype, sphere, and was composed of the followingproportions:Longlong 32 sphere 178 disk 270Sphere DiskPropose an explanation for these results, and showthe genotypes of the P, F1, and F2 generations.Assume the height in a particular plant is determined by two pairs of unlinked polygenes, each effective allele contributes 5cm to a base height of 10cm. What are the heights of each parent? What height is to be expected in the F1 if there are no environmental effects? What is the expected phenotypic ratio in the F2
