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- Fruit weight is controlled by 4 gene pairs. The homozygous dominant weighs 200 g while the homozygous recessive weighs 80 g. 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 B1B1B2B2B3B3b4b4 x b1b1b2b2b3b3B4B4 a.) Compute for the contribution of the dominant allele. b.) Compute for the fruit weight of parental 1 and parental 2. c.) Provide the genotype and phenotype of the F1. d.) Compute for the frequencies of the different individuals in the F2.A pea plant that is (RrYy) is allowed to self-fertilize. Round seed(R) is dominant to wrinkled (r), and yellow seed (Y) is dominant togreen (y). What is the probability of producing the following groupof five seeds: two round, yellow; one round, green; one wrinkled,yellow; and one wrinkled, green?Round (R) seed shape is dominant to wrinkled (r) seed shape in pea plants. If an RR plant is crossed with an rr plant, what is the frequency of phenotypes in the F2 generation? 1)1/4 RR, 1/2 Rr, 1/4 rr 2) All round seeds 3) 3 roundseeds, 1 wrinkled seed 4) All wrinkled seeds
- GgTt pea plants can produce _____ type(s) of gamete(s), but a ggtt plant can produce _____ type(s) of gamete(s). explain. In nature, the plant Plectritis congesta is dimorphic forfruit shape; that is, individual plants bear either winglessor winged fruits, as shown in the illustration.Wingless fruit Winged fruitPlants were collected from nature before floweringand were crossed or selfed with the following results:Number of progenyPollination Winged WinglessWinged (selfed) 91 1*Winged (selfed) 90 30Wingless (selfed) 4* 80Winged × wingless 161 0Winged × wingless 29 31Winged × wingless 46 0Winged × winged 44 0*Phenotype probably has a nongenetic explanation.Interpret these results, and derive the mode ofinheritance of these fruit-shaped phenotypes. Usesymbols. What do you think is the nongeneticexplanation for the phenotypes marked by asterisks inthe table?3)The inability to synthesise chlorophyll is a recessive characteristic in tobacco plants. If a tobacco plant know to be heterozygous is self pollinated, and 600 of its seed were germinated, a) how many would you expect to produce albino (completely unpigmented) plants? b) how many would be expected to have the parental genotype? c) why is this cross the only way to produce albino tobacco plants
- A pea plant is heterozygous for three genes (Tt Rr Yy), whereT = tall, t = dwarf, R = round seeds, r = wrinkled seeds, Y = yellowseeds, and y = green seeds. Tall, round, and yellow are the dominanttraits. What is the probability that an offspring from self-fertilizationof this plant will be tall with wrinkled, yellow seeds?If you cross heterozygous spherical (S) yellow (Y) seeded plant with dented(s), green(y) the offsping and the ratio would be: A)F:SsYy, ssyy; ration 1:1B)F: SSYy, Ssyy, ssYy, ssyy; ration 1:1:1:1C)F: SSYy, Ssyy, ssYy,ssyy; ration 1:1:1:1D)F: SsYy, Ssyy, ssYy, ssyy ration 1:1:1:1Fruit weight is controlled by 4 gene pairs. The homozygous dominant weighs 200 g while thehomozygous recessive weighs 80 g. Assume that the alleles have equal contribution and havecumulative effects. The F1 and F2 of the two parents below were studied. Parent 1 Parent 2B1B1B2B2B3B3b4b4 x b1b1b2b2b3b3B4B4 a. Compute for the contribution of the dominant allele.b. Compute for the fruit weight of parental 1 and parental 2.c. Provide the genotype and phenotype of the F1.d. Compute for the frequencies of the different individuals in the F2.e. Provide the phenotypes of the F2 and their frequencies.
- Fruit weight is controlled by 4 gene pairs. The homozygous dominant weights 200 g while the homozygous recessive weighs 80 g. 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 B1B1B2B2B3B3b4b4 x b1b1b2b2b3b3B4B4 Provide the genotype and phenotype of the F1.1. The allele G for yellow stigma is completely dominant to green (g). Supposingtwo strains of autotetraploid plants are available and their genotypes are as follows:GGgg – in this plant the gene is close to the centromereGggg – in this plant the gene is far from the centromere If these two plants are crossed:a) provide the gametes that can be obtained from the two plants;b) provide the genotypic and phenotypic ratios of the offspring. 2. Consider the illustration below. Diagram the configuration you would observe at Anaphase I if crossing-over happens within the inversion. (IMAGE ATTACHED)Identify the genotypes of the pea flower plants for each generation P (Guadretieding parents) Phenotypes Purple flowers : Genotypes White flowers 1 2 F1 Gerleyatids) Phenotypes All purple flowers : Genotypes 3 Self- or cross-pollination F2 Generation Phenotypes Purple flowers : Genotypes White flowers 4