Given :The given plant is triploid. Symbol For purple = P For yellow seed = Y For round seed = R Answer1. The Probability of PpYyRr = 8/64 = 1/8 2. The Probability of PPyyRr = 4/64 = 1/16 3. Probability of at least exhibiting wrinkled seeds = 32/64 = 1/2 4. Probability of at least exhibiting purple flowers = 48/64 = 6/8 = 3/4QUESTION:Answer the remaining question (5, 6, 7) Combination of the multiplication and addition rules to solve even morecomplex problems in Mendelian geneticsComplicated cross would be calculation of the probability on the inheritance ofthree or more characters. An example would be a cross of a trihybrid with purpleflowers and yellow round seeds (heterozygous for all three genes) with a plantwith purple flowers and green wrinkled seeds (heterozygous for flower color buthomozygous recessive for the other two characters). Using symbols, our cross isPpYyRr * Ppyyrr. The question would be what is the fraction of offspring fromthis cross are predicted to exhibit the recessive phenotypes for at least twoof the three characters?To answer this question, begin by listing all genotypes that fulfill the condition:ppyyRr, ppYyrr, Ppyyrr, PPyyrr, and ppyyrr. (Because the condition is at leasttwo recessive traits, it includes the last genotype, which shows all threerecessive traits.)Next, compute the probability for each of these genotypes resulting from thePpYyRr * Ppyyrr cross by multiplying together the individual probabilities for theallele pairs, just like the previous example. Take note that in a cross involvingheterozygous and homozygous allele pairs (for example, Yy * yy), the probabilityof heterozygous (Yy) offspring is 1/2 and the probability of homozygous (in thiscase, yy) offspring is 1/2.Finally, we use the addition rule to add the probabilities for all the differentgenotypes that fulfill the condition of at least two recessive traits resulting fromour PpYyRr * Ppyyrr cross, as shown below:ppyyRr = 4 (probability of pp) x ½ (yy) x ½ (Rr) = 1/16ppYyrr = ¼ (pp) x ½ (Yy) x ½ (rr) =1/16Ppyyrr = 2 (probability of Pp) x ½ (yy) x ½ (rr) = 2/16PPyyrr = 4 x ½ x 2 =1/16ppyyrr = ¼ x ½ x ½ =1/16Chance of at least two recessive traits6/16 or 3/8 SUNJECT - BIOLOGY (Genetics) × PROBABILITY.Based on the give question (PpYyRr * Ppyyrr) (SEEIMAGE), compute the following probabilities. Showyour solutions.1. Probability of PpYyRr2. Probability of PPyyRr3. Probability of at least exhibiting wrinkled seeds4. Probability of at least exhibiting purple flowers5. Probability of white flower, yellow and round seeds6. Probability of having purple flowers and green seedsand round seeds7. Probability of having white flower and yellow seed,or, white flower and wrinkled seed.

Genetics may be a branch of biology that investigates genes, genetic variations, and living beings'…

Given : The given plant is triploid. Symbol For purple = P For yellow seed = Y For round seed = R Answer 1. The Probability of PpYyRr = 8/64 = 1/8 2. The Probability of PPyyRr = 4/64 = 1/16 3. Probability of at least exhibiting wrinkled seeds = 32/64 = 1/2 4. Probability of at least exhibiting purple flowers = 48/64 = 6/8 = 3/4 QUESTION: Answer the remaining question (5, 6, 7)

Given : The given plant is triploid. Symbol For purple = P For yellow seed = Y For round seed = R Answer 1. The Probability of PpYyRr = 8/64 = 1/8 2. The Probability of PPyyRr = 4/64 = 1/16 3. Probability of at least exhibiting wrinkled seeds = 32/64 = 1/2 4. Probability of at least exhibiting purple flowers = 48/64 = 6/8 = 3/4 QUESTION: Answer the remaining question (5, 6, 7)

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter3: Transmission Of Genes From Generation To Generation

Section: Chapter Questions

Problem 31QP

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Based on the previous question (PpYyRr * Ppyyrr), compute the following probabilities. SHOW YOUR SOLUTIONS
Would appreciate help with this question please include explanation ty
ybrid Cross - Parental (P) vill be observing the F, offspring of the cross shown in nage. The purple colof (P) of the kernel is the result of ment called anthocyanin, which is dominant and not nked. /hat are the genotypes of the F, offspring if both nts shown are homozygous? F1 Mendel's dihybrid cross always showed a the same ratio ONLY for traits that are not linked on the same mosome and inherited together. If the genes are linked the ratio will not follow Mendel's dihybrid ratio. What is atio seen in dihybrid crosses that are not linked? F2 EXPERIMENTAL QUESTION: Are the genes for color (P) and shape (R) linked (on the same chromosome)?
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Given two strains of autotetraploid plants AAaa – in this plant the gene is close to the centromere Aaaa – in this plant the gene is far from the centromere provide the gametes of the two plants, and the genotypic and phenotypic ratios of the offspring when crossed
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Two plants in a cross were each heterozygous for two gene pairs (AB/ab) whose loci are linked and 10 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part D If the two genes are 15 mu apart and the plant is (Ab/aB), what proportion of gametes from a signal plant will be ab? Part E What proportion of the offspring of two plants ( both (Ab/aB)) will be A_B_ if the genes are 15 mu apart? Part F What proportion of the offspring of two plants ( both (Ab/aB)) will be A_bb if the genes are 15 mu apart? Part G What proportion of the offspring of two plants ( both (Ab/aB)) will be aaB_ if the genes are 15 mu apart? Part H What proportion of the offspring of two plants ( both (Ab/aB)) will be aabb if the genes are 15 mu apart? How would I solve these?
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