SUNJECT - BIOLOGY (Genetics) × PROBABILITY.Based on the give question (PpYyRr * Ppyyrr) (SEE IMAGE), compute the following probabilities. Show your 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 seeds and round seeds7. Probability of having white flower and yellow seed, or, white flower and wrinkled seed. 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

Name: SUNJECT - BIOLOGY (Genetics) × PROBABILITY.Based on the give question
Uploaded: 2024-03-20T06:43:04.000Z
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Description: SUNJECT - BIOLOGY (Genetics) × PROBABILITY.Based on the give question (PpYyRr * Ppyyrr) (SEE IMAGE), compute the following probabilities. Show your solutions. 1. Probability of PpYyRr2. Probability of PPyyRr3. Probability of at least exhibiting wrink