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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 Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
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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)
Combination of the multiplication and addition rules to solve even more complex problems in Mendelian genetics Complicated cross would be calculation of the probability on the inheritance of three or more characters. An example would be a cross of a trihybrid with purple flowers and yellow round seeds (heterozygous for all three genes) with a plant with purple flowers and green wrinkled seeds (heterozygous for flower color but homozygous recessive for the other two characters). Using symbols, our cross is PpYyRr * Ppyyrr. The question would be what is the fraction of offspring from this cross are predicted to exhibit the recessive phenotypes for at least two of 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 least two recessive traits, it includes the last genotype, which shows all three recessive traits.) Next, compute the probability for each of these genotypes resulting from the PpYyRr * Ppyyrr cross by multiplying together the individual probabilities for the allele pairs, just like the previous example. Take note that in a cross involving heterozygous and homozygous allele pairs (for example, Yy * yy), the probability of heterozygous (Yy) offspring is 1/2 and the probability of homozygous (in this case, yy) offspring is 1/2. Finally, we use the addition rule to add the probabilities for all the different genotypes that fulfill the condition of at least two recessive traits resulting from our PpYyRr * Ppyyrr cross, as shown below: ppyyRr = 4 (probability of pp) x ½ (yy) x ½ (Rr) = 1/16 ppYyrr = ¼ (pp) x ½ (Yy) x ½ (rr) = 1/16 Ppyyrr = 2 (probability of Pp) x ½ (yy) x ½ (rr) = 2/16 PPyyrr = 4 x ½ x 2 = 1/16 ppyyrr = ¼ x ½ x ½ = 1/16 Chance of at least two recessive traits 6/16 or 3/8
Transcribed Image Text:Combination of the multiplication and addition rules to solve even more complex problems in Mendelian genetics Complicated cross would be calculation of the probability on the inheritance of three or more characters. An example would be a cross of a trihybrid with purple flowers and yellow round seeds (heterozygous for all three genes) with a plant with purple flowers and green wrinkled seeds (heterozygous for flower color but homozygous recessive for the other two characters). Using symbols, our cross is PpYyRr * Ppyyrr. The question would be what is the fraction of offspring from this cross are predicted to exhibit the recessive phenotypes for at least two of 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 least two recessive traits, it includes the last genotype, which shows all three recessive traits.) Next, compute the probability for each of these genotypes resulting from the PpYyRr * Ppyyrr cross by multiplying together the individual probabilities for the allele pairs, just like the previous example. Take note that in a cross involving heterozygous and homozygous allele pairs (for example, Yy * yy), the probability of heterozygous (Yy) offspring is 1/2 and the probability of homozygous (in this case, yy) offspring is 1/2. Finally, we use the addition rule to add the probabilities for all the different genotypes that fulfill the condition of at least two recessive traits resulting from our PpYyRr * Ppyyrr cross, as shown below: ppyyRr = 4 (probability of pp) x ½ (yy) x ½ (Rr) = 1/16 ppYyrr = ¼ (pp) x ½ (Yy) x ½ (rr) = 1/16 Ppyyrr = 2 (probability of Pp) x ½ (yy) x ½ (rr) = 2/16 PPyyrr = 4 x ½ x 2 = 1/16 ppyyrr = ¼ x ½ x ½ = 1/16 Chance of at least two recessive traits 6/16 or 3/8
SUNJECT - BIOLOGY (Genetics) × PROBABILITY. Based on the give question (PpYyRr * Ppyyrr) (SEE IMAGE), compute the following probabilities. Show your solutions. 1. Probability of PpYyRr 2. Probability of PPyyRr 3. Probability of at least exhibiting wrinkled seeds 4. Probability of at least exhibiting purple flowers 5. Probability of white flower, yellow and round seeds 6. Probability of having purple flowers and green seeds and round seeds 7. Probability of having white flower and yellow seed, or, white flower and wrinkled seed.
Transcribed Image Text:SUNJECT - BIOLOGY (Genetics) × PROBABILITY. Based on the give question (PpYyRr * Ppyyrr) (SEE IMAGE), compute the following probabilities. Show your solutions. 1. Probability of PpYyRr 2. Probability of PPyyRr 3. Probability of at least exhibiting wrinkled seeds 4. Probability of at least exhibiting purple flowers 5. Probability of white flower, yellow and round seeds 6. Probability of having purple flowers and green seeds and round seeds 7. Probability of having white flower and yellow seed, or, white flower and wrinkled seed.
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