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

See similar textbooks

Similar questions

Mendel performs a cross using a true-breeding pea plant with round, yellow seeds and a true- breeding pea plant with green, wrinkled seeds. What is the probability that offspring will have green, round seeds? Calculate the probability for the F1 and F2 generations.
The text outlines some of the problems Frederick William I encountered in his attempt to breed tall Potsdam Guards. a. Why were the results he obtained so different from those obtained by Mendel with short and tall pea plants? b. Why were most of the children shorter than their tall parents?
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
Variations on a Theme by Mendel Pea plants usually have white or red flowers. A strange pea-plant variant is found that has pink flowers. A selfcross of this plant yields the following phenotypes: 30 red flowers 62 pink flowers 33 white flowers What are the genotypes of the parents? What is the genotype of the progeny with red flowers?
More Crosses with Pea Plants: The Principle of Independent Assortment Determine the possible genotypes of the following parents by analyzing the phenotypes of their children. In this case, we will assume that brown eyes (B) is dominant to blue (b) and that right-handedness (R) is dominant to left-handedness (r). a. Parents: brown eyes, right-handed brown eyes, right-handed Offspring: 3/4 brown eyes, right-handed 1/4 blue eyes, right-handed b. Parents: brown eyes, right-handed blue eyes, right-handed Offspring: 6/16 blue eyes, right-handed 2/16 blue eyes, left-handed 6/16 brown eyes, right-handed 2/16 brown eyes, left-handed c. Parents: brown eyes, right-handed blue eyes, left-handed Offspring: 1/4 brown eyes, right-handed 1/4 brown eyes, left-handed 1/4 blue eyes, right-handed 1/4 blue eyes, left-handed
Variations on a Theme by Mendel A characteristic of snapdragons amenable to genetic analysis is flower color. Imagine that a true-breeding red- flowered variety is crossed to a pure line having white flowers. The progeny are exclusively pink-flowered. Diagram this cross, including genotypes for all P1 and F1 phenotypes. What is the mode of inheritance? Let F = red and f = white.
Pedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the following questions. a. Is the proband male or female? b. Is the grandfather of the proband affected? c. How many siblings does the proband have, and where is he or she in the birth order?
Mendelian Genetics The presence of a dimple on the cheek is governed by a dominant gene. A couple had already 4 children with dimpled cheeks. What would be the probability of having the next child as another dimpled if both parents are heterozygous for this character? Show your solution using Punnett Square
Mendel crossed two Pea plants for plant height and flower color Tall plant (T) is dominant to Short Plant (t). Purple Flower (P) is dominant to white flower (p). Using the following information perform the dihybrid cross using punnett squares that will predict all possible genotypes of the offspring and list the number and description of the phenotypes of the offspring. A. One plant homozygous dominant for plant height and flower color crossed with another plant homozygous recessive for plant height and heterozygous for flower color.
The accompanying pedigree shows a very unusual inheritance pattern that actually did exist. All progeny areshown, but the fathers in each mating have been omittedto draw attention to the remarkable pattern.a. Concisely state exactly what is unusual about thispedigree.b. Can the pattern be explained by Mendelianinheritance?
Mendelian Genetics and Non-Mendelian Genetics: Huntington’s disease, a neurodegenerative genetic disorder that typically becomes noticeable in middle age, is due to an autosomal dominant allele. Sickle cell anemia, on the other hand, is a genetic blood disorder due to a recessive allele. Jillian is a carrier of the allele for sickle cell anemia but has no sign of any neurodegenerative disorder in her family. She married Jacobwhose father died of Huntington’s disease. His mother, however, is not inflicted with that condition. Neither of his parents exhibit sickle cell anemia. 1. Give the genotypes of Jillian and Jacob. Assuming that they will have 4 children, what is the probability that: 2. all their children will be normal? 3. they will have a son with Huntington’s disease? 4. they will have a daughter inflicted with both conditions
When Gregor Mendel was working in the mid 1800s, scientists had not yet discovered chromosomes or meiosis. However, we now understand how Mendel's principles are rooted in the events of meiosis. As an example of this, state Mendel's principle of independent assortment and explain how it relates to independent assortment in meiosis.