Biology 2e
2nd Edition
ISBN: 9781947172517
Author: Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher: OpenStax
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Chapter 12, Problem 30CTQ
Use the probability method to calculate the genotypes and genotypic proportions of a cross between AABBCc and Aabbcc parents.
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Use the forkline method to derive the gametic ratio of cross AabbDD x AABbDd. The cross are segregated independently.
Using forked line method, figure out the phenotype and genotype of the offspring between two individuals with genes: AaBbCC x AabbCc
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Directions: For the following problem, calculate the probability of the offspring between the cross using the sum and product rule.
The parent generation has genotypes of Aa Bb Cc and AA Bb CC. What is the probability the F1 generation would be AA bb cc?
Chapter 12 Solutions
Biology 2e
Ch. 12 - Figure 12.5 In pea plants, round peas (R) are...Ch. 12 - Figure 12.6 What are the genotypes of the...Ch. 12 - Figure 12.12 What ratio of offspring would result...Ch. 12 - Figure 12.16 In pea plants, purple flowers (P) are...Ch. 12 - Mendel performed hybridizations by transferring...Ch. 12 - Which is one of the seven characteristics that...Ch. 12 - Imagine you are performing a cross involving seed...Ch. 12 - Consider a cross to investigate the pea pod...Ch. 12 - A scientist pollinates a true-breeding pea plant...Ch. 12 - The observable traits expressed by an organism are...
Ch. 12 - A recessive trait will be observed in individuals...Ch. 12 - If black and white true-breeding mice are mated...Ch. 12 - The ABO blood groups in humans are expressed as...Ch. 12 - In a mating between two individuals that are...Ch. 12 - If the allele encoding polydactyly (six fingers)...Ch. 12 - A farmer raises black and white chickens. To his...Ch. 12 - Assuming no gene linkage, in a dihybrid cross of...Ch. 12 - The forked line and probability methods make use...Ch. 12 - How many different offspring genotypes are...Ch. 12 - Labrador retriever's fur color is controlled by...Ch. 12 - Which of the following situations does not follow...Ch. 12 - Describe one of the reasons why the garden pea was...Ch. 12 - How would you perform a reciprocal cross for the...Ch. 12 - Mendel performs a cross using a true-breeding pea...Ch. 12 - Calculate the probability of selecting a heart or...Ch. 12 - The gene for flower position in pea plants exists...Ch. 12 - Use a Punnett square to predict the offspring in a...Ch. 12 - Can a human male be a carrier of red-green color...Ch. 12 - Why is it more efficient to perform a test cross...Ch. 12 - Use the probability method to calculate the...Ch. 12 - Explain epistatis in terms of its Greek-language...Ch. 12 - In Section 12.3, ''Laws of Inheritance," an...Ch. 12 - People with trisomy 21 develop Down’s syndrome....Ch. 12 - A heterozygous pea plant produces violet flowers...
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- Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?arrow_forwardgive the possible genotypes and phenotypes and their ratios from the following cross between two strains of chicken:Rose comp (Rrpp) x Walnut (RrPp)arrow_forwardNotation: D = disease allele at gene d = wild-type allele at gene Let’s provide some information about each MOI. AD: Each of the F1 parents is heterozygous for the D allele. SLR: The observed F2 data are the offspring from an F1 cross where the father is affected (disease) and the mother is an unaffected carrier (wild-type, or WT). That is, the mother’s genotype has one disease allele and one WT allele in it. In Table 1, I provide the observed data counts (in dark blue). Notice that I stratify by gender. Table 1 Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 Disease, Female 267 WT, Male 285 WT, Female 301 Total 1157 DF p-value In Table 2, provide expected proportions for two different MOIs: AD and SLR. The values in this table are computed using information from the Punnett Square and the specified MOI. Once the proportions are determined, we can fill in the values in Table 3, E(xpected)…arrow_forward
- Cross a man with type B heterozygous blood with a woman with type o blood. What are the possible genotypes and phenotypes of their offspring? use punnett squarearrow_forwardExamine the following crosses among sea stars and the resulting colors of the offspring and decide what type of inheritance is involved. Then, indicate the genotypes of the parents and offspring involved in each of the crosses in the blanks provided.arrow_forwardSolve the problem using punnett square. What is the probability that each of the following pairs of parents will produce the indicated offspring? AAbbCc X AaBBCc --> AaBbCc = AaBBCcDd X AaBbCcDD --> AaBBccDD =arrow_forward
- Consider the pedigree below. Which of the individuals MUST be heterozygous carriers of the diseasse? I II III IV OQ 3 11-2 III-1 III-2 UI-3 III-4 III-5 IV-1 IV-2 IV-3 IV-4 IV-5arrow_forwardIn a cross AABbCc × aabbCc, what is the probability of producing the genotype AabbCC? (show work)arrow_forwardUsing the forked-line, or branch diagram, method, determine the genotypic and phenotypic ratios of these trihybrid crosses: (a) AaBbCc * AaBBCC, (b) AaBBCc * aaBBCc, and (c) AaBbCc * AaBbCc.arrow_forward
- Consider a cross between a true breeding purple flowered pea plant (genotype PP) and a true breeding white flowered pea plant (genotype pp) that produces all heterozygous F1 offspring. An F1 intercross is conducted between two of the F1 offspring, resulting in the F2 generation. What is the expected genotpye outcome of the F2 individuals resulting from the F1 intercross? 1PP: 2Pp: 1pp 1PP: 1Рp: 1pp 100% Pp 1PP: 1pp 100% pparrow_forwardDraw a Punnet square indicating the F2 genotypes from the mating of an F1 male (QQBbHh) with an Fi female (QQBÜHH). Note: Make sure to write all the heterozygous genotypes as Gg, not gG.arrow_forwardPerform a chi-square test of independence on the data provided to determine if the genes for flower color and pollen shape in sweet peas are assorting independently. Give the chi-square value, degrees of freedom, and associated probability. What conclusion would you draw about the independent assortment of these genes? Hint: the expected phenotype ratios are 9:3:3:1 instead of 1:1:1:1 due to the type of cross that was performed.arrow_forward
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