Answer clearly with no explanation please J. A. Moore investigated the inheritance of spotting patterns in leopard frogs. The pipiens phenotype had the normal spots thatgive leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the crosses in thetable, producing the progeny indicated. (Moore, 1943).Cross Parent phenotypes Progeny phenotypes, observed1burnsi x burnsi39 burnsi, 6 pipiens2burnsi x pipiens23 burnsi, 33 pipiens3burnsi x pipiens196 burnsi, 210 pipiensStep 1: Establish the hypothesis by assigning genotypes to the parents and progeny of the cross.Before performing a chi-square goodness-of-fit test, you need to establish the hypothesis to be tested.Hypothesize what the genotypes of the parents are. What genotypes would their progeny have? Define the burnsi allele as B andthe pipiens allele as B*.ParentalParentalProgenyProgenyphenotypesgenotypesphenotypes genotypesburnsiburnsiburnsipipiensAnswer BankBB+BBB*B*Privacy - T

Alleles An alleles is different form of same gene. These alleles are located on the same locus on a…

Answered: Hypothesize what the genotypes of the…

Hypothesize what the genotypes of the parents are. What genotypes would their progeny have? Define the burnsi allele as B and the pipiens allele as B*.

BIOLOGY:CONCEPTS+APPL.(LOOSELEAF)

10th Edition

ISBN:9781305967359

Author:STARR

Publisher:STARR

Chapter13: Patterns In Inherited Traits

Section: Chapter Questions

Problem 6GP: In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele...

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Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?
As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny
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?
. In 1919, Calvin Bridges began studying an X-linkedrecessive mutation causing eosin-colored eyes inDrosophila. Within an otherwise true-breedingculture of eosin-eyed flies, he noticed rare variantsthat had much lighter cream-colored eyes. By intercrossing these variants, he was able to make a truebreeding cream-eyed stock. Bridges now crossedmales from this cream-eyed stock with true-breedingwild-type females. All the F1 progeny had red (wildtype) eyes. When F1 flies were intercrossed, the F2progeny were 104 females with red eyes, 52 maleswith red eyes, 44 males with eosin eyes, and14 males with cream eyes. Assume that thesenumbers represent an 8:4:3:1 ratio.a. Formulate a hypothesis to explain the F1 and F2results, assigning phenotypes to all possiblegenotypes.b. What do you predict in the F1 and F2 generations if the parental cross is between truebreeding eosin-eyed males and true-breedingcream-eyed females?c. What do you predict in the F1 and F2 generationsif the parental cross is…
Professor John decided to breed bunnies during stay-at-home because why not. She performed the following cross:Parents:brown fur, gray nose x white fur, gray noseOffspring:3/8 white fur, gray nose1/8 white fur, white nose3/8 brown fur, gray nose 1/8 brown fur, white noseBased on the phenotypes of the offspring, what are the genotypes of the parents? Make sure you define your genotypes (ie B is brown fur, b is white fur)
Manx cats have no tail. This is caused by a dominantmutation M, which is embryo lethal when homozygous. An unlinked gene B controls the color of the fur onthe cat's tail. The dominant B allele produces black fur on tails, while the recessive allele b produceswhite fur on tails. A Manx cat that is B/B is crossed to a white-tailed cat. Half of the F1 progeny hadthe Manx phenotype and the other half had normal, black tails.The F1 cats with the Manx phenotype were crossed to each other, producing F2 cats. What is the ratioof the following phenotypes in the F2
Albino rabbits (lacking pigment) are homozygous forthe recessive c allele (C allows pigment formation).Rabbits homozygous for the recessive b allele makebrown pigment, while those with at least one copy ofB make black pigment. True-breeding brown rabbitswere crossed to albinos, which were also BB. F1 rabbits, which were all black, were crossed to the doublerecessive (bb cc). The progeny obtained were 34black, 66 brown, and 100 albino.a. What phenotypic proportions would have beenexpected if the b and c loci were unlinked?b. How far apart are the two loci?
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. A snapdragon plant that bred true for white petals wascrossed with a plant that bred true for purple petals, andall the F1 had white petals. The F1 was selfed. Among theF2, three phenotypes were observed in the followingnumbers:white 240solid purple 61spotted purple 19Total 320a. Propose an explanation for these results, showinggenotypes of all generations (make up and explain yoursymbols).b. A white F2 plant was crossed with a solid purple F2plant, and the progeny werewhite 50%solid purple 25%spotted purple 25%What were the genotypes of the F2 plants crossed?
J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore. 1943. Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. Parent phenotypes Progeny phenotypes burnsi × burnsi 39 burnsi, 6 pipiens burnsi × pipiens 23 burnsi, 33 pipiens burnsi × pipiens 196 burnsi, 210 pipiens a. On the basis of these results, is the burnsi phenotype most likely inherited as a dominant trait or as a recessive trait? b. Give the most likely genotypes of the parent in each cross (use B for the burnsi allele and B+ for pipiens allele). c. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes.
Given a cross between these two parents: BbCcDDEEFfGgHhjj x BbCcddEeFfGgHHjj where traits B, E, H, J exhibit Mendelian inheritance; C, and F exhibit incomplete dominance; D, and G exhibit co-dominance answer the following: How many phenotypes can be expected from the cross?
. In the tiny model plant Arabidopsis, the recessive allele hyg confers seed resistance to the drug hygromycin, and her, a recessive allele of a different gene, confers seed resistance to herbicide. A plant that was homozygous hyg/hyg ⋅ her/her was crossed with wild type, and the F1 was selfed. Seeds resulting from the F1 self were placed on petri dishes containing hygromycin and herbicide.a. If the two genes are unlinked, what percentage of seeds are expected to grow?b. In fact, 13 percent of the seeds grew. Does this percentage support the hypothesis of no linkage? Explain. If not, calculate the number of map units between the loci.c. Under your hypothesis, if the F1 is testcrossed, what proportion of seeds will grow on the medium containing hygromycin and herbicide?