Answer the multiple-choice questions based on the following crosses:
Parental generation: True-breeding flies with red eyes and long wings were crossed to flies with white eyes and miniature wings. All
The
129 red eyes, long wings
133 white eyes, miniature wings
71 red eyes, miniature wings
67 white eyes, long wings
What is/are the
a. Red eyes, long wings
b. White eyes, miniature wings
c. Red eyes, long wings and white eyes, miniature wings
d. Red eyes, miniature wings and white eyes, long wings
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Genetics: Analysis and Principles
- Please answer fastarrow_forwardUse the following information to answer the next four questions. Tomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Part A: Record the phenotypes and the phenotype ratio in lowest terms. Part B: Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these…arrow_forwardUse the following information to answer the next four questions. Tomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Part A: Record the phenotypes and the phenotype ratio in lowest terms. Part B: Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these…arrow_forward
- What is the predicted phenotypic ratio for the offspring of the following cross? Allele phenotypes: R = red, r = white, T = thin leaf, t = broad leaf Cross: RR; Tt x Rr; tt Red with thin leaves ___________ Red with broad leaves ___________ White with thin leaves ___________ White with broad leaves __________ PLEASE show work and give an explanation... Thank you!arrow_forwardGive only typing answer with explanation and conclusion A true-breeding pea plant with round and green seeds was crossed with a true-breeding plant with wrinkled and yellow seeds. Round and yellow seeds are the dominant traits. The F1 plants were allowed to self-fertilize. What are the following probabilities for the F2 generation? a. An F2 plant with wrinkled, yellow seeds. b. Two out of three F2 plants with round, yellow seeds and one with wrinkled, green seeds. c. Four F2 plants in the following order: one has round, yellow seeds; one has round, green seeds; and two have wrinkled, green seeds. d. An F2 plant will not have round, yellow seeds.arrow_forwardTable 1: F1 ebony flies - 0 F1 non-ebony flies - 560 F1 stubble flies - 560 F1 non-stubble flies - 0 The researcher collects and crosses male and female flies from the F1 generation. In the resulting offspring, F2, there are both stubble and ebony flies. Draw a Punnett Square to illustrate the F1 cross for the stubbly phenotype showing the individual gametes of each parent and the combination in the resulting offspring.arrow_forward
- Consider the data in the following table, derived from crossing a female fruit fly that was heterozygous for the sex-linked recessive loci, cut wings (ct), raspberry eyes (ra), and forked (f) bristles, with a wild-type male. Female progeny were all wild-type. Male progeny are listed below. Phenotype Number of males ct ra f 70 + ra f 76 + + + 74 ct + f 356 + ra + 345 ct ra + 3 ct + + 72 + + f 4 Total 1,000 What is the map distance between cut wings and raspberry eyes? 29.2 15.1 30.6 15.5arrow_forwardA genetic cross was made between two true-breeding parental fly strains. One parent strain displays red eyes and curly wings, while the other parent strain displays white eyes and vestigial wings. The resulting F1 generation flies were allowed to self-cross to produce F2 generation. The following data were obtained: F1 generation: all have red eyes and curly wings. F2 generation: 295 were red eyes and curly wings, 95 were red eyes and vestigial wings, 89 were white eyes and curly wings, 31 were white eyes and vestigial wings With this data above – please answer the following questions: Propose a hypothesis that explains the above inheritance pattern, test the goodness of fit between the data and your hypothesis using a chi square test. Use the chi square table from your textbook/online. Use the information obtained from the chi square table to explain what your calculated chi square results mean.arrow_forwardPhenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed to homozygous light eyed, straw bristled males: Phenotype Number light-straw 140 wild-type 160 light 360 straw 340 Total 1000 Compute the map distance between the light and straw loci. Group of answer choices 70 map units 3 map units 7 map units 0.03 map units 30 map unitsarrow_forward
- The image shows a pair of homologous chromosomes from a single parent before gamete production. M1 and M2 are maternal chromosomes, while P1 and P2 are paternal chromosomes. Two traits are shown: D represents seed color (D – green, d – yellow), while F represents flower color (F – purple, f – white). These two traits follow the patterns of basic Mendelian genetics. During crossing-over between the M2 F allele and the P1 f allele, a mutation occurred and the portion of P1 did not reattach to the chromosome. Which of the following explains what would happen to the proportion of white flowers in a population resulting from this mutation? A - There would be an increase in the proportion of white flowers because the f allele is distributed to more gametes. B - There would be a decrease in the proportion of white flowers because the f allele is not distributed to as many gametes. C - There would be an increase in the proportion of white flowers because the f allele would not be masked by the…arrow_forwardWhat would the punnett square look like for the question below? If an F1 eggplant in figure 5.2 is used in a testcross, what proportion of the progeny from this cross will be white?arrow_forwardThe images attached show a parental cross that is homozygous wild female x white male. F1 were intercrossed to produce the F2 generation as indicated below: Wild Female: 416 Wild Male: 192 White Female: 0 White Male: 192 A chi-square analysis was done and an image was also attached. Please discus what conclusions can be made based on the data/findings.arrow_forward
- Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College