Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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- A geneticist is using a three‑point testcross to map three linked Drosophila recessive mutations called a, b, and c, where a is associated with anomalous gait, b is associated with buckled wings, and c is associated with curved bristles. She first crosses homozygous anomalous, buckled flies to homozygous curved flies. Next, she testcrosses the F1 progeny to anomalous, buckled, curved flies. She obtains 1000 progeny distributed as shown. From this data, calculate the map distance between b and c. progeny curved 277 anomalous, buckled 283 wild type 5 anamalous, buckled, curved 5 buckled, curved 133 anomalous 127 anomalous, curved 86 buckled 84arrow_forwardA geneticist is using a three-point testcross to map three linked Drosophila recessive mutations called a, b, and c, where a is associated with anomalous gait, b is associated with buckled wings, and c is associated with curved bristles. She first crosses homozygous anomalous, buckled flies to homozygous curved flies. Next, she testcrosses the F₁ progeny to anomalous, buckled, curved flies. She obtains 1000 progeny distributed as shown. From this data, calculate the map distance between b and c. Testcross progeny phenotype curved anomalous, buckled wild type anamalous, buckled, curved buckled, curved anomalous anomalous, curved buckled Number 277 283 4 6 133 127 83 87 map distance: m. u.arrow_forwardA geneticist is using a three-point testcross to map three linked Drosophila recessive mutations called a, b, and c, where a is associated with anomalous gait, b is associated with buckled wings, and c is associated with curved bristles. She first crosses homozygous anomalous, buckled flies to homozygous curved flies. Next, she testcrosses the F1 progeny to anomalous, buckled, curved flies. She obtains 1000 progeny distributed as shown. From this data, calculate the map distance between b and c. Testcross progeny phenotype Number curved 281 map distance: m. u. anomalous, buckled 279 wild type 4 anamalous, buckled, curved buckled, curved 128 anomalous 132 anomalous, curved 83 buckled 87arrow_forward
- You ave established a mutant line of flies with a balancer chromosome. The balancer chromosome includes GFP and a mutation in gene X that causes flies to be sterile in the homozygous state. The recessive mutation that you are studying is in gene Y, and causes flies to develop two tail ends in the homozygous state. This mutation is lethal in the homozygous state. For each of the following phenotypes in the balanced mutant line, what alleles are present and in how many copies? a.) Glows green, fertile (Select] b.) Glows green, sterile [Select) c.) does not glow green, two tails (Select] If you crossed a fly from your balanced line with the following flies, what proportion of the offspring would glow green and be fertile? d.) wild-type [Select ) e.) fly with balancer chromosome from parental generation [ Select ]arrow_forwardDuring the study of your genetics course, you are learning the different developmental stages of a unique Tsetse fly, named defsis-timen, and you discover differences in their developmental pattern i.e. unusual and normal...... Rest of the question is in picture Answer part b and c And in part c its unlinked genes not unlinedarrow_forwardRed eyes (r), brown color (b), and curled wings (c) are 3 recessive mutations that occur in house flies. These genes have already been mapped: r-b = 12 mu b-c = 18 mu r-c = 30 mu A fly with red eyes, brown color, and curled wings is crossed with a fly homozygous for the wild-type traits. The F1 males are crossed with females with the three mutant traits and 2000 progeny are produced. Find the number of observed double crossovers if the interference is 0.38. O 43 O 62 O 27 O 16arrow_forward
- A female Drosophila fly is heterozygous for three recessive pigmentation mutations called pk, wh, and tp. pk is associated with pink eyes, wh is associated with a white body, and tp is associated with transparent wings. A geneticist crosses this fly to a male pink, white, and transparent fly and obtains the 1000 progeny given in the table. Phenotype Number Phenotype Number white 47 transparent 199 wild type 200 white, transparent 49 pink, transparent 46 pink, white, transparent 206 48 pink, white 205 pink Based on the table, which genes are linked? O pk and wh wh and tp O pk and tp Calculate the distance between the linked genes.arrow_forwardThe data set attached presents the results of a testcross using female flies heterozygous for three traits and male flies, which are homozygous recessive. For simplicity, mutant alleles are shown with letters a, b, and c and wildtype alleles are indicated by a “+” symbol. For this part of the report do the following in order: a) Determine the gene order (which gene is in the middle?)d) Construct a genetic map for the three genes, including the map distances between them. Clearly indicate the logic you followed and show all your calculations. Include the full distance calculations for the two most distanced genes (do not just add the other 2 distances). Ensure the work is neat and clear and does not contain spelling or grammatical errors so that it is understandable. Make sure to double check the solution provided.arrow_forwardWhite fruit color in summer squash is dependent on a dominant gene (W) and colored fruit to the recessive gene (w). In the presence of ww a dominant gene D produces yellow color, but when D is not present, the color is green. Upload your solution in the google classroom. Give the: F1_phenotypes_expected_from_crossing_white_fruited_WWDD_with_green_fruited_wwdd_plant F2_phenotypes_expected_from_crossing_the_F1 F2_phenotypic_ratio_expected_from_crossing_the_F1 F2_genotypic_ratio_expected_from_crossing_the_F1 Type_of_epistatis Write all phenotypes in CAPITAL letters SEPARATED BY COMMA. Write all ratio in numerical values without spacesarrow_forward
- Vestigial wing is a recessive autosomal mutation in the fruit fly Drosophila melanogaster. Wild-type flies have red eyes, but another mutation, an X-linked recessive mutation causes white eyes. You cross a parental generation of males with vestigial wings with females that have white eyes. 1) What phenotypic ratio would you expect in the F1 generation from such a cross? What phenotypes do you observe in these flies? (e.g. males with red eyes and wild-type wings). 2) What phenotypic ratio would you expect in the F2 generation? What phenotypes do you observe in these flies?arrow_forwardIn Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1arrow_forwardSuppose a geneticist uses a three-point testcross to map three linked Drosophila recessive mutations called f, z, and n. Gene fis associated with abnormally fast movement, z is associated with a zigzag movement pattern, and n is associated with narrow wings. The geneticist first crosses homozygous narrow flies to homozygous fast, zigzag flies. Next, he testcrosses the F₁ progeny to fast, zigzag, narrow parents and obtains the results reported in the table. Based on the data, select the order of the three genes. fzn zfn nzf fnz znf nfz F₁ Testcross Progeny Phenotype narrow fast, zigzag zigzag fast, narrow fast zigzag, narrow fast, zigzag, narrow wild type Number 627 621 209 229 215 223 6 7arrow_forward
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