Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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- a. In Drosophila, crosses between F1 heterozygotes ofthe form A b / a B always yield the same ratio ofphenotypes in the F2 progeny regardless of the distance between the two genes (assuming completedominance for both autosomal genes). What is thisratio? Would this also be the case if the F1 heterozygotes were A B / a b? (Hint: Remember that inDrosophila, recombination does not take placeduring spermatogenesis.)b. If you intercrossed F1 heterozygotes of the formA b / a B in mice, the phenotypic ratio among the F2progeny would vary with the map distance betweenthe two genes. Is there a simple way to estimate themap distance based on the frequencies of the F2phenotypes, assuming rates of recombination areequal in males and females? Could you estimatemap distances in the same way if the mouse F1heterozygotes were A B / a b?arrow_forwarddd-ons Help в I U A Calibri 12 三 三1 |:三 6. Consider a guinea pig with a homozygous genotype and a white fur color phenotype. a. What is the probability this parent will produce a gamete with the dominant allele? b. What is the probability this parent will produce a gamete with the recessive allele? C. If 31 sperm cells are collected from this guinea pig, how many would you expect to have the recessive allele (as determined by sequencing the gene)? !!!arrow_forwarduh ec CV + + 10.5 SC 9.1 scute bristles echinus eyes 9.2 ct + crossveinless wings Table 1: phenotype wild-type tapdance feet crossveinless wings tapdance & crossveinless cut wings 15.9 vermilion eyes V + + 66.8 Drosophila X chromosome Use the map provided above for problems 1 & 2. Problem 1: 11.2 10.9 garnet eyes M A new gene is being investigated in fruit flies. The recessive allele of this gene (t) causes the flies' feet to grow tiny tapdance shoes, while the dominant allele (t*) permits wild-type feet to develop. Preliminary studies indicate that this new gene is located on the X-chromosome. You decided to perform a two-point testcross to determine its position relative to the well-established crossveinless wings gene (cv). You cross a female heterozygous for both genes with a testcross male fly and obtain the male offspring results shown in table 1, below. Using this information, answer the following questions: # male offspring 13 405 401 11 forked bristles a) is the original…arrow_forward
- 4) In frost moths, two alleles of one gene determine the character difference of spotted versus striped wings and two alleles of a separate, independent gene determine the character difference of orange wing background versus white wing background. The results for four matings of moth phenotypes are shown in the image attached. a) Assign the letter “s” to the wing pattern gene and letter “w” to the background color gene. Write in the capital letter for the dominant phenotype and the lower case letter for the recessive phenotype. Also, write whether each allele is dominant (D) or recessive (R). Wing pattern gene: Spotted - letter assignment:____ - dominant or recessive:____ Striped - letter assignment:____ - dominant or recessive:____ Background color gene: orange: - letter assignment:____ - dominant or recessive:____ white: - letter assignment:____ - dominant or recessive:_____ b) Based on the four matings in Question 4: What are the genotypes of each parent in each cross? If more…arrow_forward7. In humans, the alleles for blood type are designated I^ (A-type blood), I ^ B (B-type blood) and i (O-type blood). What are the expected frequencies of phenotypes in the following matings? Draw a Punnett square showing the results for a). a) heter A x heter B: %A %B %0 %AB b) I ^ A * I ^ B * I ^ A_{i} : c) I ^ A * I ^ A * I ^ B * I ^ B : d) ABO :arrow_forward6.4 Write genotypes correctly to indicate if two genes are linked or not and if linked, the arrangement of the alleles on homologues.arrow_forward
- 53arrow_forward3.9 How many different kinds of F, gametes, F, genotypes, and F, phenotypes would be expected from the following crosses: (а) АА Х аа; (b) АА ВВ Х aa b; (c) АA ВВ СС Х аа bb cc? (d) What general formulas are suggested by these answers? ANS: F, Gametes F, Genotypes F, Phenotypes (a) 2 2 (b) 2 × 2 = 4 3 X 3 = 9 2 X 2 = 4 (c) 2 × 2 × 2 = 8 3 × 3 × 3 = 27 2 × 2 × 2 = 8 2", where n is the number of genes (d) 2" 3" 3.arrow_forwardThank you.arrow_forward
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