OMutation-selectionbalanceSuppose that one allele A₁ mutates to another allele A2 at some rate, μ.Suppose as well that A₁ is dominant over A2 such that A₁A1 and A₁A2 bothhave the same fitness, but that individuals that are homozygous recessive(A2A2) for the mutant allele A2 are less fit than the dominant genotype bysome amount s, the selection coefficient. In this case, A2 mutant alleles comeinto the population at rate µ, and are removed from the population onlywhen the show up in homozygous genotypes. The gory mathematical proofcan be found in Box 7.8, which tells us that:O■Example: Suppose A₁ mutates to A2 at rate 0.005, but A₂A2 homozygousrecessives are 50% less fit (s = 0.5) than either A₁A1 or A₁A2. What are theexpected equilibrial abundances of A₁ and A2?μ = 0.005S = 0.5p* = 1-sqrt(µ/s) = 1-sqrt(0.005/0.5) = 0.9= sqrt(µ/s) = sqrt(0.005/0.5) = 0.1●*p = = 1-sqrt(µ/s)q* = sqrt(μ/s)●Question: Suppose A₁ mutates to A2 at rate 0.01, but A₂A2 homozygousrecessives are 40% less fit (s = 0.4) than either A₁A1 or A1A2. What are theexpected equilibrial abundances of A₁ and A₂?8

Hardy Weinberg Equilibrium explains how allele and gene frequency remain same in the population from…

Answered: O ■ Example: Suppose A₁ mutates to A2…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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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?
Product Rule Given the following traits that the parents have, assign the letters H for hair type, E for eyelids, and L for lip color. Parent 1: Homozygous curly hair Homozygous short eyelids Homozygous pale lips (dominant) Genotypes Probability (fraction) (Parent 1) Parent 2: Heterozygous straight hair Heterozygous long eyelids Homozygous red lips (recessive) Probability (fraction) (Parent 2) Probability (fraction) 3. What is the probability (in percentage) that their child will inherit the following genotypes? Homozygous curly hair, Homozygous short eyelids, Heterozygous pale lips Probability= %
dd-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)? !!!
uh 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…
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…
7. 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 :
6.4 Write genotypes correctly to indicate if two genes are linked or not and if linked, the arrangement of the alleles on homologues.
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3
What type of linkage is shown here? CH₂OH H K ОН H ОН H O Alpha (1,4) O Beta (1,2) Beta (1,6) O Alpha (1,2) O Alpha (1,6) ان O Beta (1,4) ОН CH₂OH ОН ОН H H CH2OH
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3.9 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.

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