Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 20, Problem 5P
Thinking creatively about evolutionary mechanisms, identify at least two schemes that could generate allelic polymorphism in a population. Do not include the processes described in the answer to Problem
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Albinism is a rare genetic condition, occurring in only one in every 17,000 to 22,000individuals in the world (Campbell et al. 2003; Gronskov et al. 2007). Conduct a library oronline research and answer these questions:
A. What can you conclude about the allelic frequency of the a allele globally?
B. Where do you think the a allele would be more commonly found, among theheterozygotes (Aa) or the homozygote recessives (aa)? Why do you say so?
C. What happens to an individual who is albino (aa) or who has very low or lacksmelanin pigments in the hair, skin and eyes?
D. Based on your answer in c, how would you explain the frequency of the a allele?
Albinism is a rare genetic condition, occurring in only one in every 17,000 to 22,000individuals in the world (Campbell et al. 2003; Gronskov et al. 2007). Conduct a library oronline research and answer these questions:a. What can you conclude about the allelic frequency of the a allele globally?b. Where do you think the a allele would be more commonly found, among theheterozygotes (Aa) or the homozygote recessives (aa)? Why do you say so?c. What happens to an individual who is albino (aa) or who has very low or lacksmelanin pigments in the hair, skin and eyes?d. Based on your answer in c, how would you explain the frequency of the a allele?
Explain why you need both the multiplication and addition rules to calculate the
probability that an MN offspring will be produced in a human population
The multiplication rule gives the probability that a particular mating will occur
and that mating will produce the target genotype, and the addition rule gives the
total probability of each target genotype, because the matings that produce the
target genotype are mutually exclusive.
The multiplication rule gives the probability that a particular mating will occur,
and the addition rule gives the probability that the mating will produce the
target genotype.
The addition rule gives the probability that a particular mating will occur and
that mating will produce the target genotype, and the multiplication rule gives
the total probability of each target genotype.
The multiplication rule gives the probability that a particular mating will produce
the target genotype, and the addition rule gives the probability of obtaining the
particular…
Chapter 20 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 20 - 20.1 Compare and contrast the terms in each of the...Ch. 20 - In a population, what is the consequence of...Ch. 20 - 20.3 Identify and describe the evolutionary forces...Ch. 20 - Describe how natural selection can produce...Ch. 20 - Thinking creatively about evolutionary mechanisms,...Ch. 20 - 20.6 Genetic drift, an evolutionary process...Ch. 20 - Over the course of many generations in a small...Ch. 20 - Catastrophic events such as loss of habitat,...Ch. 20 - 20.9 George Udny Yule was wrong in suggesting that...Ch. 20 - 20.10 The ability to taste the bitter compound...
Ch. 20 - Figure 20.6 illustrates the effect of an ethanol ...Ch. 20 - 20.12 Biologists have proposed that the use of...Ch. 20 - 20.13 Two populations of deer, one of them large...Ch. 20 - 20.14 Directional selection presents an apparent...Ch. 20 - 20.15 What is inbreeding depression? Why is...Ch. 20 - 20.16 Certain animal species, such as the...Ch. 20 - Genetic Analysis 20.1 predicts the number of...Ch. 20 - 20.18 In a population of rabbits, and . The...Ch. 20 - Sickle cell disease (SCD) is found in numerous...Ch. 20 - 20.20 Epidemiologic data on the population in the...Ch. 20 - The frequency of tasters and nontasters of PTC...Ch. 20 - Tay-Sachs disease is an autosomal recessive...Ch. 20 - 20.23 Cystic fibrosis (CF) is the most common...Ch. 20 - 20.24 In the mouse, Mus musculus, survival in...Ch. 20 - 20.25 In a population of flowers growing in a...Ch. 20 - Assume that the flower population described in the...Ch. 20 - 20.27 ABO blood type is examined in a Taiwanese...Ch. 20 - 20.28 A total ofmembers of a Central American...Ch. 20 - 20.29 A sample offield mice contains individuals...Ch. 20 - Prob. 30PCh. 20 - Albinism, an autosomal recessive trait...Ch. 20 - 20.32 The frequency of an autosomal recessive...Ch. 20 - 20.33 Evaluate the following pedigree, and answer...Ch. 20 - Evaluate the following pedigree, and answer the...Ch. 20 - The following is a partial pedigree of the British...Ch. 20 - Draw a separate hypothetical pedigree identifying...Ch. 20 - Prob. 37PCh. 20 - 20.38 Achromatopsia is a rare autosomal recessive...Ch. 20 - 20.39 New allopolyploid plant species can arise by...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Based on this information (picture) A. What is the probability that a randomly sampled individual from the population has two copies of the a allele (that is, that it has an aa genotype)? B. What is the probability that both members of a randomly sampled married couple (man and woman) are aa at the asparagus-smelling gene? C. What is the probability that both members of a randomly sampled married couple (man and woman) are heterozygotes at this locus (meaning that each person has one allele A and one allele a)? D. Consider the type of couple described in (c). What is the probability that the first child of such a couple also has one A allele and one a allele (is a heterozygote)? Remember that the child must receive exactly one allele from each parent.arrow_forwardAssume an ideal diploid population of size 2N=32 (Just like the Buri Drosophila drift experiment with 16 individuals, or 32 gene copies, every generation). a) What is the probability that a neutral allele present in exactly 16 copies will CHANGE BY FEWER than 2 copies in the next generation? Be sure you understand that this includes cases where the number of copies remains unchanged, as well as the cases (plural!) where the magnitude of change between generations is one. b) What is the ultimate probability of fixation for this allele by drift? c) In the same population, what is the probability that a neutral allele present in exactly 2 copies will change by fewer than 2 copies (same meaning as above) in the next generation? d) What is the ultimate probability of fixation for this allele by drift?arrow_forwardWhich of the following statements does NOT apply to the Hardy-Weinberg expression: p2 + 2pq + q2? Group of answer choices p2 is the frequency of individuals with the homozygous recessive genotype. 2pq is the frequency of individuals with the heterozygous genotype. It can be used to determine the genotype and allele frequencies of the previous and the next generations. Knowing either p2 or q2, you can calculate all the other frequenciesarrow_forward
- What evolutionary factors can cause allele frequencies to change and possibly lead to a genetic polymorphism? Discuss the relative importance of each type of process.arrow_forwardA total of 1000 members of a Central American population are typed for the ABO blood group. In the sample, 421 have blood type A, 168 have blood type B, 336 have blood type O, and 75 have blood type AB. Part A Use this information to determine the frequency of ABO blood group alleles in the sample. Recall that when considering genes with three alleles whose frequencies are represented by the variables p, q, and r, the sum of genotype frequencies resulting from trinomial expansion is: (p+q+r)² =p² + 2pq+q2+2pr+r²+2gr = 1arrow_forwardln a population of turtles, there are yellow-green shells and green shells. The yellow shells are caused by a homozygous recessive gene and the green shells are caused by the dominant gene. Given the following data:AA = 340Aa = 270aa = 120 a) Calculate p and q. b) Use a chi square test to determine if these alleles are in Hardy-Weinberg equilibrium. Submit your answer as a pdf or doc file. Show your workarrow_forward
- What is the mathematical expression of the genetic equilibrium for genes with two alleles? Is this statistical distribution the same as the statistical distribution of the respective phenotypes?arrow_forwardTransmission of genetic information from parent to offspring follows the rule of probability. The possible genotypes of offspring are dependent on the alleles present in parent organism. The lake could of these possibilities can be represented as a fraction(1/4), decimal (0.25), or percentage (25%). Probability ranges between 0% and 100%. When predicting the likelihood of a combination of possibilities the individual probabilities may be added together and multiplied depending on the type of combination of calculating the probability of possibly A or possibility A and  possibility B The two probabilities are multiplied together. These rules continue to apply for calculating for more than two possibilities.  A) given the probabilities of the events listed above what is the probability of X and Y and Z? P(X)= 0.5, P(Y)= 0.25, P(Z)= 0.25. B) what is the probability of X and Y but not Z? arrow_forwardplease tell us the expected average time to fixation of an allele at frequency p = 0.5 in a population of 100 individuals, we can safely assume that the allele does become fixed?arrow_forward
- Imagine a locus with two alleles. Mutation at this locus changes one allele to the other (i.e. it does not create a new allele). If the rate of mutation from allele 1 to allele 2 is 0.00005 and the rate of mutation from allele 2 to allele 1 is 0.01, what is the equilibrium frequency of allele 2 (Give your answer to 5 decimal places)?arrow_forwardIn the F2 generation, 306 rabbits with red eyes and 71 with a white eye phenotype suppose the calculated x2 value is 0.35. Find the x2 range using the distribution chart. What is the p-value range? using these information do you accept or reject the null hypothesis? The distribution chart is attached below.arrow_forwardSuppose there is an autosomal locus of 2 alleles, A1 and A2, with probabilities (frequencies) p1 and p2, and the genotype probabilities (frequencies) are P(A1A1) = p1*p1, P(A1A2) = 2*p1*p2, and P(A2A2) = p2*p2, respectively. Prove the Hardy-Weinberg Law, i.e., after one generation of random mating, the genotype probabilities (frequencies) in the offspring are also P(A1A1) = p1*p1, P(A1A2) = 2*p1*p2, and P(A2A2) = p2*p2. Hint: List all possible combinations of random mating. Then list the probabilities of the resulting genotype probabilities (frequencies) in the offspring. Combine the probabilities of random mating and resulting genotype probabilities (frequencies) in the offspring.arrow_forward
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