Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 21, Problem 14P
The equation p2 + 2pq + q2> = 1 representing the Hardy-Weinberg proportions examines genes with only two alleles in a population.
a. | Derive a similar equation describing the equilibrium proportions of genotypes for a gene with three alleles. [Hint: Remember that the Hardy-Weinberg equation can be written as the binomial expansion (p + q)2 .] |
b. | A single gene with three alleles (IA, IB, and i) is responsible for the ABO blood groups. Individuals with blood type A can be either IA IA or IA i; those with blood type B can be either IB IB or IB i; people with AB blood are IA IB, and type O individuals are ii. Among Armenians, the frequency of IA is 0.360, the frequency of IB is 0.104, and the frequency of i is 0.536. Calculate the frequencies of individuals in this population with the four possible blood types, assuming Hardy-Weinberg equilibrium. |
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The equation p2+ 2pq + q2= 1 representing theHardy-Weinberg proportions examines genes withonly two alleles in a population.a. Derive a similar equation describing the equilibrium proportions of genotypes for a gene withthree alleles. [Hint: Remember that the HardyWeinberg equation can be written as the binomialexpansion (p + q)2.]b. A single gene with three alleles (IA, IB, and i) isresponsible for the ABO blood groups. Individualswith blood type A can be either IA IAor IA i;those with blood type B can be either IB IBor IB i;people with AB blood are IA IB, and type O individuals are ii. Among Armenians, the frequency of IAis0.360, the frequency of IBis 0.104, and the frequencyof i is 0.536. Calculate the frequencies of individuals in this population with the four possible bloodtypes, assuming Hardy-Weinberg equilibrium.In Problems 15–17, you will see that because matingbetween individuals within populations at Hardy-Weinbergequilibrium is random, it is possible to predict…
What is the expected genotype frequency of the heterozygous genotype under the Hardy-Weinberg equation P = 0.7?
"he equation p + 2pq + q° = 1 representing the
lardy-Weinberg proportions examines genes with
nly two alleles in a population.
. Derive a similar equation describing the equilib-
rium proportions of genotypes for a gene with
three alleles. [Hint: Remember that the Hardy-
Weinberg equation can be written as the binomial
expansion (p + q)*.]
A single gene with three alleles (r^, 1", and i) is
responsible for the ABO blood groups. Individuals
with blood type A can be either rr or ^ i; those
with blood type B can be either 1" 1 or 1" i; people
with AB blood are A ", and type O individuals are
ii. Among Armenians, the frequency of is 0.360,
the frequency of " is 0.104, and the frequency of i
Chapter 21 Solutions
Genetics: From Genes to Genomes
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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
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- Pretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem. So you hypothesize that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations). If you carry out chi-square goodness of fit test and calculate a total chi-square value of 0.03 with 1 degree of freedom (see table), what does this mean?arrow_forwardin the Hardy-Weinberg equation, what do the terms p2, q2, and 2pq represent, in terms of the genetic structure of a population?arrow_forwardExample 3.5.7 Fish Vertebrae Consider the distribution of vertebrae given in Table 3.5.1. In Exam- ple 3.5.5 we found that the mean of Y is µy = 21.49. The variance of Y is VAR(Y) = of = (20 – 21.49) x Pr{Y = 20} 21) %3D + (21 – 21.49)? x Pr{Y + (22 – 21.49)? x Pr{Y = 22} + (23 – 21.49) x Pr{Y = 23} = (-1.49) x 0.03 + (-.49) x 0.51 + (0.51) x 0.40 + (1.51)² × 0.06 = 2.2201 x 0.03 + .2401 x 0.51 + .2601 x 0.40 + 2.2801 x 0.06 = 0.066603 + 0.122451 + 0.10404 + 0.136806 = 0.4299. The standard deviation of Y is ay = V0.4299 - 0.6557. %3Darrow_forward
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- 1. Consider this graph on running speed in huskies Midosspring running speed (m/s) 16 14 12 10 8 + 2 Husky running speed 4 6 8 Midparent running speed (m/s) 10 12 14 16 a. Approximately what is the heritability of running speed in this kennel of huskies? (You can approximate by eyeballing from the graph, no need to calculate the actual slope) b. If the breeder where to selectively breed the dogs, will the dogs run substantially faster in the next generation? c. What else can the breeder do to increase running speed?arrow_forward(1 point) Humans with the genotypes DD and Dd show the Rh+ blood phenotype, whereas those with the genotype dd show the Rh- blood phenotype. In a sample of 400 Basques from Spain, 230 people were Rh+ and 170 people were Rh-. Assuming that this population is in Hardy-Weinberg proportions, what is the allele frequency of the allele D? (a) (a) 0.348 (answer) (b) (b) 0.652 (c) (c) 0.425 (d) (d) 0.575 (e) (e) 0.288 2. (2 points) In the Basque population mentioned above, what proportion of the Rh+ individuals would be expected to be heterozygote? (a) (a) 0.454 (b) (b) 0.789 (answer) (c) (c) 0.516 (d) (d) 0.250 (e) (e) 0.500 How is the answer for #2, b? please explainarrow_forwardIn a large, randomly mating human population, the frequencies of the IA, IB, and i alleles are 0.7, 0.2, and 0.1, respectively. Calculate the expected frequencies for each blood type. (show solution)arrow_forward
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