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Allele frequency question

Allele frequency question


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You are heterozygous at locus A. Both of your alleles, A1 and A2, have a 10% allele frequency in the population. If you are one of 22 people in a room, how many people do you expect to share a locus A allele with?

I come up with 6.8, working like so:

22 people = 44 locus A alleles in the room = 4.4 A1 alleles and 4.4 A2 alleles at 10% frequency.

I have one of each, leaving 3.4 of each allele to the rest of the room. So, looking at both alleles, there are 6.8 shared alleles out there.

My answer assumes that these alleles are all in different people - I don't know how to account for the probability of someone else having two of them. So I know I am wrong already. Is that last step all I'm missing, or have I gone wrong in a more fundamental way?

(Context: This was a question on a grad-level immunology exam. None of us students could agree on the answer afterwards, and the confused wording (I rephrased it above) made us strongly suspect the professor doesn't have the right answer either, so we wanted to see what the answer actually is for when the exams are graded.)


Same as @bpedit but stated a little bit differently.

You are heterozygous at locus A. Both of your alleles, A1 and A2, have a 10% allele frequency in the population. If you are one of 22 people in a room, how many people do you expect to share a locus A allele with?

There are 5 possible genotypes that could share at least one allele with "you"A1|A1,A1|A2,A2|A2,A1|A3andA2|A3, whereA3is the allele (or set of alleles) which frequency is $0.8$. Under Hardy-Weinberg assumptions, the frequencies of each of these genotypes are

  • $f(A1|A1) = 0.1^2$
  • $f(A1|A2) = 2 cdot 0.1^2$
  • $f(A2|A2) = 0.1^2$
  • $f(A1|A3) = 2cdot 0.1cdot 0.8$
  • $f(A2|A3) = 2cdot 0.1cdot 0.8$

which sums to $4cdot(0.1^2) + 2(2cdot 0.1cdot 0.8)=0.36$.

As the frequencies of bothA1andA2are $0.1$, the frequencies of each of the three mentioned genotypes are all $0.1^2$. Therefore, the expected number of individuals that share at least one allele with "you" at this locus is $21cdot 0.36 = 7.56$ people.

Why did we need Hardy-Weinberg assumption to make this calculation?

This section comes in reaction to @Atwo comment below.

To give you the intuition for why you need H-W for this calculation. Imagine allA1where in heterozygousA1|A2(and therefore same forA2), then the only other genotype that would share an allele with you areA1|A2genotypes that would be present at frequency $0.2$. And therefore the expected number of people the share at least one allele with you in the room would be $21cdot 0.2 = 4.2$ people.

For exercise, I would like to invite the reader you to repeat the same calculation for a case where allA1andA2alleles are only found in heterozygousA1|A3andA2|A3.


My first inclination for a problem like this is to address it by enumerating all the possible permutations the two alleles a random individual could have, and multiplying the two allele probabilities (frequencies) to get the combination's probability. Using the given frequencies, the remaining allele (set), call it $A3$, has freqency 0.8. Because the two alleles of an individual are independent, there are nine possible permutations: one of ${A1,A2,A3}$ for the first allele and one of ${A1,A2,A3}$ for the second. One can then just sum the probabilities for the cases of interest.

But for this particular problem, there is an easier approach. What we are interested in is all individuals who have at least one A1 or A2 allele. This is the same, though, as all individuals except those who have no A1 nor A2 alleles, or, stated another way, those having A3 and A3. Therefore:

  • $f(A3|A3) = 0.8cdot 0.8 = 0.64$
  • $f( not(A3|A3) ) = 1 - f(A3|A3) = 0.36$
  • $expected( not(A3|A3) ) = 21cdot f( not(A3|A3) ) = 7.56$

This is of course the same as noting that in the first exhaustive enumeration that there is just one case excluded from the group of interest, calculating its probability, and knowing that all the others must sum to the complement of this probability. Sometimes using the complementary set out of the full enumeration is an easier way to the answer.

Note: Remi's answer shows a slightly consolidated form of the exhaustive enumeration approach. The earlier numerical discrepancy in that answer was just due to arithmetic or typos.


I've never done one like this, here goes.

First, remove yourself from the room, just 21 people to consider. You're right that you have to account for the homozygotes. Let's represent the A1 allele with p. All other allele frequencies represented by q. (Sorry I don't yet know how to format properly)

p = 0.1, q = 0.9

The genotype frequencies in the population are represented by:

p^2 + 2pq + q^2 = 1

We're interested in the p^2 + 2pq part. It represents the genotype probabilities of having either one or two of the A1 allele.

0.01 + 2(0.1)(0.9) = 0.19

That's the frequency of individuals in the population with at least one A1 allele.

0.19 * 21 people = 3.99 people

Ditto for the number of people with the A2 allele. These should be additive giving you 7.98 people. BUT, Some people would have both the A1 and the A2 allele. I think we need to subtract that from the 7.98. If this is true, the calculation is easy.

2(0.1 * 0.1) = 0.02 chance of a person having both A1 and A2

The 2 is because there are two ways of achieving this (A1/A2 and A2/A1).

0.02 * 21 people = 0.42 people

7.98 - 0.21 = 7.56 other people in the room with at least one A1 of one A2 allele.


AP Biology : Population Genetics

Which of the following is NOT an assumption required for Hardy-Weinberg equilibrium?

Population size must fluctuate

No selection is occurring

Population size must fluctuate

Hardy-Weinberg states that for a population to be in equilibrium, it must not be experiencing migration, genetic drift, mutation, or selection. By this definition, population size cannot fluctuate.

Example Question #1 : Population Genetics

According to Hardy-Weinberg calculations, a population's allele frequency will remain the same from generation to generation as long as evolution is not occurring. There are five conditions that must be met for equilibrium to remain in effect in a population.

Which of the following is not a condition for Hardy-Weinberg equilibrium to remain in effect?

Nonrandom mating must occur

The population must be large

Nonrandom mating must occur

Random mating must occur in the population in order for the equilibrium to remain. If nonrandom mating occurred, allele frequency in the population would change. The alleles frequency of those mating the most would increase, while that of those mating less would decrease.

Large populations must be used to minimize the effects of genetic drift. Mustations cannot occur, as these could introduce new alleles.

It is important to note that no natural populations exist in Hardy-Weinberg equilibrium. This is simply a theoretical tool.

Example Question #1 : Population Genetics

Imagine that a population is in Hardy-Weinberg equilibrium. A certain gene presents as two different alleles, and 49% of the population is homozygous dominant.

What percentage of the population is homozygous recessive?

Further information is needed to solve the problem

When a population is in Hardy-Weinberg equilibrium, we can quantitatively determine how the alleles are distributed in the population. P 2 is equal to the proprtion of the population that is homozygous dominant based on the equation p 2 + 2pq + q 2 = 1. We also know that p + q = 1.

Since P 2 = 0.49 in this case, we know that p is equal to 0.7. Since there are only two alleles for this gene, we know that the other allele, q in this case, is 0.3. Since homozygous recessive is referred to as q 2 in the equation, we can plug in the value of 0.3 and determine that q 2 = 0.09. As a result, we confirm that 9% of the population is homozygous recessive.

Example Question #4 : Population Genetics

In a population of fruit flies, the allele for red eyes is dominant to the allele for white eyes. If 50% the population is heterozygous and 25% is homozygous for white eyes, what is the frequency of the allele for red eyes?

We must remember our two equations for allele frequency, according to Hardy-Weinberg equilibrium.

We know that, in the first equation, each term represents a total percentage of homozygotes or heterozygotes. represents the allele for red eyes and represents white.

Using the information from the question, we can solve for and .

The frequency of each allele is 0.50.

Example Question #5 : Population Genetics

The allele frequencies for a population displaying Hardy-Weinberg equilibrium were found to be dominant and recessive. What percentage of the population is homozygous dominant?

For this question we are going to need to make use of the Hardy-Weinberg equilibrium equations. The equation we need to use is:

These numbers represent the percentages of each genotype found in a given population. We were given the values of and in the question.

After plugging the numbers into the equation, we can find the value of . This value will give us the frequency of homozygous dominant individuals.

Example Question #2 : Population Genetics

A population of snails is in Hardy-Weinberg equilibrium. The snails come in two different colors: red, the dominant phenotype, and white, the recessive phenotype. There are sixteen homozygous dominant, forty-eight heterozygous, and thirty-six homozygous recessive snails.

What are the allele frequencies for this population?

We can solve this question using the Hardy-Weinberg equations:

In the second equation, corresponds to the frequency of homozygous dominant individuals, corresponds to the heterozygous frequency, and corresponds to the frequency of homozygous recessive individuals. We are given enough information to find each of these values from the question.

We can find the values of and by taking the square root of their squares.

Example Question #3 : Population Genetics

A population of snails is in Hardy-Weinberg equilibrium. The snails come in two different colors: red, the dominant phenotype, and white, the recessive phenotype. The population consists of sixty-four red snails and thirty-six white snails.

Assuming that the population is in Hardy-Weinberg equilibrium, what is the value of ?

We can solve this question using the Hardy-Weinberg equations:

is equal to the recessive allele frequency, while in the second Hardy-Weinberg equation corresponds to the frequency of the recessive phenotype.

The question tells us the number of dominant red snails and the number of recessive white snails. Using these values, we can find the frequency of the recessive phenotype.

From here, take the square root to find the value of .

Example Question #4 : Population Genetics

A population of snails is originally in Hardy-Weinberg equilibrium. The snails come in two different colors: red, the dominant phenotype, and white, the recessive phenotype. The original population has a dominant allele frequency of and a recessive allele frequency of . A new predator is introduced to the habitat that is particularly fond of the red snails. After a few years the dominant allele frequency has been reduced to .

What is the recessive allele frequency after the introduction of this predator?

Most of the information in the question is actually superfluous because we are given the final dominant allele frequency. The dominant allele frequency corresponds to the variable in the Hardy-Weinberg equations.

The question tells us that the dominant allele frequency after introduction of the predator is . Use this value in the first Hardy-Weinberg equation to solve for the recessive allele frequency, .

Example Question #9 : Population Genetics

A population is in Hardy-Weinberg equilibrium. In the population, 1% of individuals show the recessive trait for blue eyes. What is the value of in this situation?

For a population in Hardy-Weinberg equilibrium, every trait follows the equations:

In these formulas, represents the frequency of the dominant allele and represents the frequency of the recessive allele. represents the frequency of the homozygous dominant genotype, represents the frequency of the heterozygous genotype, and represents the frequency of the homozygous recessive genotype.

In this case, the individuals with blue eyes would be represented by the homozygous recessive genotype. Using this data, we can solve for the frequency of the recessive allele.

Use the frequency of the recessive allele to find the frequency of the dominant allele, .

Example Question #10 : Population Genetics

A population of beetles exists in which black coloration is dominant to white. If there are 64 black beetles in the population, what is the dominant allele frequency?

More information is required to solve

More information is required to solve

It is impossible to determine the allele frequency from the given information.

The problem only tells the number of black beetles, but does not give any information that would allow us to find the total number of beetles in the population. We do not have the homozygous recessive population or the distribution of heterozygotes and homozygous dominant beetles. Given information on the number of white beetles would allow us to calculate the recessive allele frequency, and subsequently the dominant allele frequency.

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Science Practice Challenge Questions

Consider a polymorphic gene with three alleles: A, B, and C.

A. If the frequencies of the alleles A and B are 0.2 and 0.3, the frequency of allele C is closest to ___.

Consider a gene with only two alleles: dominant A and recessive a. In a population of 1,000 organisms, the fraction expressing the homozygous recessive phenotype is 0.37.

B. The calculated allele frequencies p and q have values that are closest to ___.

C. The calculated number of individuals in this population that are heterozygotes is closest to ___.

Mountain pine beetles (Dendroctonus ponderosae) were collected from a one-acre tract of lodge pole pine trees (Pinus contorta) in a region of British Columbia where the forests are under temperature stress. The beetles were crushed, and a cellulase enzyme was extracted. Three polymorphs of the enzyme were observed when separated by gel electrophoresis. The three proteins observed correspond to alleles labeled C1, C2, and C3. The numbers of beetles with each allele are shown in the following table.

D. The calculated allelic frequencies pC1, pC2, and pC3 are closest to ___.

  1. pC1 = 0.57 pC2 = 0.57 pC3 = 0.59
  2. pC1 = 0.29 pC2 = 0.29 pC3 = 0.42
  3. pC1 = 0.61 pC2 = 0.80 pC3 = 0.59
  4. pC1 = 0.31 pC2 = 0.40 pC3 = 0.29

E. In order to investigate the presence of selection at the cellulase locus due to changing temperature, a biologist should:

  1. calculate the values of the sums pC1 + pC2 + pC3 and (pC1 + pC2 + pC3) 2 . If these numbers are not equal to 1, the gene is not in Hardy-Weinberg equilibrium, and the gene is evolving.
  2. return next year and repeat this examination of the enzyme, calculating frequencies of each allele each year. Then calculate the values of the sums pC1 + pC2 + pC3 and (pC1 + pC2 + pC3) 2 . If these numbers are not the same each year, the gene is not in Hardy-Weinberg equilibrium, and the gene is evolving.
  3. return each year for several years and repeat this examination of the enzyme, calculating frequencies of each allele each year. If the allele frequencies are changing, the gene is not in Hardy-Weinberg equilibrium, and temperature is exerting a selection pressure.
  4. return each year for several years and repeat this examination of the enzyme, calculating frequencies of each allele each year. If the allele frequencies are changing, the gene is not in Hardy-Weinberg equilibrium. Analysis of the dependence of allele frequencies on temperature could indicate selection.

Calamus finmarchicus is the dominant copepod in the Gulf of Maine. The polymorphic aminopeptidase locus, Lap-1, has been shown to be useful for the genetic differentiation of populations of this organism. By examining the population dynamics of copepods, the dynamics of the fin fish on which they feed can be predicted. The aerial photograph shows a landmass separating two coastal estuarine habits, the mud flats of Egypt Bay and the Mount Desert Narrows. For the past 40 years, transport between the two habits has been hindered by a dam over the Carrying Place Inlet. However, small volumes of water occasionally crest the dam.

To evaluate the geographic isolation of invertebrate populations in these two habitats, copepods are sampled at the points labeled 1 and 2 on the photograph. These points lie at either ends of the Carrying Place Inlet. Enzymes encoded by three alleles, labeled A, B, and C, were determined by gel electrophoresis of equal numbers of the organisms collected at the two sites. Numbers of each genotype are given in the following table:

A. Calculate the frequencies, f, of each allele and complete the following table:

This table of critical p values is also provided on the AP Biology Exam.

Degrees of Freedom
p 1 2 3 4 5 6 7 8
0.05 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.51
0.01 6.64 9.32 11.34 13.28 15.09 16.81 18.48 20.09

C. Based on these data, predict , with justification, changes over time in the aminopeptidase enzyme for these populations.

D. The B form of this aminopeptidase is slightly more efficient at extracting nutritional leucine from a protein than the A and C forms but slightly less efficient at extracting valine and serine. Describe an investigation of the two habitats that could suggest a causal relationship between changes in allele frequency and characteristics of the environment.

E. Single-nucleotide mutations are neutral when they encode changes in proteins that result in no significant differential selection. If differences in environmental factors between sites 1 and 2 are not observed, predict what other factors could result in departures from Hardy-Weinberg equilibrium for aminopeptidase.

Bioluminescence is an example of convergent evolution 30 distinct lineages have acquired this characteristic, and all involve some form of a class of molecules called luciferins. Sexual selection pressures are strong for light-emitting organisms. Ellis and Oakley (Curr Biol, 2016) examined the number of species that lack luminosity in groups of closest evolutionary relation (sister linear) with those species that are luminous. Similarly, scientists made the same comparison between groups that use luminosity for concealment (counter-illumination) and their sister lineages. The graphs summarize their results, comparing the natural logarithm of the number of species in each lineage.

Based on the data shown in the graphs, describe a model that can account for the increased speciation of bioluminescent lineages, including the mechanism of speciation.

A biologist is using a simulation to model populations of African hornbills (Bycanistes spp. and Ceratogymna spp.), a keystone species of the savanna. Populations of the birds are declining due to habitat loss. The hornbill’s diet consists primarily of termites and fruit. A critical component of termite digestion is chitin deacetylase, an enzyme whose mutation rate is a model parameter. The other model parameter is population size, N. In the results of the simulation study shown above, there is no selection, and the mutation rate is fixed. Although both population size and mutation rate are fixed, randomness results in the five different outcomes shown in each graph above.

A. Select the graph displaying the results that are closer to Hardy-Weinberg equilibrium. Justify the selection of the graph.

B. Based on these simulations, predict the future heterozygosity, 2pq, of the smaller populations, as shown in graph A.

C. Justify the use of a simulation study with no selection under environmental conditions in which the availability of both termites and fruit is high.

D. If a change in the environment occurs suddenly, such as an increase in average temperature, where fruit production declines, analyze the effect of the change on allele frequency in the large and small populations.

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    Allele frequency refers to the fraction of individuals with a particular version of a given gene. what effect does genetic drift have on the allele frequency of a population? a. it causes the allele frequency to resemble that of a small number of individuals that became separated from the rest of the population. b. it causes random changes and the allele frequency of certain traits may increase or decrease. c. it increases the frequency of alleles that improve a species' survival in a particular environment. d. it greatly reduces the total population, which increases the effects of natural selection on allele frequency.

    it increases the frequency of alleles that improve a species' survival in a particular environment.

    B. It increases the frequency of alleles that improve a species' survival in a particular environment.

    Natural selection encourages the survival and reproduction of individuals who are especially well-adapted to their environment. This means that natural selection causes increases in the frequency of alleles that improve the species' survival in a particular environment. (got it right on study island)


    Introduction

    In an earlier paper (Smith & Baldwin, 2015), we identified nine shortcomings common to available Hardy-Weinberg exercises. These exercises often: (1) fail to focus on understanding (2) are unclear about which specific gene is involved in the problem (e.g., “tasters”) (3) are unclear about the characteristics of the population being studied (especially size) (4) assume that the HWeq exists in the population but do not say so explicitly (5) assume students have certain biological knowledge about the gene involved in the problem (6) make assumptions that are contradicted in fact or are likely impossible (7) ask for judgments about populations that are constituted by members of multiple generations (8) ask for calculations that are meaningless in the given context and/or (9) ask for solutions that have no apparent value/are not related to genuine research questions (i.e., fail the “So what?” test).

    A particularly prevalent shortcoming is the presentation of an activity with colored buttons or other manipulatives that fails to focus on transferring the activity into biological meaning and thus into understanding of basic HWeq concepts and of population genetics and evolution. Also frequently missing is a focus on the conditions required for HWeq and, more often, on understanding why these conditions are required. Most importantly, students all too often are “doing a thing with buttons” instead of learning about HWeq. This is a transfer issue, but it is also a failure to communicate to students why the HWeq was and is of such importance, that is, how it relates to the rest of biology, how it can be relevant to their personal lives, and even more important, how it can be interesting!

    We describe here a proposed lesson plan that aims to avoid these shortcomings. It is based on a three-stage learning cycle model designed by Atkin and Karplus (1962) in the Science Curriculum Improvement Study (SCIS) program and expanded to the current five-stage model in 1978 by the Biological Sciences Curriculum Study (Bybee et al., 2006). The model is strongly supported by a body of literature documenting its effectiveness (BSCS, 2006, and citations therein) and is widely employed in science education, including in this journal (e.g., Chudyk et al., 2014). 5E instruction is an inquiry learning approach that “exposes students to problem situations (i.e., engages their thinking) and then provides opportunities to explore, explain, extend, and evaluate their learning,” in contrast to simply giving students information, such as answers to questions they do not have (Bybee et al., 2006, p. 4).

    The goal of the activity below is therefore to understand the “big ideas” about the HWeq that are involved in evolution (Table 1), which we derived from the previous analysis (Smith & Baldwin, 2015).

    After one generation of fully random mating, BOTH the genotype and allele frequencies are fixed until one of the Conditions 1–6 (Table 2) is violated.

    The frequencies of each genotype in a population can be predicted from the allele frequencies in that generation. (Allele frequencies can always be calculated from the frequency of the homozygous recessive individuals.) Being able to compute the likelihood that a given individual is heterozygous (a “carrier” of the trait) can be very useful information, e.g., in genetic counseling.

    Genotype frequencies in the next generation can be computed from the allele frequencies in the current generation.

    After one generation of fully random mating, BOTH the genotype and allele frequencies are fixed until one of the Conditions 1–6 (Table 2) is violated.

    The frequencies of each genotype in a population can be predicted from the allele frequencies in that generation. (Allele frequencies can always be calculated from the frequency of the homozygous recessive individuals.) Being able to compute the likelihood that a given individual is heterozygous (a “carrier” of the trait) can be very useful information, e.g., in genetic counseling.

    Genotype frequencies in the next generation can be computed from the allele frequencies in the current generation.

    Observation 1. As long as a population satisfies biological Conditions 1–5 (see Table 2), the allele frequencies (p and q) remain the same in each generation.

    It is also crucial to know the conditions that are required in order for HWeq to exist (Table 2).

    1.There is no migration (gene flow) in or out of the population.
    2.Natural selection is not occurring.
    3.Mutation is not occurring.
    4.Each member of the population is equally likely to breed. *
    5.The population is infinitely large.
    6.(Full random mating) Each pair from the population is equally likely to breed.
    1.There is no migration (gene flow) in or out of the population.
    2.Natural selection is not occurring.
    3.Mutation is not occurring.
    4.Each member of the population is equally likely to breed. *
    5.The population is infinitely large.
    6.(Full random mating) Each pair from the population is equally likely to breed.

    Often “random mating” is used to refer to both conditions 4 and 6. Random mating means that the frequency of mating of an individual or of any pair of individuals does not depend on the genotype.

    Teachers who are not familiar with this instructional design may find this lesson difficult to implement until you have had practice using the technique with less complicated content. Teacher questioning that focuses on helping biology students build their understanding step by step instead of lecturing can be challenging until you have mastered it, especially with activities in which biology students are building mathematical models. Like most inquiry-based instruction, the role of the teacher here is as a facilitator—the guide on the side, not the sage on the stage.

    Species evolve as the frequencies of various alleles change over generations. Change in allele frequency is caused by natural selection and genetic drift, among other reasons. Population genetics is the study of these changes. New species arise when these changes accumulate to the extent that breeding with earlier forms is no longer possible (Big Idea 1). Alleles are different forms of a gene some alleles confer increased reproductive success on the individual and will increase in frequency in response to natural selection (Big Ideas 2 and 3). Constancy of allele frequency requires that no factors such as mutation, natural selection, or migration are adding or deleting alleles (Big Idea 4). The Hardy-Weinberg equilibrium principle explains how genotype and allele frequencies are maintained (Big Idea 5) and can be used to identify populations in which equilibrium does not exist (Big Idea 6), which can signal to researchers that they should seek the causes of this state (Big Idea 7).

    The proposed lesson focuses on understanding these Big Ideas first by introducing students to Huntington's Disease (HD) (Engagement). Next, a set of manipulatives is introduced by which students can model the frequencies of HD alleles then students generate models to predict HD allele frequency change from one generation to the next in the absence of natural selection (Exploration). Students compare and improve their models, and compare these against the current scientifically accepted model (the HWeq) (Explanation). Finally, students apply their new understanding to a new case—the CCR5 deletion that reduces HIV binding to T-cells (Elaboration).

    The target audience is high school biology or AP biology and introductory college biology students with some experience with modeling (especially mathematical modeling) and with the 5E instructional approach. Minimal understanding of DNA structure, Mendelian genetics, and basic evolution and natural selection is assumed. For high school classes, the activity should take three to five 45-minute periods. Detailed solutions, explanations, and other teacher guide information are provided in Supplement 1: Teachers Guide, along with a copy master for the Student Worksheet. The activity targets the Next Generation of Science Standards (http://www.nextgenscience.org/next-generation-science-standards), including the disciplinary core ideas of natural selection (LS4.B) and adaptation (LS4.C) as well as the practice of “using mathematics and computational thinking.” This activity also targets AP Biology science practices 2 and 6, essential knowledge 1.A.1g–h, 4.C.3c, and learning objectives LO 1.6, 1.7, and 4.26:

    LO 4.26 The student is able to use theories and models to make scientific claims and/or predictions about the effects of variation within populations on survival and fitness. (College Board, 2015)


    AP Biology Question 47: Answer and Explanation

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    Question: 47

    5. A population in Hardy-Weinberg equilibrium consists of 300 sheep, some black and some white. If the allele frequency for the recessive white allele is 0.3, how many black sheep are there in the population?

    Correct Answer: 273

    Explanation:

    The question gives the allele frequency of the recessive (white) allele, so q = 0.3. Since the population is in Hardy-Weinberg equilibrium,

    p + q = 1 or
    p = 1 – q = 1 – 0.3 = 0.7

    Then, the frequency of the white individuals is q 2 , or 0.09, and the frequency of remaining black individuals must be 1 – 0.09 = 0.91. (The frequency of the black individuals can also be calculated from p 2 + 2 pq = 0.49 + 0.42 = 0.91.)

    To calculate the number of black sheep, multiply their frequency by the total number in the population, or 0.91 × 300 = 273.

    *AP & Advanced Placement Program are registered trademarks of the College Board, which was not involved in the production of, and does not endorse this site.


    Does any of this interest you?

    We have classified HLA data sets into "Gold-Silver-Bronce" standards. If you wish to see how many populations are classified as Gold, click here.

    Please cite this website using our last publication: Allele frequency net database (AFND) 2020 update: gold-standard data classification, open access genotype data and new query tools. Gonzalez-Galarza FF, McCabe A, Santos EJ, Jones J, Takeshita LY, Ortega-Rivera ND, Del Cid-Pavon GM, Ramsbottom K, Ghattaoraya GS, Alfirevic A, Middleton D and Jones AR Nucleic Acid Research 2020, 48, D783-8. [Full Text]
    We recommend you to also cite the original publication report of the data in your references.


    AP Biology Question 16: Answer and Explanation

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    Question: 16

    4. In a certain population of birds, the allele for a crown on the head ( C) is dominant to the allele for no crown on the head ( c ). A particularly cold and long winter favored the birds with no crown. When spring came, researchers determined that the population was currently in Hardy-Weinberg equilibrium and that the occurrence of the birds with no crown was up to 24%. What will the frequency of the no-crown allele be in 10 years? Give your answer to the nearest hundredth.

    Correct Answer: .49

    Explanation:

    According to Hardy-Weinberg protocol, q 2 = homozygous recessive, q = the recessive allele p 2 = homozygous dominant, and p = the dominant allele. The question asks about the frequency of the recessive allele in 10 years. The frequency of the allele is currently 24% = q 2 , so q = 4.89. Since the population is in Hardy-Weinberg equilibrium, the frequency will not change in 10 years.

    *AP & Advanced Placement Program are registered trademarks of the College Board, which was not involved in the production of, and does not endorse this site.


    Acknowledgements

    We thank Dr. Daniel Ramsköld for providing the mouse preimplantation embryo dataset, Dr. Christelle Borel for providing the human fibroblast dataset, and Cheng Jia, Dr. Arjun Raj, and Dr. Uschi Symmons for helpful comments and suggestions.

    Funding

    This work was supported by National Institutes of Health (NIH) grant R01HG006137 to NRZ, and R01GM108600 and R01HL113147 to ML.

    Availability of data and materials

    SCALE is an open-source R package available at https://github.com/yuchaojiang/SCALE with license GPL-3.0. Source code used in the manuscript is available via Zenodo with DOI 10.5281/zenodo.437554.

    Authors’ contributions

    NRZ and ML initiated and envisioned the study. YJ, NRZ, and ML formulated the model. YJ developed and implemented the algorithm. YJ, NRZ, and ML conducted the analysis and wrote the manuscript. All authors read and approved the final manuscript.

    Competing interests

    The authors declare that they have no competing interests.

    Consent for publication

    Ethics approval and consent to participate

    Publisher’s Note

    Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.


    AlleleA1

    AlleleA1 models evolution at a single locus in an ideal population of imaginary organisms. The locus of interest has two alleles: A1 and A2. The user enters values for parameters controlling selection, mutation, migration, genetic drift, and inbreeding. As the simulation runs, the software plots a graph showing the frequency of allele A1 over time.

    NEW: AlleleA1 is available as a Shiny web app.

    Download the AlleleA1 application:

    AlleleA1 for Mac OS X

    Please note: To run AlleleA1 for the first time, you may need to hold down the control key, click on the AlleleA1 icon, and select Open from the popup menu that appears. (This is necessary because I am not a registered developer with Apple.)

    If these instructions don&rsquot work, open System Preferences and select Security & Privacy. On the General pane, change &ldquoAllow apps downloaded from:&rdquo to &ldquoApp Store and identified developers.&rdquo Then try again.

    AlleleA1 is compiled for 64-bit processors and should run on Mac OS 10.15 Catalina.

    AlleleA1 for Windows
    • Please note: Do not delete the folder called AlleleA1 Libs.AlleleA1 needs it to work properly.
    • Please note: The first time you open the AlleleA1 application on Windows you may see a dialog box titled “This application may depend on other compressed files in this folder.” Click the Extract All button (not the Run button). If you just click the Run button, you’ll get a Runtime Error message. You may also see a dialog box that says “Windows SmartScreen prevented an unrecognized app from starting.” This is because I am not a registered developer. Click on &ldquoMore info,” then on &ldquoRun Anyway."

    Read or download the manual:

    Population genetics tutorials using the AlleleA1 application:

    Fine Print:

    AlleleA1 2.0 © 2003 by Jon C. Herron

    AlleleA1 is free. You may give it to anyone you like, so long as you distribute it as a stand-alone application and include the manual along with it. You may not sell it. You may not include it in a collection with other software, regardless of whether you are selling the collection or giving it away.

    I make no warranties or guarantees about the quality of AlleleA1, or the accuracy of the simulations it runs. Use it in good health, but at your own risk.



Comments:

  1. Yozshushakar

    You, probably, were mistaken?

  2. Leodegraunce

    Bravo, what words ..., brilliant idea

  3. Gulkis

    It doesn't make any sense.

  4. Nazir

    I absolutely disagree



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