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Here is a tutorial to perfectly understand Hardy-Weinberg Rule! If you feel like you just need a brief reminder, you can skip the text until the section In short... and try out the exercises just to check your understanding. Terms you should know a priori I will not define the following terms, so make sure you understand them locus allele (relative) ...

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Lets say there are 2 alleles. One of them is represented by B and other by b. Both will have some frequency at a specific time in a population. Now, frequency is number of that allele divided by total alleles. So, if frequency of B is, say X, the frequency of b is Y=(1-x). It would be because whatever alleles are not B are, for sure, b. That gives us our ...

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p = p(a) = 0.6 q = p(A) = 0.4 Frequency of homozygous recessive, aa = p2 = 0.36 All other genotypes have dominant phenotype therefore the frequency of the dominant phenotype is 1-0.36 = 0.64 (Frequency of homozygous AA = q2 = 0.16 Frequency of heterozygous Aa = 2pq = 0.48) And if you want a deeper explanation you really should study this answer by Remi....

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This is basically the same solution as @AlanBoyd's answer, but since he asked me to, I will post my solution as well (also a reason to try out $\mathcal{MathJax}$ a bit more). Assuming: $f(A)=p=0.3, \\f(a)=q=0.7$ and the standard formulas: $f(AA)=p^2\\ f(Aa)=2pq\\ f(aa)=q^2$ gives these genotypes in the parental generation (G1), before and after ...

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Hardy-Weinberg law makes a series of assumptions. One of them is the absence of selective effects. As you talk about disease, this assumption of neutrality is obviously not met. At the moment of fecundation Imagine for example that at the moment of the fecundation, the genotypes AA, Aa and aa are at Hardy-Weinberg equilibrium. Let x be the frequency of the ...

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Your reasoning is incorrect, because once you have the number of individuals surviving after the 50% mortality you can't just give them all one identical offspring, you have to mate them with one another. However, I can't get the answer given, instead I get 47.9% Here is my reasoning: f(A)=0.3 f(a)=0.7 calculate genotype frequencies: AA = 0.32 = 0.09 aa = ...

5

The two-allele scenario is often used in genetics teaching because of its simplicity. However, quite a few genes have more than two alleles. Some examples that readily come to mind are: The ABO gene in humans: This determines ABO blood group, and has six common alleles. Many more rare alleles have been described . The human leukocyte antigen (HLA) genes: ...

4

I make it 8%. Here is my reasoning. The gene is X-linked. 40% mutant males, so freq(mutant allele) = p = 0.4, and freq(wt allele) = q = 0.6 To get a mutant female we have to have a mutant male parent, probability = 0.4 Of these matings one half will produce a female offspring so 0.4*0.5 = 0.2 i.e. 20% of matings derive from a mutant male and produce a female ...

4

A: wild-type allele / a: color blind allele Because color blindness is recessive and X-linked your assumption $p=F(a)=4\%$ is correct as men do only have one copy of the allele. Subsequently $F(A)=q=1-p=0.96$ is also correct. Therefore: a) $F(Aa)=2pq=7.68\%$ is correct and b) is wrong, a is the color blind allele and $F(a)=0.04$ therefore it's $p^2=0.04^2=... 4 You almost said it all! Here are two ways to think of this problem Finding the missing probability Let's denote the fraction yz genotype in the population with$f(yz)$. The sum of of fraction of all genotypes must be equal to 1. In equation it means $$f(aa) + f(aA) + f(AA) = 1$$ Here, of course I don't make a difference between$f(aA)$and$f(Aa)$(... 3 In a population of mice, the presence of black spots is the result of a homozygous recessive condition. If the frequency of the allele for this condition is 0.15, what is the approximate percentage of heterozygous genotypes in this mouse population? (Assume that the population is in Hardy-Weinberg equilibrium.) We know the frequency of the allele (0.15) and ... 3 First of, let me correct your equation:$q = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1 ≠ 10 $. From allele frequency to genotype frequency Imagine you were to randomly sample an allele from a population of allele where the allele A is present at frequency$q$. What is the probability that you draw allele A? Answer:$P(A) = q$. Now, put this allele back in ... 3 Too long for comment : If you want to solve the question by Hardy Weinberg, there are certain assumptions involved :- 1.There is random mating. People do not choose their mates on the basis of whether they carry PKU gene or not & mating among close relatives is not there. 2.No new PKU mutations happen. 3.There is no difference in the reproductive ... 3 Your answer to a) is correct (standard Hardy-Weinberg equation). For b) you have to consider that you are calculating proportions of the population, which should sum to 1. Since 30% of the AA individuals and all SS individuals die, this is not the case after selection, which is why you should divide the frequencies by the total proportion that survive (also ... 3 You are almost correct, but with a few modifications/comments. The calculations in Q1 are independent of HWE, and are just calculating the current allele frequencies in the population based on genotype frequencies (i.e. would be the same irrespectively of if the population is in HWE or not). In Q2 it is unclear if the questions is asking for genotype ... 2 Hardy-Weinberg law gives the frequencies of genotypes from allele frequencies:$p^2 + 2pq + q^2$Because the allele "T" is dominant, the frequency you're directly measuring through the experiment is the addition of$p^2$and$2pq$. $$p^2 + 2pq = 150/215 ≈ 0.6976$$ solving for$p$(and replacing$q$by$1-p$) gives: $$p = 1 - \sqrt{\frac{13}{42}} ≈ 0.45$$... 2 Let's go through the possible answers. Which of the following could be predicted? Not (1), because we don't know whether selection favors or disfavors the recessive alleles. Not (2) for the same reason. Not (4) by definition. Not (5) because recessive alleles don't magically become dominant. This leaves us with (3), which also seems correct: not only do we ... 2 Revised to add complete solution Before the owls arrive, 100 out of the 2500 mice are dark-colored. This ratio provides the frequency of the recessive allele $$q^2 = \frac{100}{2500}$$ So$q$is 10/50 or 0.2. This means that the frequency of the dominant allele$p$is 1 - 0.2 or 0.8. Knowing these frequencies, it is possible to calculate the number of ... 2 Carrier females= 2pq= 2 x .08 x 0.92 Because carriers say with genotype Aa (A=dominant, a=recessive) can arise due to two types of inheritances;one when A is from father and a is from mother and the other when it is just opposite. 2 Solving Hardy-Weinberg problems First of all, you might want to have a look at the post Solving Hardy-Weinberg problems and eventually at Assumptions of Hardy-Weinberg rule After reading the first post, you should be able to answer your question yourself. I encourage you to try it before reading what follows. Hardy-Weinberg for sex-linked loci I will ... 2 Incest is a human concept defined by law and social conventions. For example, see a definition of the word: "sexual intercourse between persons so closely related that they are forbidden by law to marry". A person who was adopted may have siblings who are genetically more different from them than other people they could legally have romantic relationships ... 2 I suggest this as a starting point. Basically, if your aim is to determine the population frequency of the alleles, then you should sample from the population. A possible problem with the university is the age stratification, but also possible different proportion of ethnic groups compared to general population. Also hospital might be biased, because you ... 2 the population tends towards infinity Well yes, Hardy–Weinberg principle is based on several assumptions one of them being "population size is very(infinitely) large". Another being "Mating is random". However, even if the population is small your version of calculation does not make much sense biologically. If we talk about civets (gonochorists) the ... 2 I can't think of a more succinct way to phrase it than the explanation in your post. "...the genotype frequencies of the next generation depend only on the allele frequencies of the current generation..." The genotype frequencies still do depend on the initial generation, they just depend on the allele frequencies of the previous generation, not ... 1 The answer is logical if you understand the reasoning behind Hardy-Weinberg. I recommend that you to have a look at the post Solving Hardy Weinberg problems and try to answer the question yourself. Once you've done that, you can hover over the yellow zone below for the answer to compare against what you've understood. The answer: 1 Linkage is a physical reality Linkage refers to the physical presence of several loci on the same chromosome. Two loci that are said to be in close linkage mean that they are locataed relatively close to each other on the same chromosome. Linkage disequilibrium is a statistical reality If two alleles at two different loci are found on the same haplotype ... 1 The solution $$p^2 = 202 / 630 ≈ 0.32$$ $$p = \sqrt{202/630}$$ $$(1-p) = 1 - \sqrt{202/630}$$ $$(1-p)^2 = \left(1 - \sqrt{202/630}\right)^2 ≈ 0.188$$ There is a fraction of$202 / 630 ≈ 0.32$of red birds. There is a fraction of$\left(1 - \sqrt{202/630}\right)^2 ≈ 0.188$of white birds. There is therefore a fraction of$0.32 + 0.188 = 0.508\$ of birds that ...

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You are throwing numbers into an equation without understanding it. If you know the numbers for pp, pq and qq, you don't use HWE to recalculate them. So your last statement in Calculation 1 is nonsense. What you've done there is calculated the proportions for a population in HWE where the p=0.73, but you were already given the actual distribution in the ...

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Okay, so the Hardy-Weinberg Equilibrium is just that: an equilbrium. In physics, a system (some object) is at equilibrium when no net energy crosses its boundary with the surroundings. no energy change means no change in motion (the state of the object) due to outside influence on the system. In chemistry, a system (the beaker/reaction container/cytosol) ...

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You see a question like this, you should just go ahead and calculate if the population is in Hardy-Weinberg Equilibrium, because that's almost certainly what's going to be asked. For 10 years ago: 90+90+420/2000=0.3 BB 490+490+420/2000=0.7 YY That notation is not helping. B = 0.3, Y = 0.7 BB = 90/1000 = 0.09 BY = 420/1000 = 0.42 YY = 490/1000 = 0.49 ...

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