kalsam
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Can anyone help understand this while solving genetics problems please???
Thanks
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| peekay
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there has been previous posts by "meg" i suggest read that it might help
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| kalsam
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Thansk for the suggestion I am gonna search for it..
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| kalsam
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Yeah..i got it  !!!
Thanks Meg..
if anyone wants to have a look at it, here it is i just copied the same answer..posted by Meg.
If,
1)mutation does not occur
2)natural selection does not occur
3)population is infinitely large
4)all members of teh population breed
5)mating is random
6) there is no migration in and out of population,
or in other words, if no mechanism that can cause evolution to occur, then gene pool frequencies will remain unchanged. This is the Hardy-Weinberg equilibrium.
for a trait controlled by a pair of alleles (A and a),
let p represent the frequency of dominant allele for a gene in the pool
and q represent the frequency of its recessive allele
in other words, p= all the alleles in individuals who are homozygous dominant (AA) and half of alleles in individuals who are heterozygous (Aa)
hence p= AA+1/2Aa
likewise, q= aa+1/2 Aa
Because there are only 2 alleles in this case, the frequency of one plus the freq of the other must be 100% which means p+q =1. So if we know p, we can calculate q and vice versa
Since p+q=1
(p+q)square =1
p2(here 2 is for square!) +2pq+q2 =1
p2 is the frequency of the homozygous dominant(AA) in the population
2pq the heterozy (Aa)
and q2 is the homozyg (aa)
From USMLE step 1 point of view, most of the genetics questions related to this principle seems like they give the frequency of the diseased people for an autosomal recessive (most commonly tay sachs). So they give q2. Then they could ask what are the chances of the disease in an offspring if mating occurs with someone else from that population. so in a sense it is the mating between a q2 and 2pq. so using q2 , get q, then get p which is 1-q, then calculate 2pq. From that we go on to calculate whatever the question is asking......
for eg, if the frequency of tay sachs is 1/100 in population. What is the chances of an offspring born of a mating between a diseased individual and a heterozygous?
q2= 1/100
q=1/10
p=9/10
2pq= 2*9/10 * 1/10 = 18/100 (actually most of the time we can omit the p) that is 2*1/10 which is 1/5
Now, the chances of the offspring having taysachs will be
q2 * 2pq * 1/2 (1/2 because only Aa and aa offsprings possible)
1/100 *1/5 *1/2 = 1/1000
Hope I am right and did not confuse!
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| rida
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Thanx kalsam for posting meg's explanation, it really does clear it up, this is one concept that i never seem to understand either.
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| poddarc
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From Meg's answer, the chances of a diseased offspring (1/1000 = 0.001) from mating of a diseased and a heterozygote, is less than the chance of a having a diseased person (1/100 = 0.01), which doesn't seem to be logical.
I think the mistake is in the last step of calculating the chance of a diseased offspring. The derivation should be:
Pr(aa) / ( Pr(Aa) + Pr(aa)) = q2/(pq + q2) =1/100 / (9/100 + 1/100) = 1/10 = 0.1
this is because when a diseased (aa) mates with a heterozygote (Aa), the offspring can be either be a heterozygote or diseased.
Hope this is right!!!
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| Sakaki-
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It seems quite logical to me that the chance of finding a diseased person in the population is larger than finding a diseased person (aa), heterozygote (Aa), and then them having an affected offspring (aa) all at the same time, as these are all events that must occur simultaneously for the conditions to be fulfilled.
What bothers me is why Meg omitted the p in her calculation of "2pq"? Can someone please explain this? Other than that, the explanation is quite clear.
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| mash
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if p+q = 1
q^2 is very small(homozygous fr mutant allele)
=> frequency of mutant allele is less(almost negligible)
=> p = 1
that is why 2pq = 2*1*q (i.e. p= 1)
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| Sakaki-
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Thanks, mash. I didn't think of that. ^^
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| mash
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i hope u got dat q^2 means q square?
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| Sakaki-
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Heh--yeah. The carat typically means "raised to the power of."
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| rubensssss
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Hardy-Weinberg
Equilibrium Model
The biological sciences now generally define evolution as being "the sum total of the genetically inherited changes in the individuals who are the members of a population's gene pool." It is clear that the effects of evolution are felt by individuals, but it is the population as a whole that actually evolves. Evolution is simply a change in frequencies of alleles in the gene pool of a population. For instance, let us assume that there is a trait that is determined by the inheritance of either one of two alleles--B and b. If the parent generation has 92% B and 8% b and their offspring collectively have 90% B and 10% b, evolution has occurred between the generations. The entire population's gene pool has evolved in the direction of a higher frequency of the b allele--it was not just those individuals who inherited the b allele who evolved.
This definition of evolution was developed largely as a result of independent work in the early 20th century by Godfrey Hardy, an English mathematician, and Wilhelm Weinberg, a German physician. Through mathematical modeling, they concluded that gene pool frequencies are inherently stable but that evolution should be expected in all populations virtually all of the time. They resolved this apparent paradox by analyzing the probable net effects of evolutionary mechanisms.
Hardy, Weinberg, and the population geneticists who followed them came to understand that evolution will not occur in a population if seven conditions are met:
1. mutation is not occurring
2. natural selection is not occurring
3. the population is infinitely large
4. all members of the population breed
5. all mating is totally random
6. everyone produces the same number of offspring
7. there is no migration in or out of the population
In other words, if no mechanisms that can cause evolution to occur are acting on a population, evolution will not occur--the gene pool frequencies will remain unchanged. However, since it is highly unlikely that any of these seven conditions, let alone all of them, will occur in the real world, evolution is the inevitable result.
Hardy and Weinberg went on to develop a simple equation or formula that can be used to discover the genotype frequencies in a population and to track their changes from one generation to another. This has become known as the "Hardy-Weinberg equilibrium equation." In this equation (p² + 2pq + q² = 1), p is defined as the frequency of the dominant allele and q as the frequency of the recessive allele for a trait controlled by a pair of alleles (A and a). In other words, p equals all of the alleles in individuals who are homozygous dominant (AA) and half of the alleles in people who are heterozygous (Aa) for this trait. In mathematical terms, this is
p = AA + ½Aa
Likewise, q equals all of the alleles in individuals who are homozygous recessive (aa) and the other half of the alleles in people who are heterozygous (Aa).
q = aa + ½Aa
Because there are only two alleles in this case, the frequency of one plus the frequency of the other must equal 100%, which is to say
p + q = 1
Since this is logically true, then the following must also be valid:
p = 1 - q
q = 1 - p
There were only a few short steps from this knowledge for Hardy and Weinberg to realize that the chances of all possible combinations of alleles occurring randomly is
(p + q)² = 1
or more simply,
p² + 2pq + q² = 1
In this equation, p² is the frequency of homozygous dominant (AA) people in a population, 2pq is the frequency of heterozygous (Aa) people, and q² is the frequency of homozygous recessive (aa) ones.
From observations of phenotypes, it is usually only possible to know the frequency of homozygous recessive people, or q² in the equation, since they will not have the dominant trait. Those who express the trait in their phenotype could be either homozygous dominant (p²) or heterozygous (2pq). The Hardy-Weinberg equation allows us to determine which ones they are. Since p = 1 - q and q is known, it is possible to calculate p as well. Knowing p and q, it is a simple matter to plug these values into the Hardy-Weinberg equation (p² + 2pq + q² = 1). This then provides the frequencies of all three genotypes for the selected trait within the population.
Sample Problem
By comparing genotype frequencies from the next generation with those of the current generation in a population, one can also learn whether or not evolution has occurred and in what direction and rate for the selected trait. However, the Hardy-Weinberg equation cannot determine which of the various possible causes of evolution were responsible for the changes in gene pool frequencies.
It is important not to lose sight of the fact that gene pool frequencies are inherently stable. That is to say, they do not change by themselves. Despite the fact that evolution is a common occurrence in natural populations, allele frequencies will remain unchanged indefinitely unless evolutionary mechanisms such as mutation, natural selection, and non-random mating cause them to change.
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