po
Forum Elite
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In some African populations, the prevalence of sickle cell disease, an autosomal recessive condition, is 1/100. Based on this value, what proportion of the population would be heterozygous carriers of the sickle cell disease gene?
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| mytime
Kick my butt!
Topics: 39
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18%
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| yoga
usmlelogy professor
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20%
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we are all in the gutter but some of us looking at the stars
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| po
Forum Elite
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q = 0.1 p = 0.9 2pq = 0.18
Answer: 0.18
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| new_n_lost
Politically InCorrect
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can u plz explain i know this is too much to ask but have never understood these types of questions. Thanks
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FORUM RULES-- Those who believe in telekinesis, raise my hand. I get enough exercise just by pushing my luck --P4U World.." The pure and simple truth is rarely pure and never simple."
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| po
Forum Elite
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me too in learning phase,if wrong plz correct.
wish somebody knew simple way to learn this would post what they know.
we have to go to basics-
http://anthro.palomar.edu/synthetic/synth_2.htm
http://www.phschool.com/science/biology_place/lab...
http://science.nhmccd.edu/BioL/hwe.html
Its just pain,wonder why they were born
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| po
Forum Elite
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key to understand this equation lies in how these people could get here-ie focus on math
Hardy and Weinberg went on to develop a simple equation that can be used to discover the probable 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 a population. 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 correct:
p = 1 - q
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 predicted frequency of homozygous dominant (AA) people in a population, 2pq is the predicted frequency of heterozygous (Aa) people, and q² is the predicted 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 predict 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 predicted frequencies of all three genotypes for the selected trait within the population.
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| edie
I can, and I will.
Topics: 26
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20%, as yoga stated, should be the answer. 1/100 = 0.01, not 0.1.
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| po
Forum Elite
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Posts: 356
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I think it goes like this edie,
qsquare=1/100
ie q=1/10=0.1
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| edie
I can, and I will.
Topics: 26
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Oh snap! I forgot to take sqrt, thanks 
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| edie
I can, and I will.
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So given the allelic frequency, we must not forget that it is already squared, right?
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| po
Forum Elite
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yes I guess,and trick to these kind of Qs is finding out which is what;I think it formula changes depending on the type of disease AD,AR,XlinkedD&XlinkedR
key to understand this Equation is that-
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 a population. 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
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| edie
I can, and I will.
Topics: 26
Posts: 1,235
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Po...you ROCK Thanks again 
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| po
Forum Elite
Topics: 39
Posts: 356
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no rock,just trying to understand the concept;it really makes more sense to me when I am explaining it to somebody.
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| new_n_lost
Politically InCorrect
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Really A Good Description Po I honestly cudnt have understood it with much ease.
___________________
FORUM RULES-- Those who believe in telekinesis, raise my hand. I get enough exercise just by pushing my luck --P4U World.." The pure and simple truth is rarely pure and never simple."
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| po
Forum Elite
Topics: 39
Posts: 356
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I am glad you found it usefulwish kaplan another half page while explaining this Equation.
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| edie
I can, and I will.
Topics: 26
Posts: 1,235
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po wrote:no rock,just trying to understand the concept;it really makes more sense to me when I am explaining it to somebody.
Oh I get it..ok, you rock AND you roll And yes, if you can teach it, then you know it. Excellent work, Po, keep it up!
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