po wrote:
I hav lost the site link;dont remember which had this q,will get back while doing revision.
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| clavicle
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hi, new to genetics - how did u get that equation. studied genetics in kaplan. puzzled a little bit. will be greatful if you could help me.
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| sataleees
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Guys.. don't forget that the penetrance in the X linked dominant diseases in females is 60% so 0.02 multilpied by 0.6 will equal 0.012 is the freguency of the disease (phenotype = proportion of females would be affected with the X-linked dominant disorder) same as 1.2 %.
Best of Luck for all
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| drdido
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I am too confused about this one, can someone break it down in steps please.thx
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| drgho
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gene frequency 1/100 and affected females approx. 2/100???
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| usmlesavvy
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Let me give it a shot
The Q requires the use of HARDY WEINBERG Equation to X-Linked disorders.
HARDY WEINBERG Equation---------> p square + 2pq + q square=1
Note:This equation is most commonly applied for AR diseases.
In X-Linked diseases, this equation can only be applied fully to females, but not to males becoz males have only one copy of X chromosome(and the HW Equation works for 2 copies of a gene)
For males, The Allele frequency(or Gene frequency) for the mutation is equal to the number of affected males i.e. = q(Becoz males have only one X, if it is mutated ,he will have the disease)
Now to solve the Q:
Given that:An XD disorder affect 1/100 males in a population
=> The Allele Frequency(or Gene frequency)i.e q = 1/100
Now applying the full HW Equation to females.........
The number of affected females will be = 2pq + q square
Where 2pq= Heterozygous affected
q square=homozygous affected
We know from the first part of the solution that q=1/100
pi.e fraction of Normal allele, can be assumed to be close to 1(Becoz XD diseases are rare,normal allele much more common that mutant allele)
Thus The number of affected females= 2 times q + q square
i.e roughly equal to 0.02 or 1/200
( Not the exact value but close to it, exact value=0.02001)
This also points out the fact we know that XD disorders are twice as common in females as males
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| usmlesavvy
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Can someone else discuss this please, wanna know if i am right
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| dyana
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I got the same answers, but still not sure if it is correct because of what sataleees said:
"Guys.. don't forget that the penetrance in the X linked dominant diseases in females is 60% so 0.02 multiplied by 0.6 will equal 0.012 is the frequency of the disease (phenotype = proportion of females would be affected with the X-linked dominant disorder) same as 1.2 %.
Best of Luck for all "
For the first part of the question :male XY=only one X and we know 1/100 male affected=>gene frequency is q=1/100.
For the second part of the question:q=1/100 and we know p+q=1 =>p=99/100
From HW equation => genotype frequency (females affected) is 2pq+q square=1-p square=> 0.02 or 1/200
I had the same reasoning as usmlesavvy and used kaplan biochemistry section 2, chapter 2, but what about what sataleees said. I don't know were to find in the books that the penetrance in the X linked dominant diseases in females is 60%.If I could find this I could say that usmlesavvy is wright since the question asks about female affected which means phenotype.
Please post any other opinions I would like to find out the wright answer.
Thanks.
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| spinal shock
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since females are XX. the chance for been a carrier is the same as males but the chance for been affected is 1/1000
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| taipei817
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gene frq: 1/ 100 = 0.01
proportion female = female with one mutation + female with 2 mutation
= 1/100 + 1/100*1/100= 0.0101
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| taipei817
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taipei817 wrote:
gene frq: 1/ 100 = 0.01
proportion female = female with one mutation + female with 2 mutation
= 2*1/100 + 1/100*1/100= 0.0201
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