mjl1717 Forum Hero

Topics: 955 Posts: 5,451
| | 03/30/04 - 07:47 PM  
 
   
 
|   #1 |
Can anyone explain and give an example of the Hardy Weinberg Law in simple terms? :shock:
___________________ Smell the coffee! "Is That an Osler move??"
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| bright_nnn Forum Senior
Topics: 17 Posts: 97
| | 03/31/04 - 08:41 AM  
 
   
 
|   #2 |
This is a very simple Principle It gives a correlation between Gene Freq. and Genotype Freq. So if there r 2 genes p and q If the gene freq of p = 0.4 then q would be =0.6 because they add up tp be one. Now to correlate with the Genotype Freq. we use a Simple eq. p2+2pq+q2 p= o.4 square gives 0.16 q= 0.6 square gives 0.36 2pq =0.48 This also adds up to be 1 OK so with gene freq. u got Genotype Freq. Whats the use.??? Suppose if we want to find out the in a pop. how many r Heterzygotes and how many r Homozygotes for the dis..We cannot find the Heterozygotes for an autosomal recessive Dis. Because Clinically the Heterozygotes would be Indistinguishable from the Normal.So These Heterozygotes are calculated as the 2pq of the previous equation. eg if q is 0.01 p is 0.99 Of course Then 2pq is the heterozygote pop. of the autosomal Recessive dis. Hope it makes sense.Glad to help u more if I can
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| bims Forum Senior
Topics: 26 Posts: 118
| | 04/04/04 - 05:22 AM  
 
   
 
|   #3 |
to understand this u have to know that genotype frequency represents the percentage of people in a population that have a specific genotpye in relation to a disease.eg for an aut recessive dx,the freq of those who are homozygous normal,heterozygous carriers or homozygous mutant.i.e the genotype freq for carriers is the percentage of pple out of the total popn that are heterozygous for the disease. the gene frequency on the other hand using the Hb gene for example,means that of all Hb gene,what percentage of pple carry the normal gene,or what percentage of people carry the sickle[mutant] gene. hardy weinberg simply says p2+2pq+q2=1 where,for an autosomal recessive dx[which is the most likely to get in the exam] p2=freq of homozygous normal 2pq=freq of heterozygous,carriers q2=freq of homozygous mutants[those who are affected with disease] an easy shortcut to apply in the exam is to assume that p=1.u can safely do this because these diseases are very rare so 'approx 100%' of people are normal.this makes the calculation much easier and faster under exam conditions. hope this helps.cheers.
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| mistral92 Forum Senior
Topics: 4 Posts: 185
| | 04/04/04 - 03:01 PM  
 
   
 
|   #4 |
thnaks a lot for your explanations!!!! 
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| mjl1717 Forum Hero

Topics: 955 Posts: 5,451
| | 04/04/04 - 03:29 PM  
 
   
 
|   #5 |
Thx to both you guys! Heres some trivia -that Hardy Weinberg Law was discovered separately, (not by two guys on a joint project) two guys in separate countries get to go down in history for that. What gets me is sometimes if you have to change units.(especially under exam circumstances) Its incredible, a lot of people I know were making believe The Law doesnt exist. Thx again large population random mating no immigration no mutations no selection :idea:
___________________ Smell the coffee! "Is That an Osler move??"
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| asmi Forum Hero
Topics: 1043 Posts: 4,609
| | 04/04/04 - 04:25 PM  
 
   
 
|   #6 |
Thanks to all 
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