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#1




Hardy Weinberg Equation, Please Help
pls I need help on how to apply hardy Weinberg equation. I don't understand how to use it to solve questions in genetics. pls i need explanation. my exam is quite near.

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drzamzam (08172015) 
#2




Quote:
I assume you know what Hetero, Homo is? 
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#3




thanks, i know what homozygous and hetero means but i dont know how to apply it using HW equation. i will really appreciate if u can help me here.

#4




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^2 = to the power of 2 Equation: (p+q=1)^2 or p^2 +2pq+q^2=1 AA=p^2 Aa=2pq aa=q^2 The best way to learn how to use the equation is by following examples. Autosomal Dominant Inheritance  Achondroplasia Dwarfism (AD Lethal) Study of Achondroplasia documented 7 new cases out of 250 000 in the USA. IN achondroplasia Homoqygensity is genetic lethat. What is the Allele Frequency of Acondroplasia in the USA population? Solution 7/250 000 = .000028 Aa = 2pq 2pq=.000028 Allele Frequency p of the disease is small in AD allele frequency q is ~1 in AD so since q is about 1, then really the 2pq can just become 2p. 2p=.000028 / 2 (divide 2p and the .00028 by 2) p=.00614 or 1/71428 Huntingtons (AD nonlethal) Huntingtons occurs in the USA at a frequency of 1/10000 or .0001. Huntingtons is homoqygosity is not genetic lethal. What is the allele frequency of homo dominant (HD) gene in the USA solution P^2+2pq+q^2=1 All geno's containing HD include: AA=p^2 and Aa=2pq Thus frequency is equal to p^2+2pq=.0001 Frequency of homo (p^2) is small ~0 allele frequency q is about 1, so 2p(1) or 2(p) = .0001 2p/2 = .0001/2 p=.00005 or 1/20 000 I assume that AD diseases q is 1 all the time so just figure out p Autosomal Recessive (q^2 is equal to the disease frequency)  Autosomal recessive disorders homogenciity aa produces the disorder. ex homo (dd) for sickle cell produces the disorder where as Hetero (Dd) is a carrier. Sickle Cell A study of sickle documented 10 new cases out of 6 250 in the AA population. what is the allele frequency (p) of the sickle gene in AA population?  What is the frequency of the hetero carrier? Solution  Frequency= 10/6 250 = 1/625 or .0016 2 SS gene showing disease is homo aa=q^2 disease show is q^2=.0016 take square root of q^2 of .0015 q=.04 or 1/25 Allele frequency p of sickle is 1q p= 1.00.04=.96 Allele frequency (p) of ss normal is .96 or 1/1.04 Frequncey of Hetero Carrier is 2pq 2(.96)(.04)=.0768 or 1/13 XLink Dominant (q is equal to disease frequency)  Observed twice the # of females than of males. Males usually die EX: rhetts syndrome Study of Rhetts documented 10 new cases out of 180 000 over the last 10 years. What is the Allele Frequency (q) of rhetts in the usa what is the allele frequency (p) of rhetts in the usa Solution 1 10/180 000 = 1/18 000 or .00006 2 In Xlink dominant Allele frequency (q) is equal to the disease frequency. This means q=.00001 or 1/ 100 000 3 Allele frequency (p) of rhetts normal gene is 1q meaning p=1.00  .00001 = .999 or ~1 Allele frequency )p_ of rhetts normal gene is .999 or 1/1.0001 XLink Recessive (q is equal to the disease frequency)  Xlink recessive is only observed in males ex hunters Hunters study in the usa documented 10 new cases out of 1 000 000 over 5 years. What is the allele frequency (q) of hunters in the USA what is the allele frequency (p) of hunters in the USA what is the female Hetero carrier? solution 1 10/100 000  1/10 000 or .00001 2 x link recessive allele frequency (q) equals disease frequency q=.00001 3 allele frequency (p) of hunters normal is 1q p=1.00.00001 = .9999 or ~1 Allele frequency (p) of hunters normal gene is .9999 4 Frequency of female geterocarrier is 2pq p=~1 2pq = 2(1) (.00001) = .00002 or 1/50 000 I hope this helps some. The best thing you can do, is print this out and do other questions and use this as a guide. Out of a handful of people I've talked to, only maybe 1 has gotten a question on these types of quesitons If you need an clarification, let me know. I type 2 quick sometimes and mispell or my piece of of **** 1 year old dell laptop keyboard doesn't work sometimes. never buy a dell computer 
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#5




thank u Peewee for making out time to type and explain all these for me i honestly appreciate . thank u very much.

#6




reply....for hardyweinbeg equation
[QUOTE=Peewee;11259]....hey!p2 is taken for homozygous....i.e:complete homozygous n dominant traits....n q2 for homogyous recessive...n 2pq for heterozygous individuals......if the frequency is given u can find out the total population....n if the population is given u can find out the frequency of the homozygous [dominant or recessive]and heterozygous populations.....hope now u can solve the problems....if u still hav doubt u can mail me a question to my gmail account....ill solve that n ll send u...all the best for u r exam.....bye...

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achistikbenny (07032010) 
#7




thank u very much but u didnt leave ur gmail address.

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