 1 - 7 of 7 Posts

#### achistikbenny

·
##### Registered
Joined
·
234 Posts
Discussion Starter · ·
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.

• Usmle16Forall and drcuddles

#### Peewee

·
##### Registered
Joined
·
365 Posts
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.
what do you need help with? What the equation means? what the p^2+2pq+q^2=1 means?

I assume you know what Hetero, **** is?

#### achistikbenny

·
##### Registered
Joined
·
234 Posts
Discussion Starter · ·
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.

#### Peewee

·
##### Registered
Joined
·
365 Posts
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.
Ok, sorry if this is going to be a long post.

^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 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 **** 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 **** (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- **** (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 **** 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 1-q
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

X-Link 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

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 1-q
meaning p=1.00 - .00001 = .999 or ~1

Allele frequency )p_ of rhetts normal gene is .999 or 1/1.0001

X-Link Recessive (q is equal to the disease frequency)
----------------
X-link 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 1-q
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.

#### achistikbenny

·
##### Registered
Joined
·
234 Posts
Discussion Starter · ·
thank u Peewee for making out time to type and explain all these for me i honestly appreciate . thank u very much.

·
##### Registered
Joined
·
1 Posts

....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...

• achistikbenny

#### achistikbenny

·
##### Registered
Joined
·
234 Posts
Discussion Starter · ·
thank u very much but u didnt leave ur gmail address.

1 - 7 of 7 Posts