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#1
08-10-2011
 USMLE Forums Scout Steps History: Not yet Posts: 99 Threads: 29 Thanked 89 Times in 37 Posts Reputation: 99
genetics-Hardy weinberg question

Q. A man is a known heterozygous carrier of disease Y, which is an autosomal recessive condition. Disease Y has a prevalence of 4% in the population. What are the man's chances of mating with a heterozygous carrier female and having a child affected with disease Y?

A. 2/25
B. 3/50
C. 1/100
D. 3/100
E. 1/400

Explanation must !!!!

 The above post was thanked by: savleem (08-10-2011), shaarang_93 (09-20-2014), surez18 (08-10-2011), Usmle16Forall (11-26-2016)

#2
08-10-2011
 USMLE Forums Veteran Steps History: Step 1 Only Posts: 208 Threads: 30 Thanked 183 Times in 74 Posts Reputation: 193

in AR allele freq q2=4%, q=0.2
carrier freq=2*0.2=0.4
chance of child gettin disaes in AR= 1/4
mother s bcum a carrier =0.4
=1/4*0.4
1/10,
YA ME TOO getting 1/10.
its nt in ans choice...

Last edited by surez18; 08-10-2011 at 06:36 AM.
 The above post was thanked by: Usmle16Forall (11-26-2016)
#3
08-10-2011
 USMLE Forums Guru Steps History: Not yet Posts: 324 Threads: 70 Thanked 238 Times in 144 Posts Reputation: 248

Quote:
 Originally Posted by surez18 in AR allele freq q2=4%, q=0.2 carrier freq=2*0.2=0.4 chance of child gettin disaes in AR= 1/4 mother s bcum a carrier =0.4 =1/4*0.4 1/100
I calculated by the same method..but the answer comes to be 1/10..
 The above post was thanked by: drgsarunprasath (08-10-2011), surez18 (08-10-2011), Usmle16Forall (11-26-2016)

#4
08-10-2011
 USMLE Forums Veteran Steps History: Step 1 Only Posts: 208 Threads: 30 Thanked 183 Times in 74 Posts Reputation: 193
ans A

i can explain tis
q2=4%
q=0.2 (1/5)
in AR p+q=1
p=1-0.2
p=0.8
carrier freq=2pq=2*0.2*0.8
chance of child gettin disease in AR =1/4
mother chance f being carrier=2*0.2*0.8
==2*0.2*0.8*1/4
=2/25.

 The above post was thanked by: drgsarunprasath (08-10-2011), ibukun (08-10-2011), shaarang_93 (09-20-2014), Usmle16Forall (11-26-2016), USMLE2011m (08-10-2011)
#5
08-10-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,268 Times in 881 Posts Reputation: 3278

Man Aa
Female = probability Aa =
p^2 + 2pq + q^2 = 1
q^2 = 0.04 = q = 0.2 AND p = 0.8 (p is only to 1 when q->0 )
2pq = 2*0.2*0.8 = 0.32 = 32/100 = 8/25

autosomal recessive condition = aa = probability of 1/4
1/4*8/25 = 8/100 = 2/25
 The above post was thanked by: dr-ahmed (11-20-2013), drgsarunprasath (08-10-2011), shaarang_93 (09-20-2014), SuperGirl28 (09-11-2013), Usmle16Forall (11-26-2016)

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