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#1
10-16-2012
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Genetics Kaplan 2012 Question

Hi guys,
was going over the questions Kaplan Lecture Notes has at the end of each chapter and came across question #6 and 8 at the end of chapter 2 (Population Genetics) on page 342 and 343 (2012 edition). Does anyone understand the explanation they have? I'm totally lost. Help please!!

6. A man is known hetro carrier of a mutation causing hyperprolinemia, autosomal recessive conidtion. phenotypic expression is variable and ranges from high urinary excretion of proline to seizures. suppose that 0.0025% (1/40,000) of the population is homozygoous for the mutation causing this condition. if the man mates with somebody from the general population, what is the probability that he and his mate will produce a child who is homozygous for the mutation involved?

1) 1% (1/100)
2) 0.5% (1/200)
3) 0.25% (1/400)
4) 0.1% (1/1000)
5) 0.05% (1/2000)

The answer is 0.25% but i don't get how they got it.

8. A man who is known hetro carrier of oculocutaneous albinism marries his half-cousin (they share one common grandparent). This trait is transmitted as a fully penetrant autosomal recessive. what is the probability that this couple will produce a with this disorder?

1) 1/2
2) 1/4
3) 1/8
4) 1/16
5) 1/64

The answer is 1/64, again completely clueless!
Your help will be greatly appreciated.

Sarah

#2
10-16-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 1,039 Threads: 189 Thanked 570 Times in 323 Posts Reputation: 580
.

each person of this couple shares 1/4 of their genes with their grandparent.
Since they only have 1 grandparent in common they share 1 copy of gene only, so we'll assume that the wife is also a heterozygous (if they had 2 common grandparents, we would also calculate the possibility of the wife being a homozygous). ok, going back to the question:
so we calculate the probability that each one of them inherited the same mutated gene copy from this common grandparent: 1/4 x 1/4
since they're heterozygous the probability of having a mutated homozygous offspring is 1/4
so all together it is 1/4 x 1/4 x 1/4 = 1/64
#3
10-16-2012
 USMLE Forums Master Steps History: Step 1 Only Posts: 864 Threads: 62 Thanked 374 Times in 225 Posts Reputation: 384

As casandra explained 2 nd ques. I will try explaining 1 st one.

6. A man is known hetro carrier of a mutation causing hyperprolinemia, autosomal recessive conidtion. phenotypic expression is variable and ranges from high urinary excretion of proline to seizures. suppose that 0.0025% (1/40,000) of the population is homozygoous for the mutation causing this condition. if the man mates with somebody from the general population, what is the probability that he and his mate will produce a child who is homozygous for the mutation involved?

1) 1% (1/100)
2) 0.5% (1/200)
3) 0.25% (1/400)
4) 0.1% (1/1000)
5) 0.05% (1/2000)

The answer is 0.25% but i don't get how they got it.

as according to question. In AR disease
q^2= 1/40,000 ( homozygous )
So, q= 1/200
Hetrozygous would be 2q( as p~1 in AR)= 2 x 1/200= 1/100

As father is Hetrozygous so, possibility that his child would be homozygous =
1/100 x1/4= 1/400

I am not able to explain exactly but hope atleast u can get an idea.

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#4
10-17-2012
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Q1:
1- As this disease is AR so this person most likely to mate with someone who's carrier of the population as the majority are hidden in such state.
2- The probability of having homozygous infant = ( probability of the father transmitting his mutant allele x probability of a female carrier in the population x her probability transmitting the mutant allele )
3- Probability of a female carrier in population ( 2pq ) according to Hardy's equation
4 - You're giving in the stem the homozygous ratio ( q x q )
5- q = 1/200 and 2pq = 1/100 which represent now the carrier probability among the population. ( Why p = 1 cuz disease prevalence is less than 1/100)
6- you calculate all probabilities = 1/100 x 1/2 x 1/2 as the two parents are carriers everyone has a 50% chance to transmit the mutant allele
7- Answer = 1/400 or 0.25%

Q2:
1- Probability of having son with such disorder = probability of mom with mutant allele x probability of dad with mutant allele x probability of two heterozygous yielding son with AR disease
2- it = 1/4 x 1/4 x 1/4 = 1/64
3- why 1/4 for each parent cuz as you go from grandparent to parent 1/2 chance of transmitting mutant allele and then another 1/2 chance to transmit the same mutant allele to his son so now chance of son getting his grandparent mutant allele = 1/2 x 1/2 = 1/4 which is the same for the cousin as the grandparent is common.

I hope it helped
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Last edited by Easy; 10-17-2012 at 03:08 AM.
 The above post was thanked by: nandish_m (10-17-2012), venky2600 (10-17-2012)
#5
10-21-2012
 USMLE Forums Scout Steps History: 1 + CS Posts: 82 Threads: 10 Thanked 76 Times in 31 Posts Reputation: 86
Another way to do equation 2

If you can recall from genetics there is a degree of consanguinity. If you look at the pedigree you count the lines that separate the 2 individuals (see picture for example in this case). The number of genes they have in common is equal to (1/2)^n where n=the number so in this case n=4. (1/2)^4 = 1/16. Each parent will pass the gene on 1/2 times since they are heterozygous.
1/16 x 1/2 x 1/2 = 1/64
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#6
10-21-2012
 USMLE Forums Addict Steps History: 1+CK+CS+3 Posts: 155 Threads: 19 Thanked 43 Times in 36 Posts Reputation: 53
ill make this LONG and easy

hi guys..ill do my bit to make this simple

Question 6.

you have to use the Hardy Weinberg equation here and since this is an autosomal recessive condition therefore

frequency of normal allele=p
frequency of mutated or affected allele=q
frequency of homozygous normal=p (square)
frequency of carrier=2pq
frequency of affected=q (square)

so in the question the number of homozygous effected people are 1/40000 which is equal to q square.

to remove the square and for finding out q we need to take the under root of 1/40000 which would come out as 1/200

so q=1/200 which is the allele frequency

generally as a rule p the normal allele frequency is relatively close to 1. so 2pq which is the carrier frequency would be equal to 2q. right. so 2 times q would be 1/100.

for the child to be born as a homozygous affected he must receive a mutated allele from each parent. that is 1 out of 2 or 1/2 from each parent.

therefore the probability of a homozygous affected child would be the product of all three event occuring together.

1/100 (carrier frequency of the mate) x 1/2 (passing of the affected allele by the mate) x 1/2 (passing of the affected allele by the father)

which is equal to 1/400 or 0.25%
#7
10-21-2012
 USMLE Forums Addict Steps History: 1+CK+CS+3 Posts: 155 Threads: 19 Thanked 43 Times in 36 Posts Reputation: 53
question number 8

so here we go.

Guys remember this by heart!

we share 50% of our genes with each of our parents and 50% with our siblings.

but half siblings share 25% of their genes (half sibling=step bro/sis)

also 1st cousins share 12.5% or 1/8 of their genes

but 1st half cousins share 6.25% or 1/16 of their genes
. <----question

so since the question is about a autosomal recessive condition.

there will be 1in 4 chance of a homozygous child born.

therefore the probability for this couple to produce a homozygous affected child would be the product of the two events.

first event : 1st half siblings 1/16

2nd event :autosomal recessive condition 1/4 chance

answer: 1/16 x 1/4 = 1/64

 Tags Genetics-, Kaplan-Medical, Step-1-Questions

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