Basic Genetics question

[tbkb2412]

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Hey guys I am refreshing my genetics knowledge by doing questions for the MCAT using my old materials from biology. I am stuck on this question and keep getting it wrong. Can someone please explain it to me! I really appreciate your help.

 

[deleted783484]

The only explanation I could come up with is that

1/2 = the frequency of parent 1 being heterozygous and same for parent 2
then 1/4= frequency of child being affected by the disease

1/2 *1/2 *1/4= 1/16

 

[cocomos]

deleted

 

[Spectar]

Hey guys I am refreshing my genetics knowledge by doing questions for the MCAT using my old materials from biology. I am stuck on this question and keep getting it wrong. Can someone please explain it to me! I really appreciate your help.
View attachment 216184

I feel as though there is something missing from the slide like knowing if the parents are affected or carriers.

 

[5words]

I feel as though there is something missing from the slide like knowing if the parents are affected or carriers.

Nothing is missing, there is 50% (because it doesnt show, otherwise we would assumed that they had the disease) chance that the parents are carriers because their offspring has the disease. Thus, the disease is probably autosomal recessive, And the question is asking for the prob of the child having the disease or prob. of two events happening at the same time, which @ys2327 did above.

Probability of A and B happening at the same time is = PA * PB = 1/2 (probability that parent A is carrier) *1/2 (probability that parent A will pass on the diseases causing allele which is recessive in this case) * 1/2 (Probability of parent two being a carrier) * 1/2 (Prob of B of passing recessive allele)

 

[Spectar]

Nothing is missing, there is 50% (because it doesnt show, otherwise we would assumed that they had the disease) chance that the parents are carriers because their offspring has the disease. Thus, the disease is probably autosomal recessive, And the question is asking for the prob of the child having the disease or prob. of two events happening at the same time, which @ys2327 did above.

Probability of A and B happening at the same time is = PA * PB = 1/2 (probability that parent A is carrier) *1/2 (probability that parent A will pass on the diseases causing allele which is recessive in this case) * 1/2 (Probability of parent two being a carrier) * 1/2 (Prob of B of passing recessive allele)

You can't assume that the parents carry the disease allele because their offspring 100% has the disease and then use that to determine the probability of the same offspring having the disease.

 

[5words]

You can't assume that the parents carry the disease allele because their offspring 100% has the disease and then use that to determine the probability of the same offspring having the disease.

Well, where did the offspring get the disease then? And that's actually not an accurate statement, The Question stems says that the disease is autosomal recessive, thus not expressed in hetero-zygotes. Hence, the only way for the offspring to have the disease is to receive one diseased copy from each parent, or to be homozygous recesive. So, in this case, you assume that both parent are heterozygotes.

 

[freak7]

The only explanation I could come up with is that

1/2 = the frequency of parent 1 being heterozygous and same for parent 2
then 1/4= frequency of child being affected by the disease

1/2 *1/2 *1/4= 1/16

Except the allelic frequency affects the chances that the parents are heterozygous. The parents having a 1/2 chance of being heterozygous would only be true if the allelic frequency was fifty-fifty since:
2ab=1/2⇔ a=1/(4b) (Interaction term of Hardy Weinberg)
and
a+b=1⇔a=1-b (Sum of probabilities must equal 1)
only intersect at (1/2,1/2)

You can't assume that the parents carry the disease allele because their offspring 100% has the disease and then use that to determine the probability of the same offspring having the disease.

This is correct. That would be affirming the consequent, I believe.

Nothing is missing, there is 50% (because it doesnt show, otherwise we would assumed that they had the disease) chance that the parents are carriers because their offspring has the disease. Thus, the disease is probably autosomal recessive, And the question is asking for the prob of the child having the disease or prob. of two events happening at the same time, which @ys2327 did above.

Probability of A and B happening at the same time is = PA * PB = 1/2 (probability that parent A is carrier) *1/2 (probability that parent A will pass on the diseases causing allele which is recessive in this case) * 1/2 (Probability of parent two being a carrier) * 1/2 (Prob of B of passing recessive allele)

You calculated the probability of two heterozygotes mating when nothing states that diseased individuals are unable to mate. You are neglecting Rrxrr and rrxrr. You are also assuming an allelic frequency of .5 which is not what the question states.

Well, where did the offspring get the disease then? And that's actually not an accurate statement, The Question stems says that the disease is autosomal recessive, thus not expressed in hetero-zygotes. Hence, the only way for the offspring to have the disease is to receive one diseased copy from each parent, or to be homozygous recesive. So, in this case, you assume that both parent are heterozygotes.

NOPE NOPE NOPE. I think you may be confusing the red letter "B" for an indication that the child is affected when it's merely a label. Additionally, as I stated above, you are assuming diseased individuals cannot mate and thus the only way to get a homozygous diseased is to mate heterozygotes, which isn't stated in the question.

If the offspring were assumed to have the disease, the answer would be 1 and this would be a trick question. The question never states that the child has the disease. The question asks what is the probability that they have it without giving us direct information about the zygosity of the parents. There are no labels saying that either the parents or the child are affected.

Let R = Normal allele
Let r = Diseased allele

Since we know the allelic frequency in the population (R=2/3, r=1/3) and we assume Hardy-Weinberg equilibrium, we know that each parent has a 4/9 chance to be RR homozygous normal, 1/9 chance to be rr homozygous recessive, and 4/9 chance to be Rr heterozygous.

PROBABILITY THAT DAD DONATES DISEASED ALLELE:

• 0% chance from RR
  • (Chance of r from RR)*(Weight of RR branch)=0*4/9=0
• 50% chance from Rr
  • (Chance of r from Rr)*(Weight of Rr branch)=1/2*4/9=2/9
• 100% chance from rr
  • (Chance of r from rr)*(Weight of rr branch)=1*1/9=1/9

0+2/9+1/9=3/9=1/3 (We didn't actually have to do this calculation since given a random allele in the population, it'd be r 1/3 of the time as stated in the question, but I wanted to be thorough)

PROBABILITY THAT MOM DONATES DISEASED ALLELE:

The problem states nothing about sex-linked so it is the same as Dad: 1/3

PROBABILITY THAT B HAS rr GENOTYPE:

1/3*1/3=1/9. The answer is incorrect.

---------------------

Another "well duh" way to do this problem is that given a HW equilibrium population, any individual has a (1/3)^2 chance of having the disease, so without knowing the parents' genotypes of course the chance of the kid having the disease would be 1/9.

---------------------

Another "well duh" way to see that the answer in the slide is wrong is that 3 and 16 ARE FREAKING RELATIVELY PRIME (or coprime if you prefer). The only way that you could get a 1/16 out of a problem starting with a 1/3 is to at one point multiply by 3, which in single-gene genetics, basically never happens.



EDIT: I honestly think it was a typo and your prof meant to write "1 in 4 (.25)"

 

[5words]

Except the allelic frequency affects the chances that the parents are heterozygous. The parents having a 1/2 chance of being heterozygous would only be true if the allelic frequency was fifty-fifty since:
2ab=1/2⇔ a=1/(4b) (Interaction term of Hardy Weinberg)
and
a+b=1⇔a=1-b (Sum of probabilities must equal 1)
only intersect at (1/2,1/2)



This is correct. That would be affirming the consequent, I believe.



You calculated the probability of two heterozygotes mating when nothing states that diseased individuals are unable to mate. You are neglecting Rrxrr and rrxrr. You are also assuming an allelic frequency of .5 which is not what the question states.


NOPE NOPE NOPE. I think you may be confusing the red letter "B" for an indication that the child is affected when it's merely a label. Additionally, as I stated above, you are assuming diseased individuals cannot mate and thus the only way to get a homozygous diseased is to mate heterozygotes, which isn't stated in the question.

If the offspring were assumed to have the disease, the answer would be 1 and this would be a trick question. The question never states that the child has the disease. The question asks what is the probability that they have it without giving us direct information about the zygosity of the parents. There are no labels saying that either the parents or the child are affected.

Let R = Normal allele
Let r = Diseased allele

Since we know the allelic frequency in the population (R=2/3, r=1/3) and we assume Hardy-Weinberg equilibrium, we know that each parent has a 4/9 chance to be RR homozygous normal, 1/9 chance to be rr homozygous recessive, and 4/9 chance to be Rr heterozygous.

PROBABILITY THAT DAD DONATES DISEASED ALLELE:

• 0% chance from RR
  • (Chance of r from RR)*(Weight of RR branch)=0*4/9=0
• 50% chance from Rr
  • (Chance of r from Rr)*(Weight of Rr branch)=1/2*4/9=2/9
• 100% chance from rr
  • (Chance of r from rr)*(Weight of rr branch)=1*1/9=1/9

0+2/9+1/9=3/9=1/3 (We didn't actually have to do this calculation since given a random allele in the population, it'd be r 1/3 of the time as stated in the question, but I wanted to be thorough)

PROBABILITY THAT MOM DONATES DISEASED ALLELE:

The problem states nothing about sex-linked so it is the same as Dad: 1/3

PROBABILITY THAT B HAS rr GENOTYPE:

1/3*1/3=1/9. The answer is incorrect.

---------------------

Another "well duh" way to do this problem is that given a HW equilibrium population, any individual has a (1/3)^2 chance of having the disease, so without knowing the parents' genotypes of course the chance of the kid having the disease would be 1/9.

---------------------

Another "well duh" way to see that the answer in the slide is wrong is that 3 and 16 ARE FREAKING RELATIVELY PRIME (or coprime if you prefer). The only way that you could get a 1/16 out of a problem starting with a 1/3 is to at one point multiply by 3, which in single-gene genetics, basically never happens.



EDIT: I honestly think it was a typo and your prof meant to write "1 in 4 (.25)"

Just a tip, hardy Weinberg principles only applies to population, not individual. look up the definition of p and q allele frequency. The question pertains to the individual B, and sorry i am not even going to read the rest of your post. Also, only one gene is being studied here, so shouldnt be RRxx (idk how you got that) but Rr x Rr. I'll advise you to re-read your genetic textbook. And also look up the definition of recessive autosomal disorder, which is stated in the ! stem,

Another "well duh" way to do this problem is that given a HW equilibrium population, any individual has a (1/3)^2 chance of having the disease, so without knowing the parents' genotypes of course the chance of the kid having the disease would be 1/9.

You are confusing probability with allele frequency (in population) again, yeah... Genetics book my friend 🙂 . just a quick fyi:

q^2 = is the proportion or percentage of homozygous recessive individual in a population, not the probability that an unknown child C will have the disease. Don;t confuse the two.

 

[freak7]

Edit: Can't believe that you're med student, and making silly mistake such as this one.



You are confusing probability with allele frequency (in population) again, yeah... Genetics book my friend 🙂 . just a quick fyi:

q^2 = is the proportion or percentage of homozygous recessive individual in a population, not the probability that an unknown child C will have the disease. Don;t confuse the two.

1) Please do not talk down to me, I'd appreciate it.
2) The individual is a member of the population (as the question even states). The allelic frequency of the population affects the probability that a random individual will have the gene. Let's take an extreme example to prove my point. Let's say there is a 1 in a million chance of the recessive gene in the population, does that change your opinion that there's a 50/50 shot that the parent is a heterozygote?

Edit: letter

 

[5words]

Except the allelic frequency affects the chances that the parents are heterozygous. The parents having a 1/2 chance of being heterozygous would only be true if the allelic frequency was fifty-fifty since:
2ab=1/2⇔ a=1/(4b) (Interaction term of Hardy Weinberg)
and
a+b=1⇔a=1-b (Sum of probabilities must equal 1)
only intersect at (1/2,1/2)



This is correct. That would be affirming the consequent, I believe.



You calculated the probability of two heterozygotes mating when nothing states that diseased individuals are unable to mate. You are neglecting Rrxrr and rrxrr. You are also assuming an allelic frequency of .5 which is not what the question states.


NOPE NOPE NOPE. I think you may be confusing the red letter "B" for an indication that the child is affected when it's merely a label. Additionally, as I stated above, you are assuming diseased individuals cannot mate and thus the only way to get a homozygous diseased is to mate heterozygotes, which isn't stated in the question.

If the offspring were assumed to have the disease, the answer would be 1 and this would be a trick question. The question never states that the child has the disease. The question asks what is the probability that they have it without giving us direct information about the zygosity of the parents. There are no labels saying that either the parents or the child are affected.

Let R = Normal allele
Let r = Diseased allele

Since we know the allelic frequency in the population (R=2/3, r=1/3) and we assume Hardy-Weinberg equilibrium, we know that each parent has a 4/9 chance to be RR homozygous normal, 1/9 chance to be rr homozygous recessive, and 4/9 chance to be Rr heterozygous.

PROBABILITY THAT DAD DONATES DISEASED ALLELE:

• 0% chance from RR
  • (Chance of r from RR)*(Weight of RR branch)=0*4/9=0
• 50% chance from Rr
  • (Chance of r from Rr)*(Weight of Rr branch)=1/2*4/9=2/9
• 100% chance from rr
  • (Chance of r from rr)*(Weight of rr branch)=1*1/9=1/9

0+2/9+1/9=3/9=1/3 (We didn't actually have to do this calculation since given a random allele in the population, it'd be r 1/3 of the time as stated in the question, but I wanted to be thorough)

PROBABILITY THAT MOM DONATES DISEASED ALLELE:

The problem states nothing about sex-linked so it is the same as Dad: 1/3

PROBABILITY THAT B HAS rr GENOTYPE:

1/3*1/3=1/9. The answer is incorrect.

---------------------

Another "well duh" way to do this problem is that given a HW equilibrium population, any individual has a (1/3)^2 chance of having the disease, so without knowing the parents' genotypes of course the chance of the kid having the disease would be 1/9.

---------------------

Another "well duh" way to see that the answer in the slide is wrong is that 3 and 16 ARE FREAKING RELATIVELY PRIME (or coprime if you prefer). The only way that you could get a 1/16 out of a problem starting with a 1/3 is to at one point multiply by 3, which in single-gene genetics, basically never happens.



EDIT: I honestly think it was a typo and your prof meant to write "1 in 4 (.25)"

1) Please do not talk down to me, I'd appreciate it.
2) The individual is a member of the population. The allelic frequency of the population effects the probability that a random individual will have the gene. Let's take an extreme example to prove my point. Let's say there is a 1 in a million chance of the recessive gene in the population, does that change your opinion that there's a 50/50 shot that the parent is a heterozygote?

Sorry at 1, but at 2 yes it does.

becuase all, the allele frequency means is that if we have 100 individuals in the population, and the recessive allele frequency is 0.3 then 0.09 * 100 = 9 individuals in the entire population will have the disease. It doesnt , on there hand, suggest that any kids from a XX vs XX cross with 0.09 chance to have the disease. Some indivdual will have a 100% chance of having the disease, other a 0% chance.. But overall, only 9 individual in the entire people will have it. Hence, why i stated , you cant use q and p to calculate probabilty of individuals.. but sorry if i came across as a jerk.

 

[freak7]

q^2 = is the proportion or percentage of homozygous recessive individual in a population, not the probability that an unknown child C will have the disease. Don;t confuse the two.

In genetics frequency and probability are actually the same. It's actually a real pain for math people like me to work with. From Wiki:
A locus in this population has two alleles, A and a, that occur with initial frequencies f0(A) = p and f0(a) = q, respectively[1]

1. The term frequency usually refers to a number or count, but in this context, it is synonymous with probability.

 

[5words]

In genetics frequency and probability are actually the same. It's actually a real pain for math people like me to work with. From Wiki:
A locus in this population has two alleles, A and a, that occur with initial frequencies f0(A) = p and f0(a) = q, respectively[1]

1. The term frequency usually refers to a number or count, but in this context, it is synonymous with probability.

in this case, if i may, see probability as likehood.

 

[freak7]

becuase all, the allele frequency means is that if we have 100 individuals in the population, and the recessive allele frequency is 0.3 then 0.09 * 100 = 9 individuals in the entire population will have the disease.

Correct.

It doesnt , on there hand, suggest that any kids from a XX vs XX cross with 0.09 chance to have the disease. Some indivdual will have a 100% chance of having the disease, other a 0% chance.. But overall, only 9 individual in the entire people will have it. Hence, why i stated , you cant use q and p to calculate probabilty of individuals.. but sorry if i came across as a jerk.

This is a fundamental misunderstanding of probability. That is like saying "if I take a test that is pass/fail, there's a 50/50 chance that I'll pass it since I can only pass or fail". The person exists in the population. If we know nothing else about the individual, the most information we can use is the population data. Back to my 1/1,000,000 frequency of an allele example:

Say we have this same problem, but we replaced the allelic frequency. Say we know nothing about the parents and we wish to assess the the child's odds of having this extremely rare disease which is autosomal recessive. Would you honestly say that this person, without knowing anything about the parents, has a 1/16 chance of having this incredibly rare disease? Of course you wouldn't. It's not that frequent in the population. You should be able to go out on the street, point at two parents pushing a their kid in a stroller and bet your buddy a dollar that 0 of those 3 have this crazy rare disease.

There are a few cases to consider (Still using R and r, omitting duplicates):
RRxRR - EXTREMELY COMMON. Almost nobody has the allele since it's so rare. This pairing will never have the disease.
RRxRr - Extremely rare since almost nobody has this allele. 1999998/10^12% of people in a population are heterozygous. Still no diseased offspring.
RRxrr - Even more extremely rare because even fewer people are homozygous. Orders of magnitude fewer people. Still no diseased offspring.
Rrxrr - Even more even more extremely extremely rare. 50% of these offspring will have it. However this is STILL SUPER RARE.
rrxrr - This is a unicorn mating with a unicorn to create a super ultra unicorn. 100% of offspring from this pairing will have it.

Like it or not, the allelic frequency affects a random individual's alleles. Obviously not when you check etc, but with the absence of additional data we're just working with probabilities.

 

[freak7]

in this case, if i may, see probability as likehood.

That is literally what probability is. That's what I've been taking it to mean this whole time.

 

[freak7]

Basically it all boils down to this:

1/2 (probability that parent A is carrier)... 1/2 (Prob of B of passing recessive allele)

Here you are using your parentheses to clarify your 1/2's I get that. However the probability that a random person A is a carrier is not 1/2. It's based on the allelic frequency in the gene pool.

 

[5words]

Anyways, you are the med student here, if you still think i am wrong, so be it. I pretty much have nothing else to say.

 

[freak7]

Anyways, you are the med student here, if you still think i am wrong, so be it. I pretty much have nothing else to say.

Haha appeal to authority is a logical fallacy you know 😉

Yeah genetics is a pet peeve of mine since I worked in a genetics lab.
Math is also a pet peeve of mine since I majored in math.
Probability is also also a pet peeve of mine since I'm married to an actuary.

TBH It was probably just a typo on the slide though that the prof forgot to go back to correct when they changed the numbers for the new year.