Simple inheritance question

Genetics as it applies to evolution, molecular biology, and medical aspects.

Moderators: honeev, Leonid, amiradm, BioTeam

Post Reply
eth1r
Garter
Garter
Posts: 4
Joined: Thu Nov 02, 2006 2:21 am

Simple inheritance question

Post by eth1r » Thu Nov 02, 2006 2:56 am

Ok, so this is probably a simple inheritance question but I'm having a tough time grasping the concept. The problem I'm having arises out of a question from a college entrance exam, which I translated to the best of my ability.

Sickle cell anemia is a disease (...) etc. Its inheritance pattern is autosomic recessive. Two people who want to have a child consult a geneticist, since both of them have a brother who has sickle cell anemia. If you were the geneticist, even before you conducted any exams, you'd tell them that the probability that their child would have the disease is : (right answer is 1/9)

So basically here's what the question says (in my understading):
The parents of the couple were heterozygous for the gene, and since the couple was not they had at least on dominant allele. My problem is understanding the couple's probability of being heterozygous as well. Here's my biology teacher's basic explanation: "Draw a punett square and you'll have the possible pairs of alleles for each of the people in the couple:

Code: Select all

   |A    | a 
A | AA | Aa
a | Aa  | aa


Now since they obviously can't be 'aa', just cancel that out, and the probability they'll be heterezygous is 2/3"

But I just can't understand that. The way I see it, you'd have to cancel a whole column or line, because since you necessarily have one dominant allele (and thus one gamete), the chances of a person in the couple being homozygous is dependent upon the other gamete, thus being 1/2. Or even thinking this way:
If each person in the couple were to be heterozygous they could have either:
One dominant from the mother and either one recessive from the father or a dominant from the father. (1/2 chance for recessive) OR
One dominant from the father and either one recessive from the mother or one dominant from the mother. (1/2 chance for recessive)
So (1/2 + 1/2) / 2 (since order doesn't matter): 1/2.

Anyone care to explain the logistics of it being 2/3 in any other way?

(Sorry the post got so long)

Thanks in advance,
Fernando

druid
Coral
Coral
Posts: 151
Joined: Thu Oct 12, 2006 8:02 pm
Location: North Korea

Post by druid » Thu Nov 02, 2006 6:46 pm

The answer is correct.

Explanation:

1) We assume that homozygotes don't manage to reach reproductive age.

2) By 1) not man and not woman are sick.
The only way that they can bring sick child is that they are both heterozygous for disease. That is man is S/s, woman is too S/s.

2) Because both man and woman has sick brother , their parents are heterozygous S/s ( by 1) )

3) Man is healthy but his brother is sick. Probability that he is S/s is 2/3. Here you ofcoz failed and thought it's 1/2. Think deeper.

4) The same argument regards woman. Prob. she's S/s is also 2/3.

5) Probaility that two heterozygotes have homozygous ( sick ) child is 1/4

6) Finally 1/4 * 2/3 * 2/3 = 1/9

eth1r
Garter
Garter
Posts: 4
Joined: Thu Nov 02, 2006 2:21 am

Post by eth1r » Thu Nov 02, 2006 9:42 pm

druid,

While I appreciate the effort, your reply didn't answer my question to the least bit. I knew the answer was right, I knew the parents had to be homozygous, I knew the final math, etc, etc. What I didn't know was exactly what I asked, and to which you said only "think deeper": Why is it 2/3 and not 1/2?

Thanks again,
Fernando

druid
Coral
Coral
Posts: 151
Joined: Thu Oct 12, 2006 8:02 pm
Location: North Korea

Post by druid » Thu Nov 02, 2006 10:41 pm

The diagram maybe help you with school-level combi ;)

Image[/img]
Attachments
gen.JPG
The diagram
(15.19 KiB) Downloaded 86 times

eth1r
Garter
Garter
Posts: 4
Joined: Thu Nov 02, 2006 2:21 am

Post by eth1r » Thu Nov 02, 2006 10:51 pm

druid,

This is getting a little frustrating. The diagram you drew is exactly what I meant by punett square. What I don't understand is the logistics behind it, nor can I find any other way to express the probability as 2/3. So any further reasoning would be appreciated.

Thanks,
Fernando

druid
Coral
Coral
Posts: 151
Joined: Thu Oct 12, 2006 8:02 pm
Location: North Korea

Post by druid » Thu Nov 02, 2006 11:40 pm

It's really hard to explain obvious thing :(

You know that man is healthy. And he belongs to offspring with sick child. But you know for SURE he is not s/s because he is not sick. So you have to choose 2 heteerozygotes from 3 possible cases: S/S, S/s and again S/s. Thus the probability is 2/3 ( 2 account for S/s and 1 for S/S ).
It's not a case when a child is still unborn. This is case when one possibility is thrown out - possibility that the man is a/a. You throw this possibility because you already know he is NOT sick. Thus, what remained to you is to choose which kind of healthy person he is - S/S or S/s.

druid
Coral
Coral
Posts: 151
Joined: Thu Oct 12, 2006 8:02 pm
Location: North Korea

Post by druid » Thu Nov 02, 2006 11:44 pm

When child is unborn you must include s/s as an option. Thus for S/s you get 2/4.

For born child you can know if he's sick or not. In our case the child ( man ) is not sick so he is not s/s. Therefore you just exclude this case. Thus you get 2/3 instead of 2/4.

(it seems good explanation)
:)

druid
Coral
Coral
Posts: 151
Joined: Thu Oct 12, 2006 8:02 pm
Location: North Korea

Post by druid » Thu Nov 02, 2006 11:47 pm

Remember, what you want to find is what possibility is that phenotypically healthy (that is who is not s/s for sure) man has genotype S/s.

(one more addition ;))

User avatar
Dr.Stein
King Cobra
King Cobra
Posts: 3501
Joined: Thu Jul 07, 2005 7:58 am
Location: 55284 Yogyakarta, Indonesia
Contact:

Post by Dr.Stein » Fri Nov 03, 2006 3:27 am

Here I come up with a pedigree, hopefully this could help you to provide a better explanation.
Click the picture to see larger view.

Image

GI = grandparents
GII = parent who did consultation
GIII = future children (could be sons or daughters)

- Because the parent are normal but each has a brother with SCA, so the grandparents must be "Ss"

- The probability of parent to be normal is 3/4, which is 1/4 for "SS" and 1/2 for "Ss"

- The consultation is about "the probability to have SCA child". Only "Ss" parent that is possible to have SCA child. However, we do not know whether the parent is "SS" or "Ss". Thus, the probability of the parent to be "Ss" = 2/3, for both father and mother.

- The possibility for the child to be SCA is 1/4.

- Then the calculation will be = 2/3 * 2/3 * 1/4 = 1/9
Image

eth1r
Garter
Garter
Posts: 4
Joined: Thu Nov 02, 2006 2:21 am

Post by eth1r » Fri Nov 03, 2006 3:47 am

druid,

Alright, so I just reworked the problem thinking about what the punett square is supposed to represent, rather than using the square itself, and I'm finally ok with it. I am, however, still working on what becomes of the numbers resulting from the punett square (as factors of probability), and will post in this topic any further doubts or ideas I might have, and hope someone can give me feedback on them.

Thanks again for all your answers.

EDIT:

Dr. Stein,

Sorry, I guess I missed your post. Anyways, I think it's funny how you people keep assuming I know nothing of genetics is supposed to work or how you calculate probability or how I don't have a clue what the question asks for, when I (think I've) made it so clear that I was having a problem understading the logic behind a very specific part of the problem. Oh well, no biggie.

And btw, interesting use of the copyright there.

Thanks for you answer.

druid
Coral
Coral
Posts: 151
Joined: Thu Oct 12, 2006 8:02 pm
Location: North Korea

Post by druid » Fri Nov 03, 2006 12:57 pm

eth1r wrote:Alright, so I just reworked the problem thinking about what the punett square is supposed to represent, rather than using the square itself


Before you apply any method it's always smart to check if the method is applicable to the given conditions.

User avatar
Dr.Stein
King Cobra
King Cobra
Posts: 3501
Joined: Thu Jul 07, 2005 7:58 am
Location: 55284 Yogyakarta, Indonesia
Contact:

Post by Dr.Stein » Mon Nov 06, 2006 5:54 am

eth1r wrote:Dr. Stein,

Sorry, I guess I missed your post. Anyways, I think it's funny how you people keep assuming I know nothing of genetics is supposed to work or how you calculate probability or how I don't have a clue what the question asks for, when I (think I've) made it so clear that I was having a problem understading the logic behind a very specific part of the problem. Oh well, no biggie.

And btw, interesting use of the copyright there.

Thanks for you answer.

I am sorry but I never assumed you that way. You might misunderstand me. It is my habit to answer questions or solve problems from the beginning til the end. That's the way I think and I belive it is a good way. But if it made you get offended, that you thought I treated you like a beginner, well I am so sorry. I never meant to be so. Maybe I mistook your question or something like that. But I just tried to clear about what druid said "think deeper". To be there, I must start the calculation from the beginning to avoid any mistakes I did unintentionally. Genetics is not my major but I like to learn it. If you don't like the way I answer you, I won't answer you again next time. I don't want to make you offended, though I never meant to be so. Thank you and my apologies.
Image

Post Reply

Who is online

Users browsing this forum: No registered users and 3 guests