Probability of a colour blind grandson

AI Thread Summary
The discussion centers on calculating the probability of a grandson being color blind from a color blind father and a mother with normal vision and no family history of color blindness. The consensus suggests that the probability is 0.25, based on the assumption that the couple's daughters will be carriers and their sons will not inherit the color blindness gene. Some participants argue about considering various scenarios, including the genetic backgrounds of potential spouses for the children, which could alter the probabilities. However, the main argument is that without additional data on the spouses, the simplest Mendelian approach yields a probability of 0.25. Ultimately, the conclusion leans towards 0.25 as the most reasonable answer given the information provided.
Suraj M
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Homework Statement


Here is the question as it is.
A colour blind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their grandson being colour blind?
(1) 0.25
(2) 0.5
(3) 1
(4) Nil

Homework Equations


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The Attempt at a Solution


Im sorry, but i hope the images will do,
IMG_1732.JPG

Ignore the zero I've written, that branch is anyway not needed..

IMG_1733.JPG


Is this correct?
Just wanted to check, they haven't mentioned anything else, so have to consider everything, right?
But my teacher says that we don't need to consider all these cases for questions like these.
please help me out
thank you
 
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It's like none of there sons going to be colour blind, half of daughters will be normal and half carrier. From the carrier daughters,whatever husband they get only 1/2 sons will be colour blind. That's the solution in genetics. You calculate the probability using this data in mathematical equation. Maybe your answer is not correct as half of sons of the daughters are colour blind, from carrier daughters only. And carrier daughters are half of the total daughters.
 
Suppose they have four children, two daughters and two sons(only one daughter is carrier). Each of the children gets four children same way, two sons and two daughters. So, total grandsons are 4*2=8.

Again only a single son of the carrier daughter is probable to be a colour-blind.
 
I am sorry, I mixed up. Both the daughters will be carrier. If they have 4 children each, two daughters and two sons then 1 of the sons will be colour blind, no matter what husband they get. So, two daughters will have two colour blind sons. Finally there are 2 out of 8 grandsons who are colour blind. The probability is 2/8 or 0.25
 
fireflies said:
2 out of 8 grandsons
2 out of 4 right?, there are just 4 grandsons?
 
8 grandsons

In post 2, I said how. Its not 8 grandsons, its 8 probable types of grandsons
 
That's what you do in genetics. There are two types of genes coming from father, two types from mother, total makes 4 types. Since 2 of these types are female, so male types are just 2.
 
I don't understand, whatever be the genotype of the male that the carrier marries, the probability of her son to be colourblind is 1/2 ?
so shouldn't 0.5
 
You are assuming that whomever the son or daughter marries is also neither colorblind nor a carrier, right?
So this is the minimum probability. If you add in the chance of the son or daughter marrying someone with the gene, the odds go up a little bit.
 
  • #10
There should be no chance for the son to have the gene, since it is on the X chromosome, right?
 
  • #11
Yes, probability for her colour blind son is 0.5, but the question is not her probability to have colour blind sons, its her parents probability to have colour blind grandsons
 
  • #12
RUber said:
You are assuming that whomever the son or daughter marries is also neither colorblind nor a carrier, right?
So this is the minimum probability. If you add in the chance of the son or daughter marrying someone with the gene, the odds go up a little bit.

No, I am assuming about the daughters as it will be all the same then. No assumption to who the sons are married to.

Here, the sons can be married to three types of women- normal, colourblind and carrier. I just assumed it to be normal.

If yoy consider all the three cases then for each son

1) having normal wife colour blind son is 0 out of 2, Then probability 2/8

2) For carrier, it is 1 out of 2. For bolth sons its 1 out of 2.

Probability 3/8

3) colour-blind woman then it is 2 out of 2. For both sons its 4 out of 4

Probability 5/8

Then final probability should be calculated by mathematics
 
  • #13
The no.2 is a mistake. For both sons it will be 2 out of 4, so total probabilities is 4 out of 8.
 
  • #14
That's not how probability works. I am sure that the probability of a non-carrier woman is not the same as a colorblind woman. Without additional information about the population of possible wives, you should probably go for the minimal case where the son's wife would be a non-carrier.

Example: Assume 90% of potential wife population is non-carrier, 9% is carrier, 1% is colorblind
The probability of a son's son being colorblind is (.09)*(.5) + (.01)*(1) = .055.

Similarly assume potential husband population is similarly distributed with 90% non-colorblind and 10% colorblind.
Then the daughter's son's odds of being colorblind are (.9)*(.5) + (.1)*(.5) = .5.

Then the odds of a grandson being colorblind would be p(son)*p(colorblind son of son) + p(daughter)*p(colorblind son of daughter) = (.5)*(.055) + (.5)*(.5) = .2775.

Since no additional information about the population statistics are provided in the problem, you should only assume non-colorblind, non-carriers, which would give an answer just slightly smaller than .2775.
 
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  • #15
Oh sorry i forgot to give you the options, i edited the post, do see it,
so 0.25 is your answer?
I think we should assume a mendelian population.
 
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  • #16
I could get it that there was some problems in my later calculation but I couldn't understand where.

But the part with the husband is not correct. Whatever husband the daughter gets the probable colour blind son for each daughter is always 0.5 as fathers don't give the gene to the sons.
 
  • #17
0.25 is the answer just when you consider that the sons have normal wives.

I think its preferable answer if you go with your teacher you need not consider all the things (like what wives they marry to). But if you consider with all these then you need extra data and also calculate the probability of sons son to be colour blind in the process RUber did
 
  • #18
fireflies said:
colour blind son for each daughter is always 0.5 as fathers don't give the gene to the sons.
Agreed on that, but can you ignore the male? what if he marries a colourblind woman or a carrier?
 
  • #19
fireflies said:
0.25 is the answer just when you consider that the sons have normal wive
Well this question appeared in a national exam (AIPMT) on saturday, the answers aren't out yet, but the answers I've come across by experts is 0.5 and Nil.
I have seen both these answers,
 
  • #20
Then its not possible to say how they do it, or what conditions they are really asking for. Maybe they skip the part of the sons and calculated just the grandsons of the daughters as sons aren't responsible to colour blind grandsons. Then it will be 1/2. If you get the solutions then let me know.
 
  • #21
Wouldn't what I've done in the attempt make sense? My teacher thinks the answer is 0.5!
 
  • #22
With the normal sons, the blue marked genotypes, are you considering them as wives? Then why their probability 1/4, 1/2, and 1/4?

If you consider the population it will be according to RUbers calculation, otherwise taking all equal it maybe 1/3 each. Maybe you took the probability of wives same as children.
 
  • #23
In a mendelian population, 50% of women are carriers 25% homozygous( each). that's why i took 1/4 1/2 1/4.
 
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  • #24
Your math approach is correct. If Mendelian population can be considered then it maybe a solution.

Though I am not sure if Mendelian population can be considered. Of course every 1 out of 4 women are not colour blind. The population is much less.

If I were to answer my bet would have been 0.25. But Nil and 0.5 may also be answer.
 
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  • #25
Suraj, I would recommend against the Mendelian population assumption, unless it is simply an academic exercise or you have been instructed to do so.

In the event that you do make that assumption, you would have:
P(grandson is colorblind) = (.5) * ( (.25)(0) + (.5)(.5) + (.25)(1)) +(.5)(.5) = .5. So I would concur with your math and fireflies.

The true probability, based on realistic population prevalence of the condition, would be much close to .25.
 
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  • #26
RUber said:
P(grandson is colorblind) = (.5) * ( (.25)(0) + (.5)(.5) + (.25)(1)) +(.5)(.5) = .5. So I would concur with your math and fireflies.
Another thing i suggested is that, we should take half of this, because there is a 50% chance of the child being a male, and then it becomes 0.25
Now if we take the female carrier also, it becomes, we get a 0.5 assuming a 0.5 chance of marriage if effected and normal male, and then half if that because probability of a female carrier is 0.5 then that comes to 0.25, which then adds to 0.5..as i have shown in my attempt, so not just the son's son.
 
  • #27
I didn't say it would be 0.5. According to me the answer is more likely to be 0.25. But the answer can be 0.5 taking Mendelian concept, or Nil taking your concept of population.

The rest is how the question-maker made the question
 
  • #28
fireflies said:
Nil taking your concept of population.
you Can't get nil even with the concept of population because the carrier daughter has a 0.5 chance of having a colourblind son, your words...no matter who the husband is,
so how could you get nil.?
 
  • #29
Maybe you misunderstood my concept. The 0.5 chance for a carrier daughter to have colour-blind son is 0.5, So, the main couple will have colour blind grandson in a probability 0.25. Look at the posts 3,4
 
  • #30
Answer 0 is in no case. I meant nil=answer not given/none of the answers above
 
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