Simple punnet square question

[Pinkleton]

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Let's say we are dealing with an autosomal recessive disease, and 2 heterozygous parents have some children. Of course, the punnet square result is AA Aa Aa aa

I know that the probability of an unaffected child being a carrier is 2/3, BUT what if you are asked the probability that 2 children are BOTH carriers? Do you just multiply 2/3 x 2/3 and get 4/9? I could see this being feasible...but when trying to figure out the chances of EITHER being a carrier wouldn't it be 2/3 + 2/3 = 4/3? 133% chance WTF? What am I missing here?

 

[dsoz]

Being a carrier is only 1/2. AA is not a carrier and neither is aa.

 

[calvnandhobbs68]

Being a carrier is only 1/2. AA is not a carrier and neither is aa.

This is a basic and classic genetics trip up. No, being a carrier is 2/3 in UNaffected children because an autosomal recessive exhibits the phenotype.

 

[Ycut]

Let's say we are dealing with an autosomal recessive disease, and 2 heterozygous parents have some children. Of course, the punnet square result is AA Aa Aa aa

I know that the probability of an unaffected child being a carrier is 2/3, BUT what if you are asked the probability that 2 children are BOTH carriers? Do you just multiply 2/3 x 2/3 and get 4/9? I could see this being feasible...but when trying to figure out the chances of EITHER being a carrier wouldn't it be 2/3 + 2/3 = 4/3? 133% chance WTF? What am I missing here?

having a child being a carrier and having another child being a carrier are two independent events and to calculate combined probability u have to multiply

 

[Pinkleton]

having a child being a carrier and having another child being a carrier are two independent events and to calculate combined probability u have to multiply

Right, that's what I said isn't it? 2/3 x 2/3? But what I'm confused about is if you're asked the probability of child 1 OR child 2 being a carrier

 

[Ycut]

Right, that's what I said isn't it? 2/3 x 2/3? But what I'm confused about is if you're asked the probability of child 1 OR child 2 being a carrier

yep 4/9


for either healthy child being a carrier probability is 2/3

why u add up probabilities ?

 

[ScienceNerdUofU]

To get the probability of one or the other you do indeed need to add up the probability that each one is a carrier. But then you need to subtract the probability that they are both carriers so that your not above 100%.

 

[calvnandhobbs68]

Right, that's what I said isn't it? 2/3 x 2/3? But what I'm confused about is if you're asked the probability of child 1 OR child 2 being a carrier

Here just think about it this way

There are two outcomes if you have an either/or example right?

Child 1: AA, Child 2: Aa
Child 1: Aa, Child 2: AA

The probability for those events are:

AAxAa = 2/3 x 1/3 = 2/9
AaxAA = 1/3 x 2/3 = 2/9

Now you want the probability that either event could happen so NOW you add them together:

So 2/9 + 2/9 = 4/9

If you want to make sure it makes sense you can add the other two probabilities in and you get 1.

AAxAA= 1/3 x 1/3 = 1/9
AaxAa= 2/3 x 2/3 = 4/9

So all possible outcomes = 1

Edit: Okay I guess you'd have to clarify what you mean by either/or. Do you mean the probability that one would be a carrier and the other would not? Because that would be 4/9. Do you mean the probability that you randomly pick one out of either and he's a carrier? Because that would be 8/9 (you now have to add in the probability that they are both carriers because the only way NEITHER would be a carrier is if you had AAxAA which has a probability of only 1/9).

 

[Ycut]

if u randomly pick a phenotypically normal child from this family

whats the chance that he/she is a carrier ?

 

[Suncrusher]

To get the probability of one or the other you do indeed need to add up the probability that each one is a carrier. But then you need to subtract the probability that they are both carriers so that youre not above 100%.

Yep. I calculated 2/3 + 2/3 - 4/9 = 8/9. OP was essentially double-counting (well, not quite double, but you get the idea) the chances of each child being a carrier since he didn't subtract the odds that they were both carriers.

The other way to solve the problem is to do what calvinandhobbes68 did, which is to rephrase the question of "what are the odds of either child being a carrier" to "1 - the odds of neither child being a carrier." 1 - 1/9 = 8/9.

 

[Pinkleton]

Ah thanks for clearing that up guys, I knew I was leaving out something important 🙂