Bio section Bank #30 Error?

[Astra]

We are to choose 2 red beetles out of 3 and 2 are red.

The probability for choosing Red 1 is: 2/3 ( because 2 are red)

According to AAMC, the probability for choosing Red 2 is : 2/3 ( because 2 are red).

But if you already picked one beetle, the remaining total would be 2, so wouldn't the probability for picking Red be 1/2?

Nowhere in the question does it say that the experiment is done with replacement, and since the number of offspring does is constant, the probabilities would change after you remove one beetle correct?

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[NextStepTutor_3]

If this were a case where 2 of 3 beetles were red, and you removed one red beetle, you'd be on the right track! But this question is asking something different. Here's a good way to look at it:

Of the F1 generation, 31 (about 75%) are red and 9 (about 25%) are brown. We see that the next step of the problem involves crossing two red beetles, so we stop worrying about the brown ones - which all must've been rr (if we call the dominant allele R and the recessive one r).

Here's where the idea of 2/3 comes in. Of the red beetles, 1/3 are RR (homozygous dominant) and 2/3 are Rr (heterozygous), considering that the original cross was Rr x Rr. What this means is that each beetle has a 2/3 chance of being Rr. Now, what happens when we pick our first beetle to take place in our next cross? It has a 2/3 chance of being Rr, like we said.

But when we choose our second beetle, it still has a 2/3 chance of being Rr. Why? Because we didn't "remove an Rr beetle" - we chose a random beetle that might've been RR or Rr. That's what we're doing here as well. All of the beetles have a 2/3 chance of being Rr, regardless of the other individual in the cross. (Note also that there are 31 red beetles, not just 3. Even if we somehow "removed" an Rr one, leaving our population with one fewer Rr individuals, it wouldn't result in any value being 1/3.)

Finally, we just need to multiply our 2/3 * 2/3 and get our answer! Does that help?

Good luck

 

[theonlytycrane]

@NextStepTutor_3 just to double-check, we know that the F1 cross was Rr x Rr based on the 3:1 red to brown ratio that we see from the f1 generation, correct?

 

[Astra]

@NextStepTutor_3 just to double-check, we know that the F1 cross was Rr x Rr based on the 3:1 red to brown ratio that we see from the f1 generation, correct?

Yup that's right

 

[FutureDr.Jefferson]

Yup that's right

I know this is super late, but I wanted to make sure for any one else coming here for help.

The question says two red beetles are crossed from an F1 generation of 40 offspring that had both red and brown beetles. They told us that red is dominant to brown. That means the original parents had to both be heterozygous (using a Punnett square shows us it’s the only way they could have made brown + white offspring).

Next it says 2 of the red F1 were crossed and asked the possibility of getting brown + white offspring again in F2 generation. In other words: what’s the possibility that the parents chosen from F1 to produce F2 will be heterozygous like the first time.

Again using a Punnett square we know that 2 of every three red are heterozygous. Using that scale, since we had 40, we know there were 30 red. 2 out of every 3 from 30 yields 20/30 being the heterozygote we need.

So, the chance of picking the 1st heterozygous parent is 20/30 = 2/3.

Now we have 29 in our new total. That’s still ~ 30. So the chances of picking a heterozygote again are still ~2/3. This is the AAMC. Get use to doing rediculous rounding in places your instructors would never have accepted. That means use 2/3.

Finally, the product rule says to multiply probabilities that are independent. The 2nd chance doesn’t depend on the first to get 20/29 = ~ 20/30 = 2/3. That’s (2/3)*(2/3) = 4/9

It’s like rolling two three sided dice with two of the sides on each one having the desired outcome. Or rolling just one of them twice.