A variety of snail has an inherited condition that affects the thickness of the shell. The alleles are t and n.

tt have thick shells
tn have thin shells
nn do not survive

Two heterozygous snails are mated. What is the probability that a surviving snail of the next generation is a heterozygote?

My workings:

So a tn snail mates with another tn snail, and they would get a tt, tn, tn, and nn right?

That means 3 out of 4 survives.

and the heterozygotes are the tn snails, so is the answer 2/3???

If you are also Mike, please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.

If not, you are right!

Well, it seems like you have the right idea, but let's take a closer look at the possible outcomes. When two heterozygous snails (tn) mate, there are four possible combinations of alleles that their offspring can inherit:

1. tt (thick shell)
2. tn (thin shell)
3. tn (thin shell)
4. nn (does not survive)

As you correctly stated, the nn snail does not survive. So we are left with three surviving snails, out of which two are heterozygotes (tn). Therefore, the correct probability of a surviving snail in the next generation being heterozygous is 2 out of 3.

So yes, you got it right! Good job! And remember, when it comes to snail genetics, it's all about shell-ection! Keep up the great snail analysis!

Your workings are correct up to the point that you identified the genotype of the offspring. When two heterozygous snails (tn) are mated, their possible offspring genotypes are tt, tn, tn, and nn. However, it is important to note that nn offspring do not survive, as mentioned in the question.

So, out of the four possible genotypes, only three have the potential for survival:

- tt (thick shell)
- tn (thin shell)
- tn (thin shell)

This means that 3 out of 4 snails will survive to the next generation.

To calculate the probability of a surviving snail in the next generation being a heterozygote (tn), you need to determine the proportion of surviving heterozygotes to the total number of surviving offspring.

Since two out of three surviving snails are tn, the probability of a snail in the next generation being a heterozygote is 2/3 or approximately 0.667. Therefore, your answer of 2/3 is correct.

To determine the probability of obtaining a surviving snail of the next generation that is a heterozygote, we'll need to use Punnett squares.

Let's start by representing the alleles of the two heterozygous snails as:
Parent 1: Tn
Parent 2: Tn

We can now create a Punnett square to determine the possible genotypes of the offspring:

N N
----------------
T | TT | TN |
n | TN | nn |
----------------

As you correctly mentioned, the genotype tt is associated with thick shells, tn with thin shells, and nn does not survive.

From the Punnett square, we can observe that out of the four possible genotypes, three are associated with surviving snails (TT, TN, TN), and one leads to a non-surviving snail (nn).

Since we are interested in the probability of obtaining a surviving snail that is a heterozygote (TN), out of the three surviving genotypes, we have two that fit this criterion.

Thus, the probability that a surviving snail of the next generation is a heterozygote is indeed 2/3.