# Mating, how many offspring?

1. Oct 5, 2005

### Ester

Bbss with bbss, what to do?

In goats, black color (B) is domnant over red (b), and solid color (S) is dominant over spotted (s). If the offspring between Bbss and bbss individuals are mated with each other, what fraction of their offspring will be expected to be black and spotted? Assume these genes are unlinked.
A. 1/16
B. 9/16
C. 1/9
D. 3/16
E. 3/4

I think that the answer is 3/4, but I'm not sure and I don't know the right answer.
The possible gametes that I get for Bbss are Bs and bs.
The possible gametes that I get for bbss is just bs.
The possible offspring of these gametes that I get are: Bbss and bbss, which are the same as the parents. This is were I get stuck. I don't know how to do it using the punnet square or how to use probability in this problem.
How do you do this problem?

Last edited: Oct 5, 2005
2. Oct 5, 2005

### Ester

I figure, since they come out to be the same as parents, I would cross Bbss with itself because it is black and spotted. Then, the possible gametes each would have would be Bs and bs. The possible offspring that would result are: BBss, Bbss, Bbss, bbss. Of these four, there are phenotipycally, only three that are black and spotted.
Is this the right approach? Is the answer then 3/4?

3. Oct 5, 2005

### Moonbear

Staff Emeritus
Okay, setting up the Punnet square is the key to getting this question, so I'll help you with that.

For the goat of genotype Bbss, you actually have 4 possible gametes, even if some of them are duplicates. Let me illustrate using color where each gamete comes from:
With the genotype of Bbss, you get the following gametes:

Bs Bs bs bs

There are also 4 possible gametes for the goat of genotype bbss.

Can you now set up your own Punnett square with this information? I'll check back later to see how you're doing.

4. Oct 5, 2005

### Ester

but the "ss" can't be two different colors because they both represent spotted. and "bb" can't be of two different colors because they represent red. so only two gametes are possible for Bbss and one gamete for bbss. That is what I've been told. I just don't know how to get the right answer now, maybe the question isn't worded correctly?

Even if what you're saying is true, then:
the possible gametes for Bbss are: Bs, Bs, bs, bs
the possible gametes for bbss are: bs, bs, bs, bs

That would give 16 offspring which are:
Bbss, Bbss, bbss, bbss, Bbss, Bbss, bbss, bbss, Bbss, Bbss, bbss, bbss, Bbss, Bbss, bbss, bbss.

There are 8 that have B.
All of them have some type of an "s" either bold or regular.
So wouldn't that mean the fraction is 8/16 or 1/2. That is not one of the choices!

Last edited: Oct 5, 2005
5. Oct 5, 2005

### Moonbear

Staff Emeritus
The question is worded fine. I'm using the colors to represent the different alleles, not different genotypes. There are only two unique gametes possible, but if you're going to use a Punnett square, you have to account for all the possible combinations of alleles. The outcome of the Punnett square will get you the probability results you need (I'm forcing you to do it the long way so you understand the concept...once you understand the concept, you can take a shortcut, which I think you're trying to use without understanding completely how you use it).

Okay, so I don't confuse you further, here is the set-up of the Punnett square. You can fill it in and see if you then understand how to get the probability answer for your question once you have the square filled in.

__Bs Bs bs bs
bs
bs
bs
bs

6. Oct 5, 2005

### Ester

So, then how to get the probability answer for the question. I have the punnet square worked out, as you can see on top.

7. Oct 5, 2005

### Moonbear

Staff Emeritus
Okay, I see you do know how to do the Punnett square...I thought that was where you were running into trouble. You can ignore the colored part from now on (I don't think it ended up helping anyway and I'm worried now I'm confusing you more with it rather than helping).

So, you're halfway there. This is a messy question. They then want you to take the F1 offspring and cross them to determine the outcome in the F2 generation. That's going to be a BIG Punnett square.

8. Oct 5, 2005

### Ester

What is that supposed to mean? Cross each of the offspring with the remaining offspring and then see how many "B" there are?

What is the shortcut? Would you happen to know the correct answer?

I doubt that we're doing this correctly, because I would only get about 40 seconds to do the question on the test. So something isn't right here.

Last edited: Oct 5, 2005
9. Oct 5, 2005

### Moonbear

Staff Emeritus
Okay, so the quick way now is what you were trying to do in the first place for the F1 generation (before I got us sidetracked with the more difficult Punnett squares when I thought that's where you needed help).

Look at your F1 generation genotypes...it is easier ONLY because of the particular result there. You have 1/2 Bbss and 1/2 bbss. So, you only have to do 4 crosses: BbssXBbss BbssXbbss bbssXBbss and bbssXbbss.

The middle two are identical, so you can just do one and count the results double. Oh, and yes, since you can ignore the solid/spotted part of the problem because they're all the same, you only need to do a 2x2 Punnett square for Black vs red (B- vs bb).

An exam strategy suggestion is that unless it was worth more points than all the other questions, I'd skip something like this question for the end so you can take your time and think about the best approach rather than feel rushed earlier. Do everything you can that's quick, and then spend time on one like this getting as far as you can with whatever time is left.

Last edited: Oct 5, 2005
10. Oct 5, 2005

### Ester

Thanks for all your help. I really appreciate it.
I'm trying to understand what you typed here, but I don't get it. What answer do you come up with? maybe that will help me see how to derive that answer.

because (1/2)(3/4)(1/2)=3/16
the half is from Bbss x bbss and there are two = 1/4
the 3/4 is from Bbss x Bbss and there is only one = 3/4
the bbss doesn't make any, so that won't count.

Last edited: Oct 5, 2005
11. Oct 5, 2005

### pattylou

I glazed over when I saw it had two loci - but you're right Moonbear, one of the loci can be ignored because it is recessive across the board.

Ester: If you recognize that all the F2 will have ss (you can get that by scanning the original genotyps and seeing no capital S), then you can focus on the B vs. b part of the problem and cut the time you spend on the problem by at least half.

Moonbear: Did you work through the problem? I got an answer that isn't one of the options. Who's wrong: me, or the person that wrote the answer choices?

12. Oct 5, 2005

### Moonbear

Staff Emeritus
Pattylou and ester, I think it's the answer choices that are wrong. I didn't look at them closely enough before because I was busy just working through the problem. Grr!!!

Ester, do you come up with an answer that isn't one of those choices?

13. Oct 5, 2005

### Ester

By looking at: BbssXBbss BbssXbbss bbssXBbss and bbssXbbss.
I get 3/16. In the top, I showed my thinking to derive this.
Using my first method, all the way where I started from, I get 3/4.
Thinking about it, I think it should just be 1/2.
What about you? What do you get when you worked out the answer?

Last edited: Oct 5, 2005
14. Oct 5, 2005

### Moonbear

Staff Emeritus
Ester, at this point, I think it's reasonable to give you the answer and let you see if you can work backward from that.

Pattylou and I both got an answer of 7/16 (and I've just done it the long version too to make sure I didn't miss something taking a short cut considering simple addition gets challenging this hour of the night).

Ignore the solid vs spotted part of the problem (all of the possible outcomes are spotted because you don't have any dominant alleles in the parent generation to pass along).

So, just cross all combinations of your two offspring genotypes (BbXBb, BbXbb, bbXBb, bbXbb).

Remember that both genotypes BB and Bb will give you black phenotypes.

For BbXBb, you get 1-BB, 2-Bb, 1-bb
For Bbxbb, you get 2-Bb, 2bb
For bbXBb, you get 2-Bb, 2bb
For bbXbb, you get 4-bb

Adding together, you get 1-BB, 6-Bb, 9-bb (For a total of 16 offspring).

15. Oct 5, 2005

### pattylou

The crosses you have identified are correct. The answers aren't.

Imagine there are 16 offspring altogether... Four for each of the crosses you have identified. How many black goats will be in *each* set of four?

Cross #1: ____ black goats out of four

Cross #2: ____ black goats out of four

Cross #3: ____ black goats out of four

Cross #4: ____ black goats out of four

Really, genetics *is* fun --- although you probably don't believe me!! :)

Edit: Moonbear's answer came up while I was writing this. They're in agreement.

16. Oct 5, 2005

### Ester

I get it. I understand that:
For BbXBb, you get 1-BB, 2-Bb, 1-bb
For Bbxbb, you get 2-Bb, 2bb
For bbXBb, you get 2-Bb, 2bb
For bbXbb, you get 4-bb

Adding together, you get 1-BB, 6-Bb, 9-bb (For a total of 16 offspring).
I finally understand that. The only problem I'm stuck with now is how you guys figured to cross: BbXBb, For BbXBb, Bbxbb, bbXBb, bbXbb
Once I know this, I can solve any problem that is similar to this.

17. Oct 5, 2005

### Moonbear

Staff Emeritus
That was from all the way back in post #4...the offspring you worked out from the first cross were equal numbers of Bb and bb, so we just crossed all possible combinations of those offspring (including to offspring of the same genotype as themselves).