allstardentist

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Laboratory mice are to be classified based on genes A,
B, and C. How many genetically distinct offspring can
be produced from a cross of an AaBbCc individual with
an AaBBCc individual?
A. 12
B. 16
C. 32
D. 48
E. 64

I think its 18 from multiplying 3x2x3, but the answer is C. Anyone can explain?
 

714guy

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allstardentist said:
Laboratory mice are to be classified based on genes A,
B, and C. How many genetically distinct offspring can
be produced from a cross of an AaBbCc individual with
an AaBBCc individual?
A. 12
B. 16
C. 32
D. 48
E. 64

I think its 18 from multiplying 3x2x3, but the answer is C. Anyone can explain?
For AaBbCc you have 3 heterozygous genes so 2^3 = 8 ways, and AaBBCc there are only 2 heterrozygous genes so 2^2 = 4 way. 4*8 = 32
 

jorUW

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allstardentist said:
Laboratory mice are to be classified based on genes A,
B, and C. How many genetically distinct offspring can
be produced from a cross of an AaBbCc individual with
an AaBBCc individual?
A. 12
B. 16
C. 32
D. 48
E. 64

I think its 18 from multiplying 3x2x3, but the answer is C. Anyone can explain?
You have to multiply by two b/c they can be male OR female. Tricky Question
 
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A

allstardentist

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i dunno this is what i thought..
possible combination for A: AA Aa aa
possible comb. for B: BB Bb
possible comb. for C: CC Cc cc
so i just multiplied 3x2x3 since we can prolly assum independent assortment. what is wrong with my logic?
 

jorUW

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714guy said:
For AaBbCc you have 3 heterozygous genes so 2^3 = 8 ways, and AaBBCc there are only 2 heterrozygous genes so 2^2 = 4 way. 4*8 = 32
I dont know if this is the correct way to do it b/c this math counts the inheritance of 'A' mom 'a' dad Vs 'a' mom and 'A' dad as two distinctly different genotypes. I could be wrong, but i think it's b/c you have to account for female and male offspring like I said above.
 

jorUW

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allstardentist said:
i dunno this is what i thought..
possible combination for A: AA Aa aa
possible comb. for B: BB Bb
possible comb. for C: CC Cc cc
so i just multiplied 3x2x3 since we can prolly assum independent assortment. what is wrong with my logic?
I think you're right, but you miss the last step. In one of my genetics classes my prof liked to add little statements like whats the probability of an offspring with genotype blah blah blah THAT IS FEMALE (except she wouldn't highlight the tricky part of the Q). A lot of people forget to factor this in. You get so focused on the polygenic trait that you miss the easy part.
 

jorUW

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allstardentist said:
shouldnt 'A' from mom and 'a' from dad be same as 'a' from mom and 'A' from dad? because they will both yield geneticallysame offspring.
hmmm, my bad :D I'll look at it again.
 

mlle

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I agree with 714guy.

Allstar, if you do it your way you aren't accounting for all the ways each of those could combine with eachother.

If you had a simpler question like how many different gametes might form from an AaBbCc parent?

Answer: 8 (draw it out if you need to)

or how many could form from a AaBBCc parent?

Answer: 4

8*4=32.

see?
 

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mlle said:
I agree with 714guy.

Allstar, if you do it your way you aren't accounting for all the ways each of those could combine with eachother.

If you had a simpler question like how many different gametes might form from an AaBbCc parent?

Answer: 8 (draw it out if you need to)

or how many could form from a AaBBCc parent?

Answer: 4

8*4=32.

see?
Totally agree! Same logic! 8*4=32 :thumbup:
 

Uracil

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In any problem, 2 ^ # of heterozygotes = the number of genetically distinct offspring that can be produced.

In this case, you have a cross of AaBbCc X AaBBCc

Parent A (AaBbCc) is heterozygous for the genes A, B and C. So, the number of heterozygotes = 3

Parent B (AaBBCc) is heterozygous for the genes A and C only. Then here we have that the number of heterozygotes = 2

2 + 3 = 5

2^5 = 32



allstardentist said:
Laboratory mice are to be classified based on genes A,
B, and C. How many genetically distinct offspring can
be produced from a cross of an AaBbCc individual with
an AaBBCc individual?
A. 12
B. 16
C. 32
D. 48
E. 64

I think its 18 from multiplying 3x2x3, but the answer is C. Anyone can explain?
 

jorUW

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So, even though there are only 18 possible genotypes, whether a given gamet is inherited by mom or dad makes the offspring distinct? I mean if the Q asked how many possible gentoypes or phenotypes would it be 18?

Possible Gentotypes/phenotypes (assume incomplete dominance) w/o respect to how inherited, agree?

1.AABBCC 2.AABBCc 3.AABBcc 4.AABbCC 5.AABbCc 6.AABbcc
7.AaBBCC 8.AaBBCc 9.AaBBcc 10.AaBbCC 11.AaBbCc 12.AaBbcc
13.aaBBCC 14.aaBBCc 15.aaBBcc 16.aaBbCC 17.aaBbCc 18.aaBbcc

Edit Anyway, I understand the math. I've been out of genetics way too long :D
 

teefRcool

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Uracil said:
In any problem, 2 ^ # of heterozygotes = the number of genetically distinct offspring that can be produced.

In this case, you have a cross of AaBbCc X AaBBCc

Parent A (AaBbCc) is heterozygous for the genes A, B and C. So, the number of heterozygotes = 3

Parent B (AaBBCc) is heterozygous for the genes A and C only. Then here we have that the number of heterozygotes = 2

2 + 3 = 5

2^5 = 32
Yup i agree thats how I would do it
 
OP
A

allstardentist

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mlle said:
I agree with 714guy.

Allstar, if you do it your way you aren't accounting for all the ways each of those could combine with eachother.

If you had a simpler question like how many different gametes might form from an AaBbCc parent?

Answer: 8 (draw it out if you need to)

or how many could form from a AaBBCc parent?

Answer: 4

8*4=32.

see?
that problem is different cuz you don't have to worry about redundancy since it is asking for the possible different types of HAPLOID cell
 

Uracil

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There are 18 possible genotypes when the parents are crossed, you are correct.

The problem is asking about the different gametes that can possibly be formed treating genes A, B, and C individually. If the problem was asking you about the different genotypes, yes it would be 18 :thumbup:



jorUW said:
So this would mean that even though there are only 18 possible genotypes, whether a given gamet is inherited by mom or dad makes the offspring distinct? I mean if the Q asked how many possible gentoypes would it be 18?

Possible Gentotypes w/o respect to how inherited, agree?

1.AABBCC 2.AABBCc 3.AABBcc 4.AABbCC 5.AABbCc 6.AABbcc
7.AaBBCC 8.AaBBCc 9.AaBBcc 10.AaBbCC 11.AaBbCc 12.AaBbcc
13.aaBBCC 14.aaBBCc 15.aaBBcc 16.aaBbCC 17.aaBbCc 18.aaBbcc
 

rchuloholla

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oh i had a question just like this on the real DAT!
 

Uracil

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allstardentist said:
it is asking for different genotype(=offspring) not different types of gamets
yeah that's right, I was confusing myself!! :D
 

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I think everyone is reading too much into the question. There are only 18 possible genotypes for the offspring. They are not talking about possible haploid gametes, just the genotypes of the full diploid organism. The problem didn't indicate that the male/female chromosomes were involved in this either. The only possible genotypes using the alleles that was indicated in the question is given in jorUW's post.

OP where did you get this question from?
 

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allstardentist said:
Laboratory mice are to be classified based on genes A,
B, and C.
How many genetically distinct offspring can
be produced from a cross of an AaBbCc individual with
an AaBBCc individual?
A. 12
B. 16
C. 32
D. 48
E. 64

I think its 18 from multiplying 3x2x3, but the answer is C. Anyone can explain?
AaBbCc * AaBBCc

(4 gametes, 3 distinct gametes)
.....A.....a
A...AA...Aa
a...aA...aa

(2 gametes, 2 distinct gametes)
.....B.....b
B...BB...Bb

(4 gametes, 3 distinct gametes)
.....C.....c
C...CC...Cc
c...cC...cc

# of possible gametes:
4 * 2 * 4 (i.e. 2^2 * 2^1 * 2^2) = 32 genetically distinct offspring based on A, B, C and other genes.

or
3 * 2 * 3 = 18 distinct just based on A, B & C
 

dat_student

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allstardentist said:
hey DAT student, so u think the answer is correct?
I think the answer is correct. c from mother may be linked with gene Z but c from father may be linked with lower case z. cZ is not the same as cz. So genetically 32 & phenotypically it can be 18 (if A, B & C are co-dominant).
 

biocmp

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This is a simpler question than what it sounds at first. Look at each genetic set seperately.

Aa X Aa will give you 3 combinations. AA, Aa, aa
BB X Bb will give you 2 combinations. BB, Bb
Cc X Cc will give you 3 combinations. CC, Cc, cc

This gives you only 18 different possibilities. There is no discriminate difference between inheriting 'a' from mom or 'a' from dad. So, unless the question was worded different, this is the only answer. It did not ask for more than this....

If you do no agree, please explain. preferably with punant squares and good explanations. I want to make sure i understand this
 

dat_student

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biocmp said:
This is a simpler question than what it sounds at first. Look at each genetic set seperately.

Aa X Aa will give you 3 combinations. AA, Aa, aa
BB X Bb will give you 2 combinations. BB, Bb
Cc X Cc will give you 3 combinations. CC, Cc, cc

This gives you only 18 different possibilities. There is no discriminate difference between inheriting 'a' from mom or 'a' from dad. So, unless the question was worded different, this is the only answer. It did not ask for more than this....

If you do no agree, please explain. preferably with punant squares and good explanations. I want to make sure i understand this
example:

Chromosome 1 (mother) ------c-----Z (upper case)-------
Chromosome 1 (father) -------c-----z (lower case)-------

Are you saying the two chromosomes above are duplicates?
 

biocmp

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what does your Z stand for? I am not saying anything about sex chromosomes. Or perhaps I'm really tired. The question implies nothing about sex chromosomes. It simply gives you two genetic sets and asks the total number of recombinations
 

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biocmp said:
what does your Z stand for? I am not saying anything about sex chromosomes. Or perhaps I'm really tired. The question implies nothing about sex chromosomes. It simply gives you two genetic sets and asks the total number of recombinations
Z is an allele. Chromosome 1 is not a sex chromosome :) The question says the mice are classified based on genes A, B and C. Those genes can be linked to other genes (e.g. Z).
 

jorUW

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dat_student said:
Z is an allele. Chromosome 1 is not a sex chromosome :) The question says the mice are classified based on genes A, B and C. Those genes can be linked to other genes (e.g. Z).
If you look at it this way than the answer is correct (32). I just think it's worded badly. Maybe you're just supposed to assume. Personally it is difficult for me to deduce that an A from mom is different than an A from dad based on how the question is worded. Even so, i understand in real life it would be.
 

biocmp

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That's kind of disheartening if they are going to have questions worded like that thru the whole thing. I will get prepared enough for my own sake, but i would hope they would be a little more specific in their wording.

I'm trying to figure out how I shoud deduce what they want?? :oops:


So, DAT student, is that how I should aproach every genetics question on the DAT? I won't be taking it until next summer, but I am busy going through the book a few times and making sure I understand possible questions.