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Laboratory mice are to be classified based on genese A,B, and C. How many genetically distinct offsprings can be produced from a cross of an AaBbCc with an AaBBCc individual?
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Genetics Question.
- Thread starter orthosm2021
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Laboratory mice are to be classified based on genese A,B, and C. How many genetically distinct offsprings can be produced from a cross of an AaBbCc with an AaBBCc individual?
18? 3x2x3
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right. thats exactly what i came up with when I was doing this problem. However, the answer choices for this question were:
a. 12
b. 16
c. 32
d. 48
e. 64
there was no 18 as a answer choice. And the solution for this question said:
"C. The number of genetically distinct offspring is found by multiplying number of distinct gametes each parent can produce. The number of gametes can be calculated from the formula 2^n, where n=number of heterozygous genes. The AaBbCc parent can produce 8 unique gametes (2^3=8) and the AaBBCc parent can produce 4 (2^2=4). Therefore the possible number of different offspring is 8x4=32."
The process the solution used would result in some redundancy. wouldnt it.
(This is a question from a practice test called DAT/OAT natural science integration workshop A.)
a. 12
b. 16
c. 32
d. 48
e. 64
there was no 18 as a answer choice. And the solution for this question said:
"C. The number of genetically distinct offspring is found by multiplying number of distinct gametes each parent can produce. The number of gametes can be calculated from the formula 2^n, where n=number of heterozygous genes. The AaBbCc parent can produce 8 unique gametes (2^3=8) and the AaBBCc parent can produce 4 (2^2=4). Therefore the possible number of different offspring is 8x4=32."
The process the solution used would result in some redundancy. wouldnt it.
(This is a question from a practice test called DAT/OAT natural science integration workshop A.)
right. thats exactly what i came up with when I was doing this problem. However, the answer choices for this question were:
a. 12
b. 16
c. 32
d. 48
e. 64
there was no 18 as a answer choice. And the solution for this question said:
"C. The number of genetically distinct offspring is found by multiplying number of distinct gametes each parent can produce. The number of gametes can be calculated from the formula 2^n, where n=number of heterozygous genes. The AaBbCc parent can produce 8 unique gametes (2^3=8) and the AaBBCc parent can produce 4 (2^2=4). Therefore the possible number of different offspring is 8x4=32."
The process the solution used would result in some redundancy. wouldnt it.
(This is a question from a practice test called DAT/OAT natural science integration workshop A.)
Yes, I would agree. there is a lot of redundancy leading to the same offspring genotype.
aBc x aBC = aaBBCc
aBC x aBc = aaBBCc
This pair is one of the many which lead to the same genotypic offspring.
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yup i agree. thx.
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Look at this problem as if you are doing a combination of three monohybrid.
AaBbCc X AaBBCc=
for A gene locus, it is AaXAa, so 4 kinds of GENETICALLy distinct offspring .
for B gene locus, it is BbXBB, so 2 kinds of genetically offspring.
for C gene locus, it is CcXCc, 4 kinds.
Together, you are asking the total combination, so it is 4X2X4=32.
With the same method, you should also be able to answer questions for phenotypically distinct offspring, i.e. just combine those homo and heterzygous dominant.
AaBbCc X AaBBCc=
for A gene locus, it is AaXAa, so 4 kinds of GENETICALLy distinct offspring .
for B gene locus, it is BbXBB, so 2 kinds of genetically offspring.
for C gene locus, it is CcXCc, 4 kinds.
Together, you are asking the total combination, so it is 4X2X4=32.
With the same method, you should also be able to answer questions for phenotypically distinct offspring, i.e. just combine those homo and heterzygous dominant.
Look at this problem as if you are doing a combination of three monohybrid.
AaBbCc X AaBBCc=
for A gene locus, it is AaXAa, so 4 kinds of GENETICALLy distinct offspring .
for B gene locus, it is BbXBB, so 2 kinds of genetically offspring.
for C gene locus, it is CcXCc, 4 kinds.
Together, you are asking the total combination, so it is 4X2X4=32.
With the same method, you should also be able to answer questions for phenotypically distinct offspring, i.e. just combine those homo and heterzygous dominant.
Actually, for the A/a C/c genes there are only 3 distinctions.
AA, Aa, Aa, aa, CC, Cc, Cc, cc
Such is the reason for our answer of 3x2x3=18 distinct offspring.
If you want to make the case that from the AaBbCc and A'a"B'B"C'c" parents that Aa" is genetically different from aA', you'd also have to make the same distinction for BB' vs BB" and bB' vs bB". Then there would be 64 different variants.
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Actually, for the A/a C/c genes there are only 3 distinctions.
AA, Aa, Aa, aa, CC, Cc, Cc, cc
Such is the reason for our answer of 3x2x3=18 distinct offspring.
If you want to make the case that from the AaBbCc and A'a"B'B"C'c" parents that Aa" is genetically different from aA', you'd also have to make the same distinction for BB' vs BB" and bB' vs bB". Then there would be 64 different variants.
This sounds right. Aa x Aa should yield three different variations, not four.
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Sorry for my mistake. You guys are correct.
The question actually appeared in natural science subject test A, and it does have a correct choice =18, so I guess if you saw it elsewhere and it says 32, it is wrong.
The question actually appeared in natural science subject test A, and it does have a correct choice =18, so I guess if you saw it elsewhere and it says 32, it is wrong.
