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biology question
Started by demonicr
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What do you mean by 'n'?
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I'll take my time and try to explain it you guys a little better. If you want to actually work this problem out you have to be familiar with a trihybrid cross.
so we have AaBbCcXAaBBCc
1/4AA*1/2BB*1/4CC --> 1/32 AABBCC
1/4AA*1/2BB*1/2Cc --> 1/16 AABBCc
1/4AA*1/2BB*1/4cc --> 1/32 AABBcc
1/4AA*1/2Bb*1/4CC --> 1/32 AABbCC
1/4AA*1/2Bb*1/2Cc --> 1/16 AABbCc
1/4AA*1/2Bb*1/4cc --> 1/32 AABbcc
1/2Aa*1/2BB*1/4CC--> 1/16 AaBBCC
1/2Aa*1/2BB*1/2Cc--> 1/8 AaBBCc
1/2Aa*1/2BB*1/4cc --> 1/16 AaBBcc
1/2Aa*1/2Bb*1/4CC --> 1/16 AaBbCC
1/2Aa*1/2Bb*1/2Cc --> 1/8 AaBbCc
1/2Aa*1/2Bb*1/4cc --> 1/16 AaBbcc
1/4aa*1/2BB*1/4CC --> 1/32 aaBBCC
1/4aa*1/2BB*1/2Cc --> 1/16 aaBBCc
1/4aa*1/2BB*1/4cc --> 1/32 aaBBcc
1/4aa*1/2Bb*1/4CC --> 1/32 aaBbCC
1/4aa*1/2Bb*1/2Cc --> 1/16 aaBbCc
1/4aa*1/2Bb*1/4cc --> 1/32 aaBbcc
What can we conclude from this? Well as you can see 18 different genotypes are formed from the cross.
Now to figure out the gametes formed all you have to do is to take each of the gives seperately and multiply them. 2^3*2^2= 2^5 = 32
What does the 32 mean? Well you had AaBbCc*AaBBCc
so for the first set we have ABC, ABc, AbC, Abc, aBC, aBc, abC, abc
for the second set we have ABC, ABc, aBC, abc
this same general formula can be applied to us. Since we have 23 "sets" of chromosomes which contain dominant and recessive alleles, so we can use the n rule here as well. Imagine a man and woman mate. The possible gametic combinations are staggering.
2^23 (from the man) * 2^23 (from the woman) = 64 trillion gametic possibilities.
AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWw*AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWw
Hence, considering only Mandelian genetics we can assume that the likelihood that two brothers born consectively after each other from the same parents would have a 1/64 trillion chance of looking identical to each other. In other words, technically 0
so we have AaBbCcXAaBBCc
1/4AA*1/2BB*1/4CC --> 1/32 AABBCC
1/4AA*1/2BB*1/2Cc --> 1/16 AABBCc
1/4AA*1/2BB*1/4cc --> 1/32 AABBcc
1/4AA*1/2Bb*1/4CC --> 1/32 AABbCC
1/4AA*1/2Bb*1/2Cc --> 1/16 AABbCc
1/4AA*1/2Bb*1/4cc --> 1/32 AABbcc
1/2Aa*1/2BB*1/4CC--> 1/16 AaBBCC
1/2Aa*1/2BB*1/2Cc--> 1/8 AaBBCc
1/2Aa*1/2BB*1/4cc --> 1/16 AaBBcc
1/2Aa*1/2Bb*1/4CC --> 1/16 AaBbCC
1/2Aa*1/2Bb*1/2Cc --> 1/8 AaBbCc
1/2Aa*1/2Bb*1/4cc --> 1/16 AaBbcc
1/4aa*1/2BB*1/4CC --> 1/32 aaBBCC
1/4aa*1/2BB*1/2Cc --> 1/16 aaBBCc
1/4aa*1/2BB*1/4cc --> 1/32 aaBBcc
1/4aa*1/2Bb*1/4CC --> 1/32 aaBbCC
1/4aa*1/2Bb*1/2Cc --> 1/16 aaBbCc
1/4aa*1/2Bb*1/4cc --> 1/32 aaBbcc
What can we conclude from this? Well as you can see 18 different genotypes are formed from the cross.
Now to figure out the gametes formed all you have to do is to take each of the gives seperately and multiply them. 2^3*2^2= 2^5 = 32
What does the 32 mean? Well you had AaBbCc*AaBBCc
so for the first set we have ABC, ABc, AbC, Abc, aBC, aBc, abC, abc
for the second set we have ABC, ABc, aBC, abc
this same general formula can be applied to us. Since we have 23 "sets" of chromosomes which contain dominant and recessive alleles, so we can use the n rule here as well. Imagine a man and woman mate. The possible gametic combinations are staggering.
2^23 (from the man) * 2^23 (from the woman) = 64 trillion gametic possibilities.
AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWw*AaBbCcDdEeFfGgHhIiJjKkLlMmNnOoPpQqRrSsTtUuVvWw
Hence, considering only Mandelian genetics we can assume that the likelihood that two brothers born consectively after each other from the same parents would have a 1/64 trillion chance of looking identical to each other. In other words, technically 0
Tenacious said:
A question this vague won't pop up on the real DAT. I have no idea in hell how they got 6. The question, I assume, is meaning the # of different genotypes formed from the cross, or the number of gametic combinations. Those being 18 and 32 respectively.
