solving a genetic question FAST

[Smooth Operater]

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how many different genotype can you get from Aa Bb cc x Aa Bb cc cross?

Is there a quicker way to do this? Or do I need to draw the punnet square? Thanks!

 

[ernest1284]

2^n where n = # of heterozygotes.... so in this case there are 4 heterozygotes 2^4=16 possible outcomes.... i'm not sure how to tell the relative ratios without doing the punnent square unless you pre memorize them....

when are you taking your dat's i've seen you on here for a while....

 

[Ex_EE]

There are actually 9 different genotypes.
AABBcc
AABbcc
AAbbcc
AaBBcc
AaBbcc
Aabbcc
aaBBcc
aaBbcc
aabbcc

 

[GRAD]

just dont forget that when both of them are CC or cc, then you can basically cancel that letter out, you know, since it doesnt change. i dont know if that helps

 

[Ex_EE]

Here is a website that might help.
http://waynesword.palomar.edu/lmexer4.htm

 

[houstongirl]

How is it 4 hetrozygotes? Aa cc Bb? am I reading the question wrong?

 

[fbradley99]

Oops I just re-read your question. Yes ernest was right you do 2^n where n equals # heterzygotes....But here's how you can solve it quickly if you need to know the ratios 🙂

here's how i was taught. My genetics teacher never wanted us to use the punnett square!!!! He said that was for little kids. I had an awesome genetic's teacher!!!!

I am going to use a simplier one (to save space) as an example but you do the same thing.

Example:
Aa x AA

Aa x AA
1/2 A 1 A
1/2 a

so you multiply the left by right for each genotype that is possible

So .....
1/2 A x 1 A = 1/2 AA
1/2 a x 1 A= 1/2 aA

so its 1/2 AA and 1/2 Aa

Here's another example

Aa x Aa

1/2 A 1/2 A
1/2 a 1/2 a

so.....
1/2 A x 1/2 A = 1/4 AA
1/2 A x 1/2 a = 1/4 Aa
1/2 a x 1/2 A = 1/4 aA
1/2 a x 1/2 a= 1/4 aa

add them up :

1/4 AA : 1/2 Aa L: 1/4 aa

 

[Kneecoal]

lol.

 

[UCB05]

It's really simple... since the question doesn't say the genes are linked, they assort independently, so you look at each gene. Do a mini-punnet square for each gene if you have to. For A, there are three possible combinations: AA, Aa, aa. For B, there are also three: BB, Bb, bb. There is only one combination for c from two homozygotes: cc. That means the total number of different possible genotypes is 3x3x1 = 9.

 

[Kneecoal]

... this thread is 3 years old

 

[UCB05]

... this thread is 3 years old

OHAY. Maybe that's why my RC sucked.