Tuesday, July 12, 2011

Genetic problems (how to solve)

AABbCC x AaBbCc
What is the probability of having a child AABBCC?
AA ½ BB ¼ CC ½
Aa Bb Cc
AA Bb CC you need to worry only about homozygous dominant because they only want to know AABBCC
Aa bb Cc

There is only one possibility for AABBCC ½ x ¼ x ½ = 1/16
The probability of have an offspring AABBCC is 1 out of 8 or 1/8
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Parents
AABBCC x aabbcc
what is the probability of having a child AaBbCc?
Aa 4/4 =1 Bb 4/4=1 Cc 4/4= 1
Aa Bb Cc
Aa Bb Cc
Aa Bb Cc

Aa Bb Cc
1 x 1 x 1 = 1

100% of the offspring will be AaBbCc
PpYyRr x Ppyyrr
What fraction of offspring from this cross would be predicted to exhibit the recessive phenotypes for at least two of the three characters?
PP Yy Rr
Pp Yy Rr
Pp yy rr
pp yy rr

First use the information above to find all combinations that have at least two recessive phenotypes
First follow the black lines
PPYyRr, PPYyRr, PPYyrr, PPYyrr
Now follow the red lines
PPYyRr, PPYyRr, PPYyrr, PPYyrr
Continue this process until all of the combination are made with the other genotypes…Pp,Pp,pp
However, you will notice that Pp will give you the same as the other Pp so you do not have to go through those again….
Another note notice that you ignore the ones that don’t produce at least two recessive phenotypes…….
You should end up with
PPyyrr
Ppyyrr
ppyyRr
ppYyrr
ppyyrr

Now figure the probability that each one will occur

First determine the probability that each characteristic will show up…

PP ¼ Yy ½ Rr ½
Pp ½ Yy Rr
Pp yy ½ rr ½
pp ¼ yy rr

use this information to determine the probability of each individual……

PPyyrr ¼ x ½ x ½ = 1/16
Ppyyrr ½ x ½ x ½ = 1/8
ppyyRr ¼ x ½ x ½ = 1/16
ppYyrr ¼ x ½ x ½= 1/16
ppyyrr ¼ x ½ x ½ = 1/16

Now you have the individual probabilities for each of those offspring occuring…now add up those probabilities to get the probability that there will be at least two recessive phenotypes……

1/16 + 2/16 +1/16 + 1/16 + 1/16 = 6/16 = 3/8

What fraction of offspring from this cross would be predicted to exhibit the recessive phenotypes for at least two of the three characters? Answer 3/8



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