Slide 1
Phenotypes and Genotypes Probability Part 2 VoiceThread
TranscriptSo, let's work through the solution to these practice
problems. Assume you have an F1 generation that is heterozygous for 3
genes - genes A, B, and C. What is the probability of them having an
offspring with genotype aabbcc? What is the probability of them having
an offspring that is heterozygous for all three genes?
Slide 2
First, let's take time to determine how many gamete types each
parent can make.
Each parent is heterozygous for three different genes.
Slide 3
Use the equation 2^n = # different gamete types, n=#heterozygous
genes
In this example there are 3 heterozygous genes so each parent
makes 8 different gamete types (2^3 = 8)
How big would the Punnett square
be?
Slide 4
8 x 8 = 64 different squares - too big! So let's use
probability theory.
Slide 5
So what is the probability of an offspring of genotype aabbcc?
Let�s look at it gene by gene:
Gene A: aa - there is a � probability
of getting a from parent 1 and a � probability of getting a from parent 2 - so
the overall probability is � x � = �
Gene B: bb - there is a � probability
of getting b from parent 1 and a � probability of getting b from parent 2 - so
the overall probability is � x � = �
Gene C: cc - there is a � probability
of getting c from parent 1 and a � probability of getting c from parent 2 - so
the overall probability is � x � = �
The overall probability is � x � x � = 1/64. Remember, the question is asking you what is the probability of getting genotype aa AND genotype bb AND genotype cc so you need to multiply the individual probabilities together.
Slide 6
So what is the probability of an offspring of genotype AaBbCc?
Let�s look at it gene by gene:
Gene A: Aa - there are two ways to be
heterozygous - you can either get A from parent 1 and a from parent 2 (� x � =
�) OR a from parent 1 and A from parent 2 (� x � = �) . You can use a
Punnett square to visualize this idea. Notice - I am using a small 4
square Punnett square to help you visualize - not a 64 square Punnett square!
So the probability of an offspring having genotype Aa is 1/2 (� + � = �). We use both rules of probability in this example - the rule of multiplication to determine the probability of either Aa or aA and the rule of addition to calculate the overall probability of being heterozygous for A.
Slide 7
We use the same logic for genes B and C.
Gene B: Bb - there are two ways to be heterozygous - you can either get
B from parent 1 and b from parent 2 (� x � = �) OR b from parent 1 and B from
parent 2 (� x � = �) .
Gene C: Cc - there are two ways to be
heterozygous - you can either get C from parent 1 and c from parent 2 (� x � =
�) OR c from parent 1 and C from parent 2 (� x � = �) .
The overall probability is � x � x � = 1/8
If you have questions about these examples, please contact either me or your TA.