| Review of Reassigning Probabilities Based on New Information |
| 1. | Probability of being a Carrier in an Autosomal
Monohybrid Cross. |
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Let A be the dominant allele and a be the recessive allele. Under the specified conditions there are two genotypes which are phenotypically normal: AA and Aa. These two genotypes make up 3/4 of the expected genotypes from a cross of Aa x Aa. Eliminating the homozygous recessive phenotype (aa), we see that 1/3 of the normal progeny have genotype AA [(1/4)/(3/4) = 1/3] and 2/3 of the normal progeny have genotype Aa [(2/4)/(3/4) = 2/3].
The probability of appearing normal is 1. This value is obtained by dividing both sides of the equation by 3/4 1/4/3/4 (AA) + 1/2/3/4 (Aa) = 3/4/3/4 1/3 (AA) + 2/3 (Aa) = 1
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| 2. | Probabilities change with new information -a cystic fibrosis example. | |||||||||
A normal woman (Helga) has a brother with CF (aa
genotype). Her parents do not have CF. Each of these parents
had to give an a allele to their son so they must
each have the genotype Aa. Thus, the marriage
that produced Helga was
Aa x Aa
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The predicted outcome of such a marriage is theoretically:
Genotypic Array
1/4AA + 1/2Aa + 1/4aa = 1
Phenotypic Array
3/4 Normal + 1/4 CF =1
Therefore, before Helga was born, expected ratio of Normal to
CF was 3: 1. The ratio of AA to Aa genotypes was 1/4:2/4
or 1:2
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After Helga was born, her phenotype is no longer a
theoretical event! Helga does not have CF.
The probability that she does not have CF is 1,
not 3/4.
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By observation Helga's phenotype is normal. The only
question remaining is Helga's genotype (AA vs. Aa).
If Helga was a mouse, we could do a test cross to determine her genotype. With humans, however, we can only assign probabilities.
As previously stated, the expected ratio of AA: Aa is
1:2 or 1/3:2/3. If the overall probability is one, the
probability of AA is 1/3 and of Aa is 2/3.
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