Sorry, there have been a number of posts whilst I wrote this.
This worked solution using a punnet square should enable you to see where the percentages come from...
SYMBOLS: Let
N represent the allele (gene) for
normal colour
Let
a represent the allele (gene) for
albinism
PARENTS:
N a (het albino) x
N a (het albino)
CROSS:
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|-
| x
|
N
|
a
|-
|
N
|
|
|-
|
a
|
|
|-
All four possible outcomes are individually equally probable.
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|-
| x
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N
|
a
|-
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N
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N N
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N
a
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a
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N a
|
a
a
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RESULTS:
Genotypes & Likelihood Phenotypes and Likelihood
N N = 1 out of 4 Normal = 3 out of 4
N a = 2 out of 4 Albino = 1 out of 4
a a = 1 out of 4
There are three outcomes that result in normal colour. One of these three has two copies of the normal colour allele (gene). Two of these are heterozygous carrying the recessive allele for albinism as well asthe dominant allele for normal colour.
The probability a normal coloured bearded dragon is homozygous for normal colour is 1 out of 3 = 33.3%.
The probability a normal coloured bearded dragon is heterozygous for normal colour and abino (i.e. carrying the recessive allele for albinism) is 2 out of 3 = 66.7%.
What you need to understand is that probabilities are NOT predictors with small samples. They work well with very large samples. For example, we know there is a 50% chance of having a girl and a 50% chance of having a boy with each child people produce. Yet I know a family that has seven sons. However, if we check the gender of babies born in Australia in any one year, the ratio of females to males is 1 : 1.
Blue