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Lets say we cross AaBbCcDdEe with AaBbCcDdEe. How many different genotypes will occur?
The answer is 3^5 = 243 genotypes. Where do the 3 and the 5 come from?

I know that AaBbCcDdEe will produce 32 kinds of gametes because 2^5= 32. I know that in this case 5 represents the number of heterozygous pairs. Perhaps the five in 3^5 comes from the number of number of possible gametes that can be produced? I'm lost and the book did not cover this.

Thank you for any insights.

I think the 3 comes from the various types of combinations; take for instance the gene A

(1) Aa

(2) AA

(3) aa

The five comes from the different genes in question, A, B, C, D, and E.

:)

But let's say the cross is AaBb x AaBb.  That would be the usual dihybrid cross with 16 different genotypes, right?  But if we use your formula, we'd say 3^2 = 9 not 16.  

Here's how I thought the genotypes would be predicted.
n= number of heterozygous genes.
example: AaBbCc X AABbCC
(2^3) x(2^1) = 16 types of offspring. 

I do not know what I'm missing.

FREE COOKIES to the person who helps me finally understand.   ;) 

Let the number of genes in the cross = n

Dihybrid n=2
trihybrid n=3

There are two options for each allele so 2^n = gamete types, i.e. two alleles raised to the number of genes in the cross

tetrahybrid has 2^4 = 16 gametes

Next 2^2n = offspring, i.e. Two alleles raised to the # of genes x 2 for the # of parents
tetrahybrid 2^2n = 2^8 = 256 offspring {this is also 16 gametes x 16 gametes}

To calculate the number of genotypes it is 3^n
tetrahybrid 3^n = 81

look at pascal's triangle for the genotype ratios

1:8:28:56:70:56:28:8:1 add these and get 256

To find the chances of an offspring that is genotype AAbbCcDd
is (1/4)x(1/4)x(1/2)x(1/2)=1/64

In general the phenotype calculation is
let m+n = number of genes in cross. So for a tetrahybrid cross with m+n = 4
(3/4)^m x (1/4)^n

(3/4)^4 x (1/4)^0 = 81/256 all genes are dominant phenotype
(3/4)^3 x (1/4)^1 = (27/64)(1/4) = 27/256 [four times for all combinations]
(3/4)^2 (1/4)^2 = (9/16)(1/16) = 9/256 [six times for all combinations]

Phenotype ratio => 81:[27:27:27:27] : [ 9:9:9:9:9:9] : [3:3:3:3] :1 = 256

====

If you want to construct a punnett square to prove this:

Let n= # of genes
there are two alleles for each gene
2^n= the # of different gametes from a parent
2^n squared = 2^2n is the # of offspring
3^n is the # of different genotypes in the offspring

In a dihybrid cross n=2
2^2= 4 gamete types
2^4 = 16 offspring
3^2 = 9 genotypes
genotype ratio 1:4:6:4:1 = 16
phenotype ratio 9:3:3:1 = 16

Now let n = 5 as in your cross

Genotype ratio comes from pascals triangle 1:10:45:120:210:252:210:120:45:10:1

Calculate the phenotype ratios:
m is the number of dominant genes expressed
Dominant alleles are seen in 3/4 of the chances

n is the recessive genes expressed & recessive are seen in 1/4 of the chances
so that m+n = number of genes in cross   
penta = 5 = m+n

(3/4)^5 x (1/4)^0 = 243/1024 are a dominant phenotype in every gene

(3/4)^4 x (1/4)^1= 81/1024 are dominant phenotype in four genes & fully recessive in one. This repeats five times to cover each possible gene as a recessive.
5 x 81 = 405 offspring have a single homozygous recessive gene out of 5 possible genes.

Pentahybrid phenotype ratio -> 243:81:81:81:81:81:
27:27:27:27:27:27:27:27:27:27:9:9:9:9:… = 1024 offspring

(3/4)^5 x (1/4)^0 = 243/1024 243
(3/4)^4 x (1/4)^1= 81/1024 - five times - 405 offspring
(3/4)^3 x (1/4)^2= 27/1024 - 10 times ----270
(3/4)^2 x (1/4)^3= 9/1024 - 10 times -------90
(3/4)^1 x (1/4)^4= 3/1024- 5 times ---------15
(3/4)^0 x (1/4)^5= 1/1024- once - - - - - - 1

To find any given genotypes chances look at one gene at a time.
AA has 1/4, Bb has 1/2, cc has 1/4, DD has 1/4, Ee = 1/2
AABbccDDEe has 1/4x1/2x1/4x1/4x 1/2 = 1/256

To find a phenotypes chances

A_B_ccD_E_ has 3/4x3/4x1/4x3/4x3/4 = 81/1024

Where are my cookies!

 :-\ sorry confused_student no idea  :-[

Duddy, you are AWESOME!!!!!!
MUCHAS gracias.   :-*
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