Pentahyrbid cross?

Question

how do i construct a punnett square for the pentahybrid cross

AaBbCcDdEe x AaBbCcDdEe

and the phenotypic and genotypic ratios?

thanks!

Answer 1

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

Answer 2

First, determine the gametes produced by the parents (the gametes will be the same for both parents since they have the same genotype).  There are five heterozygous pairs of alleles.  If you take 2 to the fifth power, you will discover there are 32 different possible gametes.  So you will need to draw a Punnett Square with 32 horizontal spaces and 32 vertical spaces.  That will give you 1024 possible offspring!  This is going to take you a while to fill in all the possible combinations of genes in the offspring.  Then you'll need to count all the offspring with the different genotypes and phenotypes.

Hope you have enough patience to complete the task.

Answer 3

1) Start with a VERY large sheet of paper.  SERIOUSLY.  My teacher in 10th grade gave this the option of doing this as an extra credit project.  Mine ended up about 2 feet by 3 feet with the individual squares being about 1 inch.

The easiest way I can explain how to even begin is to figure out the possible combinations for the gametes (each box across the top and side).  Since each of the 5 traits is heterozygous, that makes it a little easier.

Start with each letter in caps.  In each row down switch the case of the last letter.  In every two rows, switch the case of the "D".  In every four rows, switch the case of the "C".  In every 8, the "B", and in every 16, the "A".

ABCDE

ABCDe

ABCdE

ABCde

ABcDE

ABcDe

ABcdE

ABcde

I did the first 8 so you can see how this works.  This way you'll get 32 unique combinations of letters which are your possible gametes from each parent (2 chromosomes raised to the 5th power = 32 combinations).

From there, it's just a matter of combining the alleles from the top and side and figuring out what your results are.  You'll get 1024 possible offspring, so keep close count on the genotypes and phenotypes for each one.