Albino Genetics slide presentation
No Genetics Calculator? Follow the directions below!
D+SZ X D+SZ
Clown
black X Clown black
| First Allele | Second Allele | Genotype | Phenotype |
| 25% SS | 6.25% DDSS | Black (DD) blushing | |
| 25% DD | 50% SZ | 12.5% DDSZ | Black (DD) clown black |
| 25% ZZ | 6.25% DDZZ | Black (DD) zebra | |
| 25% SS | 12.5% D+SS | Turquoise | |
| 50% D+ | 50% SZ | 25% D+SZ | Clown black |
| 25% ZZ | 12.5% D+ZZ | Zebra lace | |
| 25% SS | 6.25% ++SS | Blushing | |
| 25% ++ | 50% SZ | 12.5% ++SZ | Silver Clown |
| 25% ZZ | 6.25% ++ZZ | Zebra | |
This is the way I figure out
crosses. The first step is to
figure how many alleles you are working with.
In this cross there are just 2. If
we added a 3rd allele like
smokey there would be a third column after the zebra stripeless column.
Step 1 is to take the first
allele, in this case Dark and figure out the genotypes of just that cross.
In this case there were 3 possibilities.
This is just simple 2 by 2 Punnett square. 25% DD, 50% D+, 25% ++
| D | + | |
| D | DD | D+ |
| + | D+ | ++ |
Leave some room between
these 3 outcomes, as each one will get all of the results from the next allele
SZ X SZ
| S | Z | |
| S | SS | SZ |
| Z | SZ | ZZ |
25% SS, 50% SZ, 25% ZZ
Each of these 3
possibilities is added to each of the 3 genotypes from the first allele.
So we end up with 9 resulting genotypes.
If this second allele had 4 possibilities you would end up with 12 total.
Total percentages are just
multiplications of each one. 25% of
25% is 6.25% etc.
If there was a third allele
it would be added as multiple entries after each of the second alleles.
You could end up with a maximum of 64 different genotypes from 3 alleles.
In reality I can’t think of any other genes that share the same locus
besides the dark and stripeless locus. So
max would be 48
Hope this helps.