Cells in G0 phase are not actively preparing to divide. The cell is in a quiescent (inactive) stage that occurs when cells exit the cell cycle.
First, write out all combinations for the parents:
<u>SsBB</u>:
SB
sB
SB
sB
<u>ssBb</u>:
sB
sb
sB
sb
So now that you know what to put on as the x and y for the Punnett Square, set it up:
<u> | SB | sB | SB | sB </u>
<u>sB | SsBB | ssBB | SsBB | ssBB </u>
<u>sb | SsBb | ssBb | SsBb | ssBb </u>
<u>sB | SsBB | ssBB | SsBB | ssBB </u>
<u>sb | SsBb | ssBb | SsBb | ssBb </u>
Now, listing each possible genotype and the number of times they show up in the square:
SSBB
0 times
SSBb/SSbB
0 times
SSbb
0 times
SsBB/sSBB
4 times
SsBb/SsbB/sSBb/sSbB/
4 times
Ssbb/sSbb
0 times
ssBB
4 times
ssBb/ssbB
4 times
ssbb
0 times
Since there are only 4 genotypes are possible, set up ratios of those four:
SsBB:
4 times/16 total = .25 = 25%
SsBb:
4 times/16 total = .25 = 25%
ssBB:
4 times/16 total = .25 = 25%
ssBb:
4 times/16 total = .25 = 25%
This means that the<em> genotypes</em> have a 25% : 25% : 25% : 25% likelihood.
Now, let's figure out the phenotypes of the 4 genotypes:
SsBB:
Ss = dominant
BB = dominant
SsBb:
Ss = dominant
BB = dominant
ssBB:
ss = recessive
BB = dominant
ssBb:
ss = recessive
Bb = dominant
So we have 2/4 that are S-dominant B-dominant and 2/4 that are S-recessive B-dominant.
This means that the phenotypes are:
2/4 = 50% Thin stripes with black eyes, and
2/4 = 50% Thin stripes with gray eyes.
I hope this helped you out :)
Riding animals or mounts ☝
Both chloroplasts and cyanobacteria get their energy from performing photosynthesis which led researchers to propose that chloroplasts evolved from cyanobacteria.
