Illustrate the differences between prokaryotic and eukaryotic cells.
1 answer:
You might be interested in
I think you wrote this wrong? Is it not round rr vs wrinkled RR?? I'll answer it with that assumption:
You are crossing a rr with an RR:
You only get Rr back so they will phenotypically be all wrinkled
If you are crossing rr with an Rr (so a heterozygote wrinkled):
50% will be Rr (wrinkled)
50% will be rr (round)
See my picture for the Punnett Squares for both answers!
You are crossing a rr with an RR:
You only get Rr back so they will phenotypically be all wrinkled
If you are crossing rr with an Rr (so a heterozygote wrinkled):
50% will be Rr (wrinkled)
50% will be rr (round)
See my picture for the Punnett Squares for both answers!
The Hardy-Weinberg equation is as follows:
Where:
(convert all % to decimals)
p= homozygous dominant
q= homozygous recessive
pq= heterozygous
While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.
Using the 1st equation, we can solve for the other dominant/recessive frequency:
1-0.2=0.8
Meaning that:
p= 0.8 & q=0.2
If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.
2(0.8)(0.2)= 2(0.16)= 0.32
So, the heterozygous frequency would be:
0.32
Hope this helps!
Where:
(convert all % to decimals)
p= homozygous dominant
q= homozygous recessive
pq= heterozygous
While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.
Using the 1st equation, we can solve for the other dominant/recessive frequency:
1-0.2=0.8
Meaning that:
p= 0.8 & q=0.2
If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.
2(0.8)(0.2)= 2(0.16)= 0.32
So, the heterozygous frequency would be:
0.32
Hope this helps!