Answer:
Yellow smooth - 9
Yellow wrinkle - 3
Green smooth- 3
Green Wrinkle - 1
Explanation:
Let the green color of the seed be depicted by "G" and the yellow color of the seed be depicted by "g"
Let the smooth the seed be depicted by "R" and wrinkled seed be depicted by "r"
F1 cross -
true breeding smooth green plant ( RRGG) and true breeding wrinkled yellow (rrgg)
F1 gamete will be RG, RG, rg, rg
F1 offspring will be RrGg , Thus all F1 offspring will be heterozygous smooth and yellow.
Thus, R is dominant over r and g is dominant over G
F2 Generation –
RrGg x RrGg
Gametes will be RG, Rg, rG, rg
RG Rg rG rg
RG RRGG RRGg RrGG RrGg
Rg RRGg RRgg RrGg Rrgg
rG RrGG RrGg rrGG rrGg
rg RrGg Rrgg rrGg rrgg
R is dominant over r and g is dominant over G
Genotypes are –
RRGG - 1 (Smooth Green)
RRGg-2 (Smooth yellow)
RrGG-2 (Smooth Green)
RrGg-4 (Smooth yellow)
RRgg- 1 (smooth yellow)
Rrgg – 2 (Smooth yellow)
rrGG – 1 (wrinkled Green)
rrGg – 2 (Wrinkled yellow)
rrgg – 1 (wrinkled yellow)
Yellow smooth - 9
Yellow wrinkle - 3
Green smooth- 3
Green Wrinkle - 1
Answer:
1:1:0:0 Round yellow to Wrinkled yellow to round green to wrinkled green.
I just got this question on my test, the answer is D) glycogen. All excess glucose is stored long-term in the liver and muscle cells as glycogen when animals need it for energy.
Answer:
Explanation:
The higher temperature is the factor which is responsible for the process of denaturation of proteins in which the breaking down of hydrogen, disulphide bonds and destabilization occurs. This process breaks up the bond between the polypeptides.
The bovine pancreatic trypsin inhibitor comprises of 58 amino acids this consists of the disulphide bonds which are destroyed on heating as a result of this the inhibitor becomes inactive. But on cooling it comes to the original form as the internal structure starts making hydrogen and disulphide bonds again. Thus the activity of the BPTI again gets restored.
It’s the doctor of all systems
