They use the process known as decomposition- much like the name!
Answer:
Considering the <u>whole progeny</u> (100%), there will be
- 25% Black male kittens, XBY
- 0% Calico Male kittens
- 25% Calico female kittens, XBXR
Explanation:
<u>Available data:</u>
- The gene for Calico (multicolored) cats is an x-linked trait and codominant
- Calico Females receive a "B" and an "R" gene, and have black and orange splotches on white coats. Their genotype is XBXR.
- Males can only be black or orange, but never calico. Their genotype is XBY and XRY
Cross: a female calico cat with a black male
Parentals) XBXR x XBY
Gametes) XB XR XB Y
Punnett square) XB XR
XB XBXB XBXR
Y XBY XRY
F1) Among the whole progeny:
- 2/4 = 50% will be black (female XBXB and male XBY)
- 1/4 = 25% will be Calico (female XBXR)
- 1/4 = 25% will be Orange (male XRY)
Among females:
- 1/2 = 50% of females will be black, XBXB
- 1/2 = 50% of females will be Calico, XBXR
Among males:
- 1/2 = 50% of males will be black, XBY
- 1/2 = 50% of males will be orange, XRY
Answer:
In the genetic code, a stop codon (or termination codon) is a nucleotide triplet within messenger RNA that signals a termination of translation into protein
Explanation:
Proteins are based on polypeptides, which are unique sequences of amino acids. Most codons in messenger RNA (from DNA) correspond to the addition of an amino acid to a growing polypeptide chain, which may ultimately become a protein. Stop codons signal the termination of this process by binding release factors, which cause the ribosomal subunits to disassociate, releasing the amino acid chain. While start codons need nearby sequences or initiation factors to start translation, a stop codon alone is sufficient to initiate termination.
Answer:
When the pKa is 6.0, we can determine the fraction of protonated H is by:
pH = pKa + log [A]/[HA]
Where
A = Deprotonated imidazole side
HA = Protonated side
Given, pH = 5.0
5 = 6 + log [A]/[HA]
log [A]/[HA] = -1 (take antilog of both side)
[A]/[HA] = 0.1
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 0.1
Given, pH = 7.5
7.5 = 6 + log [A]/[HA]
log [A]/[HA] = 1.5 (take antilog of both sides)
[A]/[HA] = 31.62
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 31.62
