Answer:
A 22 to 25 amino acid sequence present in the central section of the protein, which gives rise to an alpha helix in the membrane is known as the stop-transfer anchor sequence. The sequence plays an essential function in targeting the protein towards the plasma membrane. On the other hand, it also ceases targeting of the protein towards the endoplasmic reticulum, which was started by the signal peptide.
Thus, the process of translation of the remaining of the protein occurs within the cytosol due to the tethering of the transmembrane domain. In the stop-transfer anchor sequence, the hydrophobic amino acids present are isoleucine and valine. After mutation, these amino acids get converted into arginine and lysine, thus, hydrophilic amino acids replace hydrophobic amino acids in the sequence.
Due to this, the transmembrane domain cannot be targeted towards an integral part of the plasma membrane by the short transfer anchor sequence, and therefore, now the translocation of the protein will take place towards the endoplasmic reticulum as initiated by the signal peptide at the beginning.
Answer:
The first male was bb Ee, and the second male was bb EE.
Explanation:
In Labradors coat colour is controlled by two genes. Suppose the two genes are B and E. B produces black colour and recessive form bb gives brown colour. Gene E is epistatic over gene B in its recessive form which means that ee will produce yellow colour regardless of the genotype present of B gene.
The first case is possible if the female lab is bbee (yellow) and the male lab is bbEe (brown):
bbee X bbEe
bE be
be bbEe bbee
So half of the offspring will be brown (bbEe) and half of them will be yellow (bbee)
The second case is possible if the same female bbee mates with a brown male of different genotype which can be bbEE:
bbee X bbEE
bE
be bbEe
So all offspring will be brown (bbEe)
Hence, the first male was bbEe and the second male was bbEE.
Answer:
The answer is c)oxygen is an input to acetyl CoA formation.
Explanation:
After pyruvate is produced during glycolysis, its fate depends on oxygen availability. If there is enough oxygen available then pyruvate will enter the citric acid cycle, also called Krebs cycle. Pyruvate will then be oxidized by the pyruvate dehydrogenase complex into acetyl-coA and CO2. However, if there is not enough oxygen available, then pyruvate will undergo a reduction reaction by the lactate dehydrogenase enzyme and produced lactate along NAD+ production from NADH.
Answer:
5556
Explanation:
If a DNA polymerase synthesizes in average 50 nucleotides/second, that means that in three hours (10800 seconds) it synthesizes about 540000 nucleotides.
However, if the human genome is composed of 3000000000 (3 billion) base pairs (nucleotids), the minimum number of DNA polymerases (working in the same number of origins of replication) to finish the duplication of all the genome in three hours is 5555,5. (3000000000/540000). As we know there is no half polymerase, so we round to 5556.
5556 molecules of DNA polymerases acting on 5556 origins of replication are needed.
