Answer:
0.0035 mL of oxygen per hour is metabolic rate at 20°C at cockroach.
Explanation:
From the data we can see that metabolic rate of cockroach is in direct relation with temperature.
On graph, with metabolic rate on y-axis and temperature on x axis straight line will obtained , and the equation of the line can be determined by the help of point slope:
m = Slope of the line =
As per information given , :
x = temperature
y = metabolic rate
Now , put x = 20
y = 0.0035
0.0035 mL of oxygen per hour is metabolic rate at 20°C at cockroach.
Answer:
e. The tRNA that was in the A site moves into the P site.
Explanation:
Translation comprises 3 steps :
1) Translation initiation
2) Translation elongation
3) Translation termination
Translocation step is a part of translation elongation step which is followed by the peptidyl transferase activity of ribosome. Due to the peptidyl transferase activity of ribosome, the N terminal of newly entered amino acid (amino acid attached with tRNA i.e. aminoacyl tRNA or charged tRNA) at 'A site' attacks the C terminus of amino acid which is already present in the 'P site' of complex and amide bond is formed between amino acids leading to elongation of peptide chain. Then after, the uncharged tRNA which is present in the 'P site' is expelled out the complex via 'E site' while the tRNA that was in 'A site' moves into the 'P site'.
Answer:
Damage to the villi means less lactase released, so the body can't break down dairy.
Explanation:
If the small intestine is damaged, then that can mean that the villi are damaged. The villi usually release lactase, so this process will be compromised.
Less lactase released means that the body will have less of the enzyme required to break down lactose, the sugar present in dairy. Thus, individuals with celiac disease will struggle to digest dairy products
Answer:
It will bring about a stop to the translation process
Explanation:
The mutant tRNA despite still being charged with Glu, since it's anticodon is mutated and then recognizes and reads the codon UAA which is one of the stop codons (UAA, UAG and UGA) on the mRNA strand causes the translation process of that particular mRNA strand to stop. And the growing polypeptide chain to be released if any from the ribosomes.
Answer:
The anticodon will be unable to recognize the mRNA codon that is GAA, and the translation of this protein/polypeptide will be abruptly stopped. This may result in a truncated protein which is defected and hence, will be degraded by the relevant mechanisms in place. Since, UAA is actually one of the stop codons, the ribosome will not continue onward with the translation and fall off the mRNA.
Hope that answers the question, have a great day!
Answer:
punctuated equilibrium
Answer:
The correct statement is that a slow mutation rate makes it useful for determining evolutionary relationships between ancient species.
Explanation:
The rRNA or ribosomal RNA has an application as a molecular clock because of many factors. It exhibits a very slow mutation rate, that is, around one million years because of which it stays highly conserved in all the organisms. In the ribosomal RNAs of two organisms, the degree of mutations can be associated directly with the time they were originated in the past.
In supplementation, it possesses essential biological function because of which it can be witnessed in almost all the species.
Answer:
A slow mutation rate