They are further classified into: fish, amphibians, birds, reptiles and mammals.
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Answer:
(C) Aminoacyl-tRNA synthetases have an additional active site that binds to non-cognate tRNAs. The tRNAs that bind to this second active are hydrolyzed and released from the enzyme.
Explanation:
In case of translation, proof reading is done by aminoacyl-tRNA synthetases only. Aminoacyl-tRNA synthetases have two mechanisms to avoid error during translation which are mentioned as under:
<u>(1) Chemical proof reading:</u> Incorrect amino acids rather than being hydrolyzed in catalytic pocket get hydrolyzed in editing pocket and thus they hardly get attached to tRNA.
For example: For distinguishing similar amino acids like isoleucine and valine, isoleucyl-tRNA synthetase uses a second active site which is meant for only valine not for isoleucine. In this particular site, valine which had entered the enzyme is cleaved away with the help of editing reaction after which the enzyme is well prepared to process isoleucine which is the correct amino acid for this enzyme.
<u>(2) Kinetic proof reading: </u>Even if an incorrect amino acid has entered a particular aminoacyl-tRNA synthetase, it does not cause appropriate conformational change in the enzyme because of which the incorrect amino acid loosens from the enzyme and does not get incorporated.
Note: In this example, only chemical proof reading is mentioned not kinetic proof reading.
The average length of a specific dinosaur leg bone is 60 inches, with a standard variation of 3 inches. If a z-score is 0, then the length is 67.
Divide your data point by your standard error after taking your data point and subtracting the mean from it. How is the z-score determined? Z-scores are calculated using the formula z = (x-)/, where x represents the raw score, the population mean, & the population variation. The z-score is just the raw score less the population mean, split by the population variance, as the calculation demonstrates. You now have your Z-score. For instance, a scoring of 450 would equate a scores of (450-500)/100 (= -50/100 = -0.50 on a scale with a mean of 500 as well as a standard variation of 100, indicating that the scoring is half a variance below the mean.
Learn more about variation
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Answer:
The answer is provided below
Explanation:
The man’s genotype is DD while the females genotype is Dd.
The mans phenotype is 100% on the other hand the woman’s phenotype is only 50%.
The cross comes out with results looking like this (going from left to right, then left to right again):
DD
Dd
dD
dd
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