https://www \'assessment -i c luHomel RosburyC emaining Time: 55 minutes, 09 sec
ID: 195908 • Letter: H
Question
https://www 'assessment -i c luHomel RosburyC emaining Time: 55 minutes, 09 seconds. M Course Inveitationc Microbielk uestion Completion Status: If the wild type DNA sequence reads THE CAT ATE THE BIG RAT, what type of mutation would change the sequence to THE CAT ATA ETHEBI GRA T? Missense Nonsense O Insertion Deletion O Silent QUESTION 11 "the wild type DNA sequence reads THE CAT ATE THE BIG RAT what type of mutaton would change the sequence to THE CAT ATE( O Missense O Nonsense O Insertion O Deletion Silent stop)? QUESTION 12 Eukaryotic chromosomes differ from prokaryotic chromosomes because only eukaryotes have Histone proteins O chromosomes in a nucleus O Several to many chromosomes O Elongated, not circular, chromosomes O All of the choices are conect QUESTION 13 Each nucleotide is composed of O One phosphate, one nitrogenous base, one sugar One phosphate, one nitrogenous base, two sugars O Two phosphates, one nitrogenous base, one sugar O Two phosphates, one niterogenous base, two sugars One phosphate, two ntrogenous bases, one sugar QUESTION 14 bind to the repressor proein before it can bind to the operator Repressible operons require that O The product A cofactor 0 A coenzyme O The substrate O The reactant QUESTION 15Explanation / Answer
ANSWER 10
If the wild type DNA sequence reads THE CAT ATE THE BIG RAT, the sequence would change to THE CAT ATA ETH EBI GRA T: Insertion - as the insertion of another nucleotide A results in this set up.
ANSWER 11
If the wild type DNA sequence reads THE CAT ATE THE BIG RAT, the sequence would change to THE CAT ATE (stop): Nonsense - as an insertion/deletion/substitution (transition or transversion) could have led to the occurence of a stop codon.
ANSWER 12
Eukaryotic chromosomes differ from prokaryotic chromosomes because only eukaryotes have histone proteins, chromosomes in a nucleus, several to many chromosomes, and elongated, not circular, chromosomes.
Hence: All of the choices are correct.
ANSWER 13
Each nucleotide is composed of: One phosphate, one nitrogenous base, oen sugar
ANSWER 14
Repressible operons require that the product bind to the repressor protein before it can bind to the operator. This constitutes a form of feedback loop/inhibition that helps regulate overall quantities of a unit within the cell.
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