Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Take Test Homework 11 x E BIO201 Lecture 34.pdf BIO201 Lecture 35.pdf BIO21 lear

ID: 80647 • Letter: T

Question

Take Test Homework 11 x E BIO201 Lecture 34.pdf BIO201 Lecture 35.pdf BIO21 learns buffalo.edu/webapps/assessment/take/launchjsp?course assessment ida -195566 18course ida 144973 O It depends on unwinding of DNA. O All of the above are true for both replication and transcription. Question 6 Which of the following aspects of transcription is unique to bacteria? O Transcription and translation may occur simultaneously O Only bacteria have mRNAs O RNA splicing. O A7-methyl guanosine cap at the 5 end of mRNA increases stability. O All of the above Question 7 The lac repressor in bacteria is analogous to in eukaryotes. O Genes O General transcription factors, such as TFID O specific transcription factors O The p21 Cdk inhibitor Question 8 shown below is a drawing showing the result of an experiment in which an RNA molecule is allowed hybridize. The DNA strand is presumed to be the lighter-shaded one on the top. Note that only one sto 0 0 a

Explanation / Answer

6) In bacteria there is no nuclei so both transcription and translation occured in cytoplasm.

Ribosomes can bind to mRNAs ,which is being transcribed

So Answer is transcription and translation are simultaneous in bacteria.

7) Eukaryotes transcription factors and lac repressor are homologous as transcription factors bind DNA and make it easy or hard for RNA polymerase to do its work—same as lac repressor protein of E. coli

So answer is general transcription factors.

2) In mismatch repair, mistake during DNA replication are recognized, cut out and replaced.
Cut by endonuclease
Corrected by polymerase
Pasted by ligase

So Answer is all of a above

3) Single Gene can encode multiple protein through the alternative splicing.
Single Gene encode only one Protein is not true

So answer is a

4) Answer is B

A and T as they are not tightly bound , they are bound by double bonds and G and C bound by triple bonds

So bonding between A and T is easy to break and it is for enzymes to act.