8: replication, trascription, translation

Question 1

Griffith’s experimen

Answer 1

Griffith injected live, nonlethal bacteria (“rough” cells because they didn’t have an outside capsule) and dead, lethal bacteria (“smooth” cells because they had a capsule that protected them from the host’s immune system) into the same mouse. Separately, neither of the bacteria were deadly, but when together, the mouse died. Griffith concluded that something from the dead, lethal bacteria had transformed the live, nonlethal bacteria into lethal ones.

Question 2

Transformation

Answer 2

Change in genotype and phenotype due to cell assimilating external DNA

Question 3

Purines

Answer 3

A and G

Question 4

Pyrimidines

Answer 4

C and T

Question 5

Bonds between nitrogenous bases (of 2 nucleotides across from each other)

Answer 5

hydrogen bonds

Question 6

Meselson-Stahl Experiment

Answer 6

1) E. coli cells were grown in a heavy isotope of nitrogen. This means that the cells took up the heavy nitrogen to make biomolecules like DNA.
2) After many generations, all of the nitrogenous bases in the DNA had heavy nitrogen.
3) When centrifuged, the DNA produced a single band near the bottom of the beaker (heavy nitrogen).
4) Then, the E. coli was transferred to a different medium with a light isotope of nitrogen. The cells incorporated the light nitrogen into their biomolecules as well, as this was the only nitrogen source.
5) After 1 round of DNA replication, the DNA was centrifuged, and the band that appeared was in the middle (density between the light and heavy isotope, showing both had been incorporated into the DNA).
6) Another round of DNA replication resulted in 1 band at the same position as last time (mixture of light/heavy nitrogen) and another higher up (just light nitrogen), confirming the semi-conservative model.

Question 7

Direction that DNA Polymerase travels in

Answer 7

5’ to 3’

Question 8

Topoisomerase

Answer 8

Enzyme that prevents the double helix ahead of the replication fork from getting too tightly wound as helicase unwinds the DNA (during replication)

Question 9

DNA Polymerase I

Answer 9

Removes RNA primers and replaces them with nucleotides

Question 10

DNA Polymerase II

Answer 10

Helps DNA Polymerase III by proofreading nucleotides and ensuring the bases have been paired correctly

Question 11

Bonds in deoxyribose / phosphate backbone

Answer 11

Phosphodiester (covalent) bonds between the ribose of 1 nucleotide and the phosphate of the next

Question 12

Leading strand

Answer 12

Runs 5’ to 3’
Meaning the TEMPLATE strand runs 3’ to 5’
(from the beginning of the DNA to the fork)

Question 13

Lagging strand

Answer 13

Runs 3’ to 5’
Meaning the TEMPLATE strand runs 5’ to 3’
(from the beginning of the DNA to the fork)

Question 14

Steps of DNA replication

Answer 14

1) Helicase unwinds DNA by breaking H bonds between complementary bases.
2) Single-strand stabilizers bind to the replication fork to prevent the 2 strands from re-joining.
3) DNA primase creates RNA primers.
4) RNA primers attach to the leading and lagging strand. (Only 1 is needed for the leading strand; 1 is needed per Okazaki fragment on the lagging)
5) DNA Polymerase III matches complementary nucleotides to the template.
6) Okazaki fragments are formed.
7) DNA Polymerase I removes RNA primers (on both strands) and replaces them with nucleotides.
8) Ligase forms phosphodiester bonds between Okazaki fragments.
9) DNA Polymerase II proofreads strands.

Question 15

Promoters

Answer 15

Area of DNA (in nucleus) with the promoter (lots of TATAs with easy-to-break H bonds), where the RNA Polymerase binds in order to start transcribing DNA into mRNA

Question 16

Pre-mRNA to mature mRNA

Answer 16

1) The pre-mRNA detaches from the parent DNA in the nucleus. It is not yet in the cytoplasm.
2) The 5’ cap is added. The cap is a modified guanine nucleotide.
3) The introns are removed from the mRNA, and remaining exons are stitched together.
4) At the end, the Poly-A tail (long sequences of adenines) is added to the 3’ end of the mRNA.
5) The pre-mRNA is now mature mRNA and leaves the nucleus to a (bound or free) ribosome.

Question 17

Sites on a ribosome

Answer 17

A = acceptance site (where the tRNA binds to the ribosome)
P = peptide-linking site (where amino acids are joined together)
E = exit site (releases tRNA)

Question 18

Hershey-Chase experiment (is DNA or protein the genetic information?)

Answer 18

There were 2 batches of bacteriophages (viruses that infect bacterial cells) - one grown in a radioactive sulfur isotope, and the other grown in a radioactive phosphorous isotope. Only protein (and not DNA) incorporates sulfur, so proteins in the bacteriophage were labeled with radioactive sulfur, whereas only DNA (and not protein) incorporates phosphorous, so DNA was labeled with radioactive phosphorous.

The 2 batches were injected into different cells, and after the infected cells were centrifuged, the researchers found that most phosphorous appeared in the pellet (meaning it had entered the cell), whereas most sulfur appeared in the supernatant (liquid layer at the top of the tube, meaning it had remained outside the cell). This showed that phages inject host cells with DNA (the genetic information), not protein.

Question 19

Functions of 5’ cap and poly-A tail

Answer 19

1) Protects mRNA from denaturing enzymes in the cytoplasm
2) Help mature mRNA leave nucleus
3) Help ribosomes attach to the 5’ end of the mRNA