Biochemistry 2280A Lecture Notes - Lecture 2: Alpha And Beta Carbon, Covalent Bond, Aspartic Acid

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22 Oct 2018

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Biochemistry lecture 2 september 10: the flow of information in living things can be summarized as: Dna > rna > proteins: central dogma of molecular biology. Rna is then translated into a protein. The dna can act as a template for its own replication. A gene is a unit of dna that encodes one protein: the race for the double helix. Watso(cid:374) a(cid:374)d cri(cid:272)k sol(cid:448)ed the stru(cid:272)ture of dna (cid:271)(cid:455) (cid:271)orro(cid:449)i(cid:374)g a(cid:374)d steali(cid:374)g others" ideas. The structure indicated that a only paired with t and g only paired with c. Structure and function are linked together: review: textbook panels. Panel 2-1 chemical bonds and groups. Panel 2-2 the chemical properties of water. Panel 2-7 the principal types of weak non-covalent bonds. Many foods are rich in protein because our cells are rich in protein. Enzymes catalyze covalent bond breakage or formation. Structural proteins provide mechanical support to cells and tissues. Transport proteins carry small molecules or ions.

Related Questions

In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that

	DNA is a double helix.
	DNA may be taken up by bacterial cells and be active.
	DNA is present in bacteriophage T2.
	eukaryotic cells may be genetically transformed.
	DNA binds to the cell wall of the bacterium.

In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.

	temperatures high enough to kill cells
	enzymes that hydrolyze proteins
	enzymes that hydrolyze DNA
	enzymes that hydrolyze polysaccharides
	enzymes that hydrolyze RNA

Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?

	Pure RNA from S cells
	Cell-membrane extracts from S cells
	A mixture of pure RNA and protein from S cells
	Cell-free extracts from S cells
	Pure protein from S cells

A-T base pairs in a DNA double helix

	form three hydrogen bonds with each other.
	are more heat-stable than G-C base pairs.
	are of a substantially different length than G-C base pairs.
	are not accessible to DNA binding proteins.
	are chemically distinct from G-C base pairs.

If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?

	27
	50
	0
	73
	23

The uniform diameter of the DNA structure provides evidence for

	antiparallel DNA strands.
	a right-handed helix.
	a double-stranded helix.
	3â² end phosphorylation.
	purineâpyrimidine pairing.

If a sequence of one strand of DNA is 5â²-TGACTATC-3â², what is the complementary strand?

	5â²-ACTGATAG-3â²
	3â²-TGACTATC-5â²
	3â²-ACTGATAG-5â²
	3â²-ACTGATAG-3â²
	5â²-TGACTATC-3â²

What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?

	3â² end phosphorylation
	Purineâpyrimidine pairing
	Antiparallel DNA strands
	Covalent bonds between sugars and phosphate groups
	Hydrogen bonding between paired bases

If the MeselsonâStahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only ¹⁵N and the second only ¹⁴N, this result would have indicated that replication was

	unidirectional.
	dispersive.
	conservative.
	bidirectional.
	semiconservative.

10.

The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3â² hydroxyl group (âOH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.

	The replication complex will no longer bind to origins of replication.
	The DNA polymerase will only incorporate acyclovir molecules.
	Acyclovir will bind single-strand binding proteins.
	DNA polymerase will not be able to link a successive nucleotide.
	Acyclovir will block the activity of the DNA helicase.

11.

Which of the following does not demonstrate the stability of the DNA double helix?

	Single-strand binding proteins are required to keep the strands apart.
	Paired bases form hydrogen bonds.
	High temperatures are required to denature it.
	DNA synthesis requires energy.
	DNA helicases require ATP.

12.

What effect would a primase inhibitor have on DNA replication?

	DNA polymerase would not be able to add bases to the DNA.
	Primase would be blocked from joining the DNA replication complex.
	Primase would not be able to provide primers for DNA polymerases.
	DNA polymerase would not be able to use DNA as a template.
	There would be no effect on DNA replication.

13.

To replicate their DNA in a timely manner, most eukaryotic chromosomes

	normally have uncoiled DNA.
	have multiple origins of replication.
	contain linear molecules of single-stranded DNA.
	have two replication forks that move in the same direction.
	are composed entirely of DNA.

14.

Which statement about DNA replication is false?

	Okazaki fragments are synthesized as parts of the leading strand.
	Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
	The sliding clamp increases the rate of DNA synthesis.
	Replication forks represent areas of active DNA synthesis on the chromosomes.
	Ligases and polymerases function in the vicinity of replication forks.

15.

In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.

	leading; telomerase
	lagging; telomerase
	lagging; DNA polymerase I
	leading; DNA polymerase I
	leading; replicase

16.

A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?

	DNA polymerase is constantly active.
	Primase is able to extend the telomeres.
	The mismatch repair system is extra efficient.
	The telomerase gene is being expressed.
	The cells do not need the ends of their chromosomes.

17.

If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?

	The replication complex excises the incorrect nucleotide.
	The excision repair proteins remove the incorrect nucleotide.
	The mismatch repair system scans for base-pairing mismatches.
	There is no correction, because nucleotide errors by DNA polymerases are not repaired.
	DNA ligase connects the correct nucleotide.

18.

Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.

	1, 2, 3, 4
	2, 1, 3, 4
	3, 4, 2, 1
	2, 4, 1, 3
	4, 2, 3, 1

19.

Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.

	64
	32
	12
	128
	6

20.

When double-stranded DNA is heated to temperatures above 90Â°C, it denatures. Denaturation is a process that

	breaks covalent bonds.
	hydrolyzes DNA.
	separates double-stranded DNA into two single strands.
	causes new hydrogen bonds to form.
	irreversibly destroys the double helical structure.

Second messengers play an important role in cellcommunication.

Which of the following best describes their role?

		None
		integrate, amplify and distribute messages throughout thecell
		Release messages to other cells?
		Receive information at the cell surface

What is true of disulfide bonds?

	a.	They often occur in the cytosol
	b.	They are a noncovalent bond
	c.	They are formed enzymatically in the ER
	d.	They act as reducing agents
	e.	they increase protein flexibility

Which of the following is NOT an example of a cell communicationmolecule?

		nucleases
		nitric oxide
		hormones
		kinases

What is true of binding pockets?

	a.	They only occur in the membrane
	b.	They occur in unfolded proteins
	c.	They contain electrostatic interactions
	d.	They contain variable amino acids
	e.

What is true of cell signaling?

	a.	Extracellular signals transduced by rectors always alter cellbehavior very quickly
	b.	Some cells require cell signals just to survive
	c.	Cell signals can be interpreted differently by differentcells
	d.	B and C only
	e.	All of the above (A, B and C)

Histone tails can be modified to change the shape and"tightness" of the nucleosomes. Acetylated histones result in whicheffect?

		Open DNA
		increased transcription
		closed DNA
		decreased transcription
		1+2
		1+4
		3+4
		2+3

Shifting the two bound tRNA from the A and P sites to the E andP sites of the ribosome involves:

	a.	The movement of the small ribosome subunit down the mRNAchain
	b.	The degradation of the A site on ribosome
	c.	The synthesis of the E site on the ribosome
	d.	The movement of the large subunit relative to the smallsubunit
	e.	All of the above

Intracellular signaling pathways are awesome! What is the onething they can NOT do?

	a.	relay the signal, to spread it through the cell
	b.	nothing, they can communicate any cellular need
	c.	distribute the signal to more than one effector protein
	d.	integrate the signal from more than one cellular pathway
	e.	Amplify the signal received, making the signal stronger

Specificity of protein/protein and protein/DNA binding isdetermined by what?

		Non-covalent bonds
		Hydrophobic interactions
		Covalent bonds
		shape of the binding site
		all of the above
		none of the above
		1,2,3
		1,2,4
		1,3,4

The coiling structure of the DNA into chromosomes is achieved bywhat?

		alpha helix
		beta pleated sheets
		chaperone proteins
		DNA/Protein complexing

Which regions (s) of the DNA are found in the final protein?

	a.	poly-A tail
	b.	A and B
	c.	Exons
	d.	UTRs
	e.	Introns

Metabolic processes can be turned "on" and "off" by chemicalmodifcation.

Which chemical modification do we associate with changing theactivity of an enzyme in a transient/non-permanent manner?

		Phosphorylation
		Ubiquitination
		Methylation
		Acetylation

DNA hybridization involves attaching a florescent dye towhat?

	a.	Antibody
	b.	Nuclear stain
	c.	Probe DNA
	d.	tRNA
	e.	rRNA

sapphirecrocodile239

QUESTION 10

If you can drink milk as an adult, it means that you have inherited a mutation in the promoter of your lactase gene (the gene that encodes the enzyme you need to break down lactose). Predict the effect of this mutation:

		The mutation changes the number of domains in the enzyme, which makes it work more efficiently
		The mutation changes the amino acid sequence of the lactase protein
		The mutation increases the number of copies of the lactase gene that will be found in your genome
		The mutation changes whether the lactase sequence is found in an intron or exon
		The mutation affects the expression of the lactase gene

1.2 points

QUESTION 11

A competitive inhibitor is decreasing the activity of an enzyme. Predict the effect of adding more substrate to the reaction.

		The substrate will increase the reaction rate by binding to the allosteric site
		The substrate will increase the reaction rate by competing with the inhibitor for the active site
		The reaction rate will not change unless the inhibitor can be removed
		The enzyme adjusts its shape so that the substrate, but NOT the competitive inhibitor, can bind
		The substrate will bind to the competitive inhibitor and block its ability to bind to the enzyme

1.2 points

QUESTION 12

What determines where in the genome a transcription regulator will bind?

		Transcription regulators bind to the 5' UTR region of a gene
		Regulators bind via complementary base-pairing to certain DNA molecules
		Covalent bonds form between the transcription regulator and the atoms of the DNA backbone
		Every eukaryotic gene has a different transcription regulator that will bind to the 5' end of the gene
		Transcription regulators bind to specific DNA sequences via multiple weak non-covalent interactions

1.2 points

QUESTION 13

What is the basic premise of cell theory?

		DNA -> RNA -> protein
		All cells arise from pre-existing cells
		DNA provides the complete instructions to create a cell
		The identity of a cell is determined through gene expression patterns
		All cells contain the same four basic macromolecules

1.2 points

QUESTION 14

What is the benefit of using BOTH the lac activator and the lac repressor to control gene expression?

		Using both an activator and repressor enables cells to more accurately determine the amount of lactose available in the environment
		Enzymes to digest lactose are only made when energy is low and lactose is available
		The activator can override the inhibition of the lac operon by the repressor
		The repressor can control the enhancer, while the activator can control the promoter
		When neither the lac activator or repressor is present, expression of the lac operon is too high

1.2 points

QUESTION 15

What is the histone code used for?

		Phosphorylation and acetylation of DNA affect its ability to be compacted
		Changes to the sequence of DNA change whether DNA will wrap around histone proteins
		Covalent modifications of histones affect the ability of the transcription initiation complex to form
		Histones provide the codon sequences needed for translation to occur
		The histone code affects which amino acids will get added to a polypeptide

Biochemistry 2280A Lecture Notes - Lecture 2: Alpha And Beta Carbon, Covalent Bond, Aspartic Acid

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