BIOCHEM 423 Chapter Notes - Chapter 12.1: Chief Operating Officer, Lactase, Glycolysis

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viridianhornet922

7 Dec 2018

School

UMASS-Amherst

Department

Biochemistry & Molecular Bio.

Course

BIOCHEM 423

Professor

Ma Fu

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Document Summary

An overview of atyeolysis on relation of glycolysis to. The lo reactions of glycolysis occur in two phases ( energy investment c first five reactions energy generation. E convert one molecule of glucose to two molecules of pyruvate , 2 at pt an energy - investment phase. E 2 atp t atucose l six - carbon substrate ) 2 sugar phosphates l three carbon i trio se phosphates ) [ 2 those phosphates energy - rich compounds +4 atp ) Its electrons to an electron acceptor so that a steady - state concentration is maintained . Nam by transferring the electrons to pyruvate , reducing it to lactate via the enzyme. Claiming ) adding phosphoryl groups to glucose , yielding compounds with low phosphoryl group transfer potential . Chemically converting these low phosphoryl group transfer potential intermediates into compounds with high phosphoryl group transfer potential .

Related Questions

1. C6H12O6 + 6O2 Ã 6CO2 + 6H2O Glucose is being _____ to _____ and oxygen is being _____ to _____. a. oxidized; carbon dioxide; reduced; water b. oxidized; water; reduced; carbon dioxide Incorrect c. reduced; water; oxidized; carbon dioxide d. reduced; carbon dioxide; oxidized; water

2. Which of the following stages in the breakdown of the piece of toast you had for breakfast generates the most ATP? a. the citric acid cycle Incorrect b. glycolysis c. the digestion of starch to glucose d. oxidative phosphorylation

3. Which of the following is a catabolic pathway? a. gluconeogenesis b. production of starch Incorrect c. glycolysis d. production of glycogen

4. Per molecule, the oxidation of a fatty acid results in _____ energy storage than the oxidation of glucose. a. the same b. less Incorrect c. more d. None of the above; fatty acids cannot be oxidized.

5. Of the eight reactions in the citric acid cycle, how many are oxidation-reduction reactions? a. one b. two c. three d. four e. five

6. The process of glycolysis transfers the energy in sugar molecules into the stored energy of activated carriers such as ATP and NADH. Which statement is true about glycolysis? a. Only eukaryotic organisms do glycolysis. b. The ten reactions of glycolysis are identical in all eukaryotic cells, but are different in prokaryotic cells due to their diverse metabolic strategies. Incorrect c. The ten reactions of glycolysis are identical in almost all living organisms on the planet. d. There are many ways to oxidize glucose and transfer energy into activated carriers; the diversity of organisms on the planet is reflected in the diversity of their mechanisms of glycolysis.

7. Referring to the sentence completed in #13 above, the purpose of the latter reaction (in active muscle cells) is to _____. a. accumulate lactic acid in the cell, which provides additional energy for the production of ATP Incorrect b. slow down glycolysis until oxygen becomes available again c. accelerate the Krebs cycle, which will restore the oxygen supply d. regenerate NAD+ so that glycolysis can continue in the absence of oxygen

8. What are the products of the citric acid cycle? a. carbon dioxide, NADH, FADH2 and GTP b. carbon dioxide, NAD+, FAD and GTP c. NADH, FADH2, and acetyl CoA Incorrect d. carbon dioxide and water

9. Which statement is true about enzymes? a. Enzymes catalyze chemical reactions because they make inherently endergonic reactions exergonic. b. Each enzyme in a cell can catalyze many different chemical reactions involving chemically different substrates. Incorrect c. Enzymes can only catalyze reactions in one direction. d. After catalyzing a reaction, an enzyme is permanently altered and can no longer bind substrate. e. None of the above is true. 1

10. âG measures the change in free energy in a system as it converts reactant (Y) into product (X). When [X] = [Y], âG is equal to _____. a. RT b. ÎGÂ° + RT c. ÎGÂ° d. ln [X]/[Y]

11. The âGÂ° for the hydrolysis of ATP is -7.3 kcal/mol. What conditions in the cell would lead to a âG for ATP hydrolysis of -11 kcal/mol? a. An increase in the ATP/ADP ratio b. An increase in anabolic activity in the cell c. Transport of ATP out of the compartment where the reaction is occurring Incorrect d. A decrease in the ATP/ADP ratio

12. In the first reaction of glycolysis, a phosphate transferred from ATP to a glucose molecule to produce glucose-6-phosphate and ADP. Under conditions existing in the cell, the free energy change for this reaction is -8.0 kcal/mole. Which of the following statements is true? Recall that the âGÂ° for the hydrolysis of ATP is -7.3 kcal/mol. a. Glucose concentrations in the cell are always low, making this reaction more favorable than would be expected. b. This reaction is less favorable than would be expected, given the âGÂ° of ATP hydrolysis. c. Glucose-6-P concentrations in the cell are very low relative to glucose concentrations, making the reaction more favorable than would be expected. d. The âGÂ° for the addition of a phosphate to glucose must be -0.7 kcal/mole.

13. Protein E can bind to two different proteins, S and I. The binding reactions are described by the following equations and values: E + S --> ES Keq for ES = 10 E + I> EI Keq for EI = 2 Given the equilibrium constant values, which one of the following statements is true? a. When S is present in excess, no I molecules will bind to E. b. E binds I more tightly than S. c. The binding energy of the ES interaction is greater than that of the EI interaction. d. Changing an amino acid on the binding surface of I from a basic amino acid to an acidic one will probably make the free energy of association with E more negative.

i need question number nine please

Case Study Section 2 â glycolysis
Based on âMonday at the Metabolic Clinic, Adventures in Glycolysisâ by Nancy Boury (National Center for Case Study Teaching in Science)

Background:
You have been chosen for a very competitive undergraduate clinical internship at St. Visintainerâs Hospital and Metabolic Clinic working with Dr. Saccharo, an expert in enzyme deficiency disorders related to glycolysis. Before your first day Dr. Saccharo has asked that you review the reactions of glycolysis and the concept of isoenzymes.

Recall that isoenzymes are groups of enzymes that catalyze the same reaction in that they use the same reactants to product the same products. They may have different genes, tissue expression, developmental timing of expression, and affinities for their substrates. As an example, phosphofructokinase-1 has 3 isoenzyme forms: PFK-L is expressed in the liver and the gene is found on chromosome 21; PFK-M is expressed in the muscle and the gene is found on chromosome 12; and PFK-P is expressed in platelets and the gene is found on chromosome 10.

Question 1:
If a person would be deficient in all 3 forms of PFK-1, this person would .
A. Have lower than normal glycogen stores.
B. Have more mitochondrial activity than normal.
C. Be dead.
D. Produce more energy per glucose molecule because glycolysis would be unregulated.

It is now Monday morning and you are ready to start seeing patients with Dr. Saccharo.

Patient 1:
Ann is a teenager and avid golfer who was referred to the clinic after being refused at the blood drive and tiring easily on the high school golf course during practice. Ann is examined and a blood sample was obtained for analysis of a glycolysis related metabolic panel. The results of her panel is below (levels: + = normal, - = decreased; ++= increased):
Blood Metabolic Panel
Blood Glucose +
Glucose-6-phosphate ++
Fructose-6-phosphate ++
Fructose-1,6,bisphosphate ++
Glyceraldehyde-3-phosphate ++
1,3-bisphosphoglycerate ++
Phosphoenolpyruvate ++
Pyruvate -
ATP -
Red Blood Cell Concentration -
Question 2:
Assuming that Dr. Saccharo is correctly assuming that Ann has a defect in glycolysis, what glycolytic enzyme is the most likely to have the defect based on the blood metabolic profile?
A. Hexokinase
B. Phosphofructokinase
C. Triose Phosphate Isomerase
D. Pyruvate Kinase
E. Aldolase

Patient 2:
Marie is a 32-year-old mother of three complaining of fatigue and muscle cramps with exercise. She had always blamed her intolerance to exercise on her sedentary lifestyle. However, she recently joined a gym and after a week of aerobics classes went to her physician, who ordered a series of blood tests. The blood work came back with abnormal results, leading to her muscle biopsy and referral to the metabolic clinic. Marieâs blood was subjected to the same metabolic panel used before with the following results:
Blood Metabolic Panel
Blood Glucose +
Glucose-6-phosphate ++
Fructose-6-phosphate ++
Fructose-1,6,bisphosphate -
Glyceraldehyde-3-phosphate -
1,3-bisphosphoglycerate -
Pyruvate -
ATP -
Red Blood Cell Concentration -

Marieâs muscle biopsy was also analyzed and showed an excess of glycogen.

Question 3:
What is the most likely enzyme deficiency for Marie?
A. Hexokinase
B. Phosphofructokinase
C. Triose Phosphate Isomerase
D. Pyruvate Kinase
E. Aldolase

Question 4:
If you wanted to test the ability of red blood cells to complete glycolysis what compound would you try to detect?
A. Fructose-6-phosphate
B. Aldolase
C. Pyruvate
D. Phosphofructokinase
E. Dihydroxyacetone phosphate

Question 5:
If you used a sample of Marieâs blood for this, what compound could you add to enable these cells to complete glycolysis?
A. Glucose
B. Fructose-1,6-bisphosphate
C. Insulin
D. Glucagon
E. Fructose-6-phosphate

Patient 3:
Leo is a 25-year-old accountant that is not anemic, has perfect teeth, but has experienced two fainting spells recently. One happened after accidentally drinking a non-diet soda; the other after eating ânaturally sweetenedâ fruit salad. His dentist referred him to Dr. Saccharoâs clinic. Dr. Saccharo suspects Leo may have Hereditary Fructose Intolerance (HFI) and has recommended a closely monitored fructose tolerance test.

Hereditary Fructose Intolerance is a recessive disorder where individuals lack a functional copy of one isoform of the enzyme aldolase, specifically Aldolase B (also known as fructose-1-phosphate aldolase). In the liver, fructose is phosphorylated by Fructokinase to fructose-1-phosphate, then is cleaved by Aldolase B to glyceraldehyde and dihydroxyacetone phosphate. Fructose-1-phosphate is also a known allosteric inhibitor of Glycogen Phosphorylase, the primary enzyme responsible for releasing glucose from glycogen.

Question 6:
As compared to a normal individual that consumes fructose, a person with HFI who ingested fructose would have (select all that apply):
A. High levels of blood fructose
B. High levels of blood fructose-1-phosphate
C. High levels of blood glucose
D. Low levels of blood fructose
E. Low levels of blood glucose

That afternoon Leo undergoes the fructose tolerance test with the following results:

Question 7:
Based on the results of the test, what was the effect of giving Leo the intravenous fructose?
A. A rise in blood glucose levels
B. A decrease in blood glucose levels
C. An increase in glycogen stores
D. A rapid drop in blood pressure

However there were complications: Soon after the test Leo orders and drinks a cup of tea, but after half the cup Leo passes out. Dr. Saccharo orders two sequential blood glucose checks. The first was 50 mg/dl and 15 minutes later Leoâs blood glucose was down to 40 mg/dl. With those results Dr. Saccharo injected something into Leoâs IV and within minutes Leo was back to normal.

Question 8:
From these results does it appear that Leo has HFI?
A. Yes
B. No
C. It is unclear from the results

Question 9:
What did Dr. Saccharo inject into Leoâs IV line?
A. Insulin
B. Glucagon
C. Glucose
D. Fructose
E. Phosphofructokinase

Dr. Saccharo puts Leo on a restricted diet and orders follow-up visits.

Question 10:
Which of the following foods/ingredients would Leo NOT be allowed to eat?
A. Artificial sweeteners
B. High starch foods, such as potatoes
C. Sweet tea made with cane sugar
D. Any dairy products

Dr. Saccharo puts Leo on a restricted diet and orders follow-up visits.

Case Study Section 2 â glycolysis
Based on âMonday at the Metabolic Clinic, Adventures in Glycolysisâ by Nancy Boury (National Center for Case Study Teaching in Science)

Question 9:
What did Dr. Saccharo inject into Leoâs IV line?
A. Insulin
B. Glucagon
C. Glucose
D. Fructose
E. Phosphofructokinase

indigolocust876

What are the correct answers to the parts of this quiz marked incorrect? (that is my one question. Please take your time. I value accuracy over speed. )

IncorrectQuestion 2

0 / 1 pts
Studies done by Buchner showed that ground-up yeast cells were able to convert sugar to alcohol. The components of the mixture that were responsible for this transformation were

DNA molecules.

enzymes.

lipids.

carbohydrates.

Bloom's Level: 2. Understand
Learning Outcome: 06.01
Section: 06.01
Topic: History of Microbiology

IncorrectQuestion 3
0 / 1 pts
Bacteria that can absorb light energy and convert it into ATP are commonly called

cytochromes.

lysosomes.

chemotrophs.

phototrophs.

cytochromes AND chemotrophs.

Bloom's Level: 1. Remember
Learning Outcome: 06.02
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 4
0 / 1 pts
The readily usable energy currency of cells is

electricity.

the electron transport system.

ATP.

CTP.

the electron transport system AND CTP.

Bloom's Level: 1. Remember
Learning Outcome: 06.02
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 5
0 / 1 pts
NAD in its reduced form is symbolized as

NADH.

NAD.

NAD-.

NADP.

Bloom's Level: 1. Remember
Learning Outcome: 06.03
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 6
0 / 1 pts
The most common starting pathway for the breakdown of sugars is

respiration.

fermentation.

glycolysis.

oxidation.

Bloom's Level: 2. Understand
Learning Outcome: 06.04
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 7
0 / 1 pts
Which of the following processes generates the greatest amount of energy?

fermentation

aerobic respiration

the Entner-Duodoroff pathway

glycolysis

fermentation AND the Entner-Duodoroff pathway

Bloom's Level: 2. Understand
Learning Outcome: 06.05
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 8
0 / 1 pts
Fermentation

uses an inorganic molecule as the final electron acceptor.

uses an organic molecule as the final electron acceptor.

results in the production of a large amount of ATP.

is necessary in some organisms to produce reduced electron carriers.

uses an organic molecule as the final electron acceptor AND is necessary in some organisms to produce reduced electron carriers.

Bloom's Level: 2. Understand
Learning Outcome: 06.05
Section: 06.01
Topic: Microbial Metabolism

IncorrectQuestion 9
0 / 1 pts
The use of the suffix "ase" on a word denotes a(n)

substrate.

product.

enzyme.

electron acceptor.

substrate AND electron acceptor.

Bloom's Level: 1. Remember
Learning Outcome: 06.06
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 10
0 / 1 pts
Enzymes speed up reactions by

raising activation energy.

producing heat.

reducing entropy.

lowering activation energy.

raising activation energy AND reducing entropy.

Bloom's Level: 2. Understand
Learning Outcome: 06.06
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 11
0 / 1 pts
Most enzymes

are generalists and typically recognize a number of different substrates.

are specialists and typically recognize a single substrate.

are active over a wide pH range.

have several active sites.

are generalists and typically recognize a number of different substrates AND have several active sites.

Bloom's Level: 2. Understand
Learning Outcome: 06.06
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 12
0 / 1 pts
Allosteric enzymes

may bind two substrates.

are used to bind to other enzymes.

have an additional binding site that is involved in regulating enzyme activity.

are twice as fast as single site enzymes.

may bind two substrates AND are twice as fast as single site enzymes.

Bloom's Level: 2. Understand
Learning Outcome: 06.09
Section: 06.02
Topic: Microbial Metabolism

Question 13
1 / 1 pts
Feedback inhibition

is a means of regulating the amount of product produced.

often involves the use of allosteric enzymes.

involves inhibiting the last of a series of reactions.

results in raising the activation energy for the reaction.

is a means of regulating the amount of product produced AND often involves the use of allosteric enzymes.

Bloom's Level: 2. Understand
Learning Outcome: 06.10
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 14
0 / 1 pts
Coenzymes are derivatives of

minerals.

proteins.

lipids.

vitamins.

Bloom's Level: 1. Remember
Learning Outcome: 06.07
Section: 06.02
Topic: Microbial Metabolism

Question 15
1 / 1 pts
Environmental factors that may affect enzyme activity include

temperature

pH

salt

age

temperature, pH AND salt

Bloom's Level: 2. Understand
Learning Outcome: 06.08
Section: 06.02
Topic: Microbial Metabolism

Question 16
1 / 1 pts
Most enzymes function best at

acidic pH and high salt concentrations.

basic pH and low salt concentrations.

neutral pH and high salt concentrations.

slightly above pH 7 and low salt concentrations.

Bloom's Level: 2. Understand
Learning Outcome: 06.08
Section: 06.02
Topic: Microbial Metabolism

Question 17
1 / 1 pts
Mercury in mercurochrome inhibits growth

through competitive inhibition.

through a reversible action.

by oxidizing the sulfhydryl groups in cysteine.

by changing the shape of proteins.

by oxidizing the sulfhydryl groups in cysteine AND by changing the shape of proteins.

Bloom's Level: 3. Apply
Learning Outcome: 06.10
Section: 06.02
Topic: Microbial Metabolism

IncorrectQuestion 18
0 / 1 pts
The term precursor metabolite refers to molecules that

activate cell components.

are used in biosynthesis.

result from cell activities.

are present but inactive.

Bloom's Level: 2. Understand
Learning Outcome: 06.11
Section: 06.03
Topic: Microbial Metabolism

IncorrectQuestion 19
0 / 1 pts
Glycolysis

requires oxygen.

produces 3 ATP molecules.

produces 4 molecules of NAD.

may occur under aerobic or anaerobic conditions.

produces 3 ATP molecules AND produces 4 molecules of NAD.

Bloom's Level: 2. Understand
Learning Outcome: 06.12
Section: 06.03
Topic: Microbial Metabolism

IncorrectQuestion 20
0 / 1 pts
The transition step

links glycolysis to the pentose phosphate pathway.

links the pentose phosphate pathway to the Entner-Duodoroff pathway.

links glycolysis to the TCA cycle.

takes place in the matrix of the nucleus.

Bloom's Level: 1. Remember
Learning Outcome: 06.11
Section: 06.03
Topic: Microbial Metabolism

Question 21
1 / 1 pts
The TCA cycle produces

ATP through substrate-level phosphorylation.

NADH.

FADH2.

precursor metabolites.

All of the choices are correct.

Bloom's Level: 1. Remember
Learning Outcome: 06.12
Section: 06.03
Topic: Microbial Metabolism

Question 22
1 / 1 pts
The electron transport system

requires a membrane.

utilizes the nuclear membrane of eukaryotes.

utilizes the mitochondrial membrane of prokaryotes.

generates a concentration gradient of protons.

requires a membrane AND generates a concentration gradient of protons.

Bloom's Level: 2. Understand
Learning Outcome: 06.13
Section: 06.04
Topic: Microbial Metabolism

IncorrectQuestion 23
0 / 1 pts
Which type of phosphorylation does not require a membrane?

Substrate-level phosphorylation

Oxidative phosphorylation

Photophosphorylation

All types of phosphorylation require a membrane.

Bloom's Level: 2. Understand
Learning Outcome: 06.13
Section: 06.04
Topic: Microbial Metabolism

Question 24
1 / 1 pts
Which pairing is incorrect?

A:T

G:C

A:U

A:G

Bloom's Level: 3. Apply
Learning Outcome: 07.01
Section: 07.01
Topic: Genetics

Question 25
1 / 1 pts
RNA is characterized by which one of the following features?

Deoxyribose.

Thymine.

Ribose.

Double-stranded.

Bloom's Level: 1. Remember
Learning Outcome: 07.01
Section: 07.01
Topic: Genetics

Question 26
1 / 1 pts
DNA is characterized by which of the following feature(s)?

Ribose.

Single-stranded.

Deoxyribose.

Thymine.

Deoxyribose AND thymine.

Bloom's Level: 1. Remember
Learning Outcome: 07.01
Section: 07.01
Topic: Genetics

Question 27
1 / 1 pts
Which of the following is/are true of DNA replication?

It starts at the origin of replication.

Nucleotides are added to the 3' end.

It requires an RNA primer to get started.

It utilizes polymerases.

All of the choices are true.

Bloom's Level: 2. Understand
Learning Outcome: 07.03
Section: 07.02
Topic: Genetics

Question 28
1 / 1 pts
Which is true about the RNA transcript?

It is formed using the DNA minus strand as a template.

It has the same 5'-3' orientation as the DNA positive strand.

It is made in short fragments that are then stitched together.

The template starts at the promoter region.

It is formed using the DNA minus strand as a template, it has the same 5'-3' orientation as the DNA positive strand AND the template starts at the promoter region.

Bloom's Level: 2. Understand
Learning Outcome: 07.04
Section: 07.03
Topic: Genetics

IncorrectQuestion 29
0 / 1 pts
How many nucleotides are in a codon?

1

2

3

4

5

Bloom's Level: 1. Remember
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

Question 30
1 / 1 pts
Which may be or is an RNA molecule?

AGCCTAC

GGGCCCA

GCCCUUA

AGCCTAC AND GGGCCCA

GGGCCCA AND GCCCUUA

Bloom's Level: 2. Understand
Learning Outcome: 07.01
Section: 07.01
Topic: Genetics

IncorrectQuestion 31
0 / 1 pts
Which molecule carries an anticodon?

DNA

mRNA

rRNA

tRNA

Bloom's Level: 1. Remember
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

Question 32
1 / 1 pts
AUG

is only used as the start codon.

codes for methionine.

determines the reading frame.

is one of the stop codons.

codes for methionine AND determines the reading frame.

Bloom's Level: 2. Understand
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

IncorrectQuestion 33
0 / 1 pts
The A-site

is found on the RNA polymerase enzyme.

is found on the 30S ribosome.

is found on the 70S ribosome.

is the amino acid site.

is found on the 70S ribosome AND is the amino acid site.

Bloom's Level: 1. Remember
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

IncorrectQuestion 34
0 / 1 pts
A stop codon

codes for the stop amino acid s-methyl-methionine.

forms a hairpin loop forcing the ribosome to fall off.

codes for no amino acid.

enhances the binding of the e-polymerase.

codes for the stop amino acid s-methyl-methionine AND forms a hairpin loop forcing the ribosome to fall off.

Bloom's Level: 2. Understand
Learning Outcome: 07.05
Section: 07.03
Topic: Genetics

IncorrectQuestion 35
0 / 1 pts
Some segments of the precursor mRNA in eukaryotes are non-coding and are called

exons.

introns.

integrans.

uselessans.

Bloom's Level: 1. Remember
Learning Outcome: 07.06
Section: 07.04
Topic: Genetics

Question 36
1 / 1 pts
Gene regulation may entail

turning on genes only when needed.

turning off genes when not needed.

turning on or off entire groups of genes.

All of the choices are correct.

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

IncorrectQuestion 37
0 / 1 pts
The set of genes in bacteria that are linked together and transcribed as a single unit is referred to as a(n)

operon.

regulon.

operator.

repressor.

Bloom's Level: 1. Remember
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 38
1 / 1 pts
The molecules that bind to a repressor and cause it to no longer bind to the operator are called

activators.

repressors.

introns.

inducers.

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 39
1 / 1 pts
Repressors

are involved in negative control.

are involved in positive control.

always bind to the promoter.

bind or do not bind to the operator depending on their shape (conformation).

are involved in negative control AND bind or do not bind to the operator depending on their shape (conformation).

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 40
1 / 1 pts
Activators

are involved in negative control.

are involved in positive control.

always bind to the promoter.

are allosteric proteins.

are involved in positive control AND are allosteric proteins.

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 41
1 / 1 pts
Signal transduction

is the relay of information about conditions outside a cell to inside the cell.

often relies on a two component system.

may involve phosphorylation of various molecules.

is used by certain pathogens to sense low magnesium conditions.

All of the choices are correct.

Bloom's Level: 2. Understand
Learning Outcome: 07.07
Section: 07.05
Topic: Genetics

IncorrectQuestion 42
0 / 1 pts

Quorum sensing

is used by bacteria to sense the density of cells.

is used by bacteria to sense the proximity of eukaryotic cells

is used by bacteria to limit the density of cells.

is used by eukaryotes to sense the presence of bacteria.

does not influence the expression of genes by a bacteria.

Quorum sensingBloom's Level: 2. UnderstandLearning Outcome: 07.07Section: 07.05Topic: Genetics

Question 43
1 / 1 pts
In DNA sequence analysis

the + strand of DNA is used.

the start codon is ATG.

ORFs are searched for.

codon usage is a helpful indicator for protein coding areas.

All of the choices are correct.

Bloom's Level: 2. Understand
Learning Outcome: 07.12
Section: 07.08
Topic: Genetics

IncorrectQuestion 44
0 / 1 pts
DNA replication is usually unidirectional.

True

False

Bloom's Level: 2. Understand
Learning Outcome: 07.03
Section: 07.02
Topic: Genetics

IncorrectQuestion 45
0 / 1 pts
There are 4 functional types of RNA.

True

False

Bloom's Level: 1. Remember
Learning Outcome: 07.04
Section: 07.03
Topic: Genetics

Question 46
1 / 1 pts
Antisense RNA is the complement of the plus strand and may be useful in inhibiting translation.

True

False

Bloom's Level: 2. Understand
Learning Outcome: 07.11
Section: 07.07
Topic: Genetics

IncorrectQuestion 47
0 / 1 pts
A drug that inhibits the activity of bacterial DNA gyrase ______ be a good antibiotic because _______.

would; it would inhibit all DNA synthesis

would NOT; it would inhibit all DNA synthesis, even our own cell's DNA synthesis

would; it would selectively inhibit bacterial DNA synthesis due to differences between bacterial/human gyrase

would NOT; the DNA replication machinery could still function, even with impaired DNA gyrase enzyme

Bloom's Level: 5. Evaluate
Learning Outcome: 07.03
Section: 07.02
Topic: Genetics

IncorrectQuestion 48
0 / 1 pts
The regulatory protein

binds to the promoter region of DNA.

may inhibit or enhance transcription.

may control translation of the operon.

affects the activity of the DNA polymerase.

binds to the promoter region of DNA AND affects the activity of the DNA polymerase.

Bloom's Level: 2. Understand
Learning Outcome: 07.09
Section: 07.06
Topic: Genetics

Question 49
1 / 1 pts
The source of variation among microorganisms that were once identical is

antibiotic resistance.

virulence factors.

sigma factors.

mutation.

Bloom's Level: 1. Remember
Learning Outcome: 08.02
Section: 08.01
Topic: Genetics

Question 50
1 / 1 pts
The characteristics displayed by an organism in any given environment is its

genotype.

archaetype.

mutatotype.

phenotype.

Bloom's Level: 2. Understand
Learning Outcome: 08.01
Section: 08.01
Topic: Genetics

Question 51
1 / 1 pts
Transposons

are informally known as jumping genes.

may cause insertion mutations.

may cause knockout mutations.

were first recognized in plants.

All of the choices are correct.

Bloom's Level: 2. Understand
Learning Outcome: 08.05
Section: 08.02
Topic: Genetics

Question 52
1 / 1 pts
The largest group of chemical mutagens consists of

radiation.

base analogs.

nitrous acid.

alkylating agents.

Bloom's Level: 1. Remember
Learning Outcome: 08.06
Section: 08.03
Topic: Genetics

Question 53
1 / 1 pts
Nitrous acid most frequently causes mutations by

substituting oxygen for hydrogen bonds.

converting keto groups to amino groups.

altering the bonding ratios of nitrogen bases.

converting cytosine to uracil.

Bloom's Level: 2. Understand
Learning Outcome: 08.06
Section: 08.03
Topic: Genetics

IncorrectQuestion 54
0 / 1 pts
Chemical mutagens that mimic the naturally occurring bases are called

nitrogen mustards.

alkylating agents.

base analogs.

nitrous oxide.

Bloom's Level: 1. Remember
Learning Outcome: 08.06
Section: 08.03
Topic: Genetics

IncorrectQuestion 55
0 / 1 pts
Irradiation of cells with ultraviolet light may cause

4 nucleotides to covalently bind together.

thymine dimers.

adenine complementary base pairing with cytosine.

the addition of uracil.

Bloom's Level: 2. Understand
Learning Outcome: 08.08
Section: 08.03
Topic: Genetics

Question 56
1 / 1 pts
Thymine dimers are dealt with by

no repair mechanisms.

photoreactivation repair.

SOS repair.

excision repair.

photoreactivation repair AND excision repair.

Bloom's Level: 2. Understand
Learning Outcome: 08.11
Section: 08.04
Topic: Genetics

IncorrectQuestion 57
0 / 1 pts
X-rays

have no effect on DNA.

cause thymine trimers.

cause single and double strand breaks in DNA molecules.

make the DNA radioactive.

Bloom's Level: 1. Remember
Learning Outcome: 08.08
Section: 08.03
Topic: Genetics

IncorrectQuestion 58
0 / 1 pts
DNA repair mechanisms occur

only in prokaryotes.

only in eukaryotes.

in both eukaryotes and prokaryotes.

in neither eukaryotes or prokaryotes.

Bloom's Level: 1. Remember
Learning Outcome: 08.09
Section: 08.04
Topic: Genetics

IncorrectQuestion 59
0 / 1 pts
Antibiotics

cause mutations to occur.

may act as alkylating mutagens.

provide an environment in which pre-existing mutants survive.

increase the rate of spontaneous mutation.

Bloom's Level: 2. Understand
Learning Outcome: 08.12
Section: 08.05
Topic: Genetics

IncorrectQuestion 60
0 / 1 pts
Prokaryotic cell mutations can be observed very quickly because the prokaryotic chromosome is

diploid.

polyploid.

haploid.

polysomal.

Bloom's Level: 2. Understand
Learning Outcome: 08.02
Section: 08.01
Topic: Genetics

Question 61
1 / 1 pts
The diploid character of eukaryotic cells may mask the appearance of a mutation since

this may be a frame shift.

the mutation is often reversible.

the mutation may be palindromic.

the matching chromosome may carry the dominant gene.

Bloom's Level: 2. Understand
Learning Outcome: 08.02
Section: 08.01
Topic: Genetics

IncorrectQuestion 62
0 / 1 pts
A clever technique that streamlines the identification of auxotrophic mutants is

gas chromatography.

replica plating.

direct selection.

reversion.

Bloom's Level: 2. Understand
Learning Outcome: 08.12
Section: 08.05
Topic: Genetics

IncorrectQuestion 63
0 / 1 pts
To increase the proportion of mutants in a population of bacteria one may use

direct selection.

replica plating.

penicillin enrichment.

individual transfer.

Bloom's Level: 2. Understand
Learning Outcome: 08.12
Section: 08.05
Topic: Genetics

Question 64
1 / 1 pts
Bacteria that have properties of both the donor and recipient cells are the result of

UV light.

SOS repair.

frame shift mutations.

genetic recombination.

Bloom's Level: 2. Understand
Learning Outcome: 08.13
Section: 08.06
Topic: Genetics

IncorrectQuestion 65
0 / 1 pts
The mechanism by which genes are transferred into bacteria via viruses is called

ellipsis.

replica plating.

transformation.

transduction.

conjugation.

Bloom's Level: 1. Remember
Learning Outcome: 08.14
Section: 08.07
Topic: Genetics

IncorrectQuestion 66
0 / 1 pts
In conjugation the donor cell is recognized by the presence of

an F plasmid.

a Y chromosome.

diploid chromosomes.

an SOS response.

an F plasmid AND diploid chromosomes.

Bloom's Level: 2. Understand
Learning Outcome: 08.15
Section: 08.08
Topic: Genetics

IncorrectQuestion 67
0 / 1 pts
Gene transfer that requires cell-to-cell contact is

transformation.

competency.

conjugation.

functional genomics.

Bloom's Level: 2. Understand
Learning Outcome: 08.15
Section: 08.08
Topic: Genetics

Question 68
1 / 1 pts
The transfer of vancomycin resistance from Enterococcus faecalis to Staphylococcus aureus is thought to have involved

conjugation.

transformation.

transduction.

transposons.

conjugation AND transposons.

Bloom's Level: 2. Understand
Learning Outcome: 08.17
Section: 08.09
Topic: Genetics

IncorrectQuestion 69
0 / 1 pts
Double-stranded DNA enters the recipient cell during transformation.

True

False

Bloom's Level: 2. Understand
Learning Outcome: 08.13
Section: 08.06
Topic: Genetics

Question 70
1 / 1 pts
Plasmids often carry the information for antibiotic resistance.

True

False

Bloom's Level: 2. Understand
Learning Outcome: 08.16
Section: 08.09
Topic: Genetics

Quiz Score: 31 out of 70

BIOCHEM 423 Chapter Notes - Chapter 12.1: Chief Operating Officer, Lactase, Glycolysis

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