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If radioactive glucose, labeled at carbon #6 with 14C, were incubated in a cell-free
liver homogenate under anaerobic conditions, which carbon in the lactate produced would be
labeled with 14C (refer to the carboxyl carbon of lactate as C1, the central carbon as C2, and the
methyl carbon as C3)? What if the glucose were labeled at carbon #3 instead? In both cases,
explain your answer.

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Elin HesselLv2
6 Nov 2019
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Related questions

2) To study the fate of carbon during glycolysis under aerobic conditions, cultured cells are fed with 14C-glucose labeled in various positions. a) You use a radiolabeled form of glucose where the radioactive 14C occupies position 1.

Circle the correct statement and briefly explain your choice.

I. The carboxyl group of acetyl-CoA and CO2 become radiolabeled

II. The methyl group of acetyl CoA and CO2 become radiolabeled

III. The carbonyl group of acetyl-CoA becomes radiolabeled

IV. The methyl group of acetyl CoA becomes radiolabeled

V. CO2 becomes radiolabeled

VI. I don’t have enough information to conclude

b) You realize two independent experiments. In the first experiment, you use glucose where 14C occupies position 1. In the second experiments, 14C occupies position 2 and 6 in the glucose molecules. You use for each experiment 0.2 moles of radiolabeled glucose and you assume that all the pyruvate formed is converted to acetyl-CoA.

Circle the correct statements and explain your choices.

I. When glucose is labeled on carbon #1, 0.2 mole of acetyl-CoA is radiolabeled

II. When glucose is labeled on carbon #1, 0.1 mole of acetyl-CoA is radiolabeled

III. When glucose is labeled on carbons #2 and #6, 0.2 mole of acetyl-CoA is radiolabeled

IV. When glucose is labeled on carbons #2 and 6, 0.1 mole of acetyl-CoA is radiolabeled

V. When glucose is labeled on carbons #2 and 6, the carboxyl groups of half of the acetylCoA molecules are radiolabeled

VI. When glucose is labeled on carbons #2 and 6, the methyl groups of half of the acetyl-CoA molecules are radiolabeled

Question 1


Which of the following provides the best description of adifference between prokaryotic and eukaryotic cells?.l.14.3>

The cell membrane of prokaryotic cells contains proteins, whilethe cell membrane of eukaryotic cells contains only lipids andcarbohydrates.

The DNA in eukaryotic cells is enclosed in the nucleus, whilethe DNA in prokaryotic cells is not separated from the rest of thecell’s contents.

Eukaryotic cells have cell walls that consist mainly of sugars,while prokaryotic cells do not.

Prokaryotic cells contain chloroplasts and mitochondria, whileeukaryotic cells do not contain specialized organelles.

Question 2

Cellular respiration is a process involving the breakdown ofglucose molecules in order to generate energy for cellularactivities. Given the general reaction below, which of thefollowing best describes the products labeled x andy?

C6H12O6 + O2 → _x___+ __y___ + energy.l.18.8>

The products are CO (carbon monoxide) andH2O2 (hydrogen peroxide), which thendecompose into carbon dioxide (CO2) and water(H2O).

The products are H2O (water) and CO2 (carbondioxide), which then undergo anaerobic respiration.

The products are CO2 (carbon dioxide) andH2O (water), which then become the reactants forphotosynthesis.

The products are O2 (oxygen) and H2O(water), which are then used for growth and to keep the organismhydrated.

Question 3

If a researcher added a quantity of phospholipid molecules to abeaker of water, what would be most likely to happen to theorientation of the phospholipid molecules?.l.14.2>

They would break apart, and the remaining lipid tails wouldfloat up and form a layer on the surface of the water.

Their lipid tails would align within a spherical double layerwith their phosphate heads in the water inside and outside thesphere.

They would form a single-layered sphere with the phosphate headspointing outward and the lipid tails pointing inward.

Their lipid tails would adhere to the glass of the beaker andtheir phosphate heads would float in the water.

Question 4

While looking at a cell under a microscope, a scientist is ableto see a biological molecule. This molecule is a nucleic acid witha ladder structure and contains the nitrogenous bases adenine,thymine, cytosine, and guanine. Based on this information, whatconclusion can the scientist draw regarding this nucleicacid?.l.18.1>

The nucleic acid is RNA and would be found in the cell’snucleus.

The nucleic acid is RNA and would be found in the cytoplasm.

The nucleic acid is DNA and would be found in the cytoplasm.

The nucleic acid is DNA and would be found in the cell’snucleus.

Question 5


Which of the following is needed for cellular respiration tooccur?.l.18.8>

Sodium

Oxygen

Water

Nitrogen

Question 6


What would be the most likely effect if the side chain of an aminoacid group was unable to vary its structure?.l.18.4>

The organism would not be able to make and store starches.

The organism would not be able to process glucose moleculeseffectively.

The organism would not be able to regulate the creation ofspecialized cells.

The organism would not be able to make more than one type ofprotein.

Question 7


During photosynthesis, unused oxygen atoms form oxygen gas(O2) which is released as a waste product. How did theseoxygen atoms originally enter the plant?.l.18.7>

They were part of the ATP molecules gathered by thechloroplasts.

They were part of the sugar molecules(C6H12O6) the plant brought in forfood.

They were part of the carbon dioxide (CO2) moleculesthat entered through the leaves.

They were part of the water molecules (H2O) pulled inthrough the roots.

Question 8


Which of the following organelles houses DNA in eukaryoticcells?.l.14.3>

Endoplasmic reticulum

Golgi apparatus

Nucleus

Mitochondrium

Question 9


M. J. Schleiden shifted the scientific community's attention tocellular processes with his.l.14.1>

attention to disproving spontaneous generation

publication of ideas on cell generation

theories on the division of cells

claim that cells were the basic unit of all plant matter

Question 10


While the average human is able to hold his or her breath forapproximately one minute, a whale can dive for over 30 minuteswithout returning to the surface. Which of the following correctlydescribes this difference?.l.18.10>

Whales need less energy than humans.

Whales gather energy from their environment better thanhumans.

Whales are able to tolerate the products of cellular respirationbetter than humans.

Whales have a slower cellular respiration metabolism rate thanhumans.

Question 11


What can you conclude about the advantage of the cell membranehaving hydrophobic ends in the phospholipid bilayer?.l.18.3>

The cell requires the ability to digest and eliminateenzymes.

The cell requires a barrier between itself and itsenvironment.

The cell requires a means to store starches and glucose.

The cell requires a structure for eliminating waste.

Question 12


What would be the most likely outcome if the ribosomes of a cellwere damaged and stopped functioning?.l.14.3>

The cell would lose structure and shape.

Cell growth and repair would slow or stop.

Cell waste would no longer be processed.

The cell would produce more mitochondria.


Multiple answers to each question might be possible!

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

Using an RNA probe complementary to the region not removed by the truncation.

Using an RNA probe complementary to the region removed by the truncation.

Using an DNA probe complementary to the region not removed by the truncation.

Using an DNA probe complementary to the region removed by the truncation.

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

As both are composed of nucleic acids, using either would result in identical results.

An RNA probe because RNA has uracil bases.

An RNA probe because it could also be used in a translation experiment.

A DNA probe because it is more stable than RNA.

A DNA probe because RNA cannot bind to DNA.

One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?

To collect the bacteria at the bottom of the tube.

To break open the bacteria to release the genome.

To separate the bacteria from the bacteriophages.

To be able to detect the radioactivity.

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

No, because only DNA can be labeled with radioactivity.

No, because the RNA genome would not enter the bacteria upon infection.

No, because while DNA and RNA nucleotides are similar, they are not identical.

Yes, because DNA and RNA nucleotides are similar.

Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

Random DNA mutations generally won't affect RNA and protein function.

It is faster to duplicate the genome when these are present.

The existence of introns can lead to multiple variations of proteins encoded by a single gene.

It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

Explain the 5’ to 3’ directionality of a DNA stand.

It is due to the fact that the free 5’ carbon is on one end and the free 3’ carbon is on the other

It is due to the fact that new nucleotide are added to the 5’ carbon of the previous nucleotide

It is due to the fact that there are 3 phosphate groups attached to the 5’ carbon

It is due to the fact the complementary strand is 3’ to 5’

More than one of the above explain the 5’ to 3’ directionality

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3’ carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP

It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide

It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP

It cannot be incorporated into a DNA strand.

Why does DNA polymerase utilize an RNA primer?

DNA polymerase is unable to initiate strand synthesis but RNA polymerase can

DNA polymerase can only add a dNTP to an rNTP

DNA synthesis proceeds in the 3’ to 5’ when initiating strand synthesis

Chromosomal DNA contains interspersed RNA fragments

The RNA primer increases stability of the newly synthesized strand

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