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C2006/F2402 12 Key to Review Questions for Exam #3

1. (Pax6 binds to similar DNA sequences in both genes)(Pax6 works with a different combination of TFs in len

& pancreas)

Explanation: Pax6, a TF, would recognize a similar DNA sequence in both genes, which is conferred by the amino

acids in Pax6s DNA binding domain. However, PAX6 works with a different set of TFs in the different cell types. F

example, activation of crystallin in the lens requires PAX6 and TFs A & B and activation of somatostatin in the

pancreas requires PAX6 and TFs C & D.

B-1. Pax6 could bind the DNA at (any of these).B-2. Pax6, Sox2, and L-Maf are (trans-factors)(proteins)(TFs)(nuclear).

B-3. The binding sites for Pax6, Sox2 & L-Maf in the lens are (DNA) (nuclear)(a cis-element)

Diagram for Problem 6-B:

Explanation: You had to draw a diagram and label the following to get full credit:

1. Enhancer or cis element2. TFs or trans-factors labeling Pax6, Sox2 & L-Maf wasnt enough

3. +1 or transcription initiation site

C. The histone acetlytransferase (i) switches the chromatin configuration (loose euchromatin to loosest euchromatiAND (ii) (increases transcription).

Explanation: The enhancer, when Pax6, Sox2 & L-Maf bind, switches from loose to loosest euchromatin, increasing

transcription. Regions of the DNA that are sometimes transcribed are always euchromatin, never heterochromatin.

D. The region would (represent a hypersensitive site) & (be more digested than the crystallin coding region).

Explanation: The enhancer while bound by Pax6, Sox2 & L-Maf has a paucity of nucleosomes and so is relatively

exposed to DNase. Treatment of this enhancer with DNase, results in a hyper sensitive site. In contrast, the transcribe

region of the crystallin gene (looser) has less nucleosomes than when this gene isnt actively transcribed (loose) but h

more nuclesomes than the enhancer (loosest).

E. For somatic nuclear transfer to succeed (all of the above) must occur.

Explanation: In order for somatic nuclear transfer to work, the heart cells DNA needs to reverse the differentiation

process, so that the DNA expresses the correct genes for totipotent and can progress through the ensuing stages of

potency (pluripotency, multipotency, etc.) to form the entire organism. For this to happen all of the above choices mu

be reset or changed. Reseting of the potency is performed by the various components present in the egg cytoplasm.

+1(or transcription initiation site)

Enhancer(or cis-element)

TFs:(or trans-factors)

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C2006/F2402 12 Key to Review Questions for Exam #3

2. A-1. The hypoglycemic infant is probably producing (too much insulin) .

A-2. In the pregnant mother, the levels of blood sugar were probably (elevated)

Explanation: The baby was exposed to high (maternal) blood sugar in utero, and made high levels of insulin in

response. When the child is born, there is too much insulin for a normal level of sugar intake and/or blood sugar,

and the excess insulin causes too much glucose uptake and therefore hypoglycemia. It takes a short while for the

infant to readjust the levels of insulin production to match the levels of blood sugar.

B-1. Hypoglycemia shortly after feeding probably results primarily from abnormally (high glucose uptake).B-2. The major effector(s) causing the problem are (muscle) (liver).

Explanation: The problem occurs right after feeding, when the infant is in the absorptive state, not later in the

post-absorptive. Therefore the hypoglycemia must be caused by too much uptake (due to the high insulin) right

after feeding, not too little release between feedings (due to low glucagon or persistent insulin).

3. A. A normal person can compensate for low blood sugar by (glycogen breakdown) &

(gluconeogenesis).

Explanation: Liver is the primary effector for releasing glucose to correct for low blood sugar. The glucose can

come from glycogenolysis (depolymerization of glycogen), or from gluconeogenesis -- manufacture of glucose

from smaller molecules, such as lactate. Both occur in the liver. Muscles also breakdown glycogen, but they dont

release glucose they release lactate, which the liver either breaks down further for energy generation, or

converts to glucose for release or storage (as glycogen).

B. All of these should be able to compensate for high blood sugar.

Explanation: To compensate for high blood sugar, a person has to be able to take sugar up from the blood; if

there is extra, beyond immediate energy needs, the person must be able to store the glucose by synthesizing

glycogen (or fat). All these people can do that. The GSD types have problems with breaking down glycogen, not

with making or storing it.

C. In GSD 1, the primary problem is with the (effector(s)) & there has been a change in the (controlled

process(es)).

Explanation: The problem here is glucose release (the controlled process) by the liver (the effector). The

regulated variable (glucose) and the desired value (the set point) remain unchanged. Glucose release is

controlled by adjusting the rate of Glucose-phosphate hydrolysis, but the physiologically important process is

glucose release, not hydrolysis.

D. & E. A person with GSD type (1) is most likely to develop hypoglycemia; a person with GSD type (5) is

least likely.

Explanation: To get full credit, you had to explain the situation in people with each of the three types of GSD.).

Note that phosphorylase breaks down glycogen (a glucose polymer) to glucose phosphate, and phosphataseremoves the phosphate from the glucose.

People with GSD1 can not release glucose from the liver at all both glycogen breakdown and

gluconeogenesis produce glucose phosphate, and the glucose is trapped in side the cell if the phosphate cannot be

removed. Therefore these people are very likely to develop hypoglycemia.

People with GSD6 can generate (& release) some glucose from gluconeogenesis in liver. They cant break

down liver glycogen, but they can breakdown muscle glycogen to lactate, convert the lactate to glucose phosphate

in liver, remove the phosphate, and release the glucose. So they can develop hypoglycemia, but its not as severe a

problem as with type 1.

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C2006/F2402 12 Key to Review Questions for Exam #3

Explanation of 3 D & E, cont.

People with GSD5 have no problems with release of glucose from the liver. They may generate less lactate (from

muscle) for the liver to use for gluconeogenesis, but they have normal liver function. Since the muscle is not

directly involved in release of glucose, and their problem is with muscle, they are least likely to develop

hypoglycemia.

4. A. When worms are first exposed to heat, you expectA-1. Ca++ to (go out of the cell) (go in to the cell) (stay put no significant movement across the membrane).

A-2. Na+ to (go out of the cell) (go in to the cell) (stay put no significant movement across the membrane).A-3. Cl- to (go out of the cell) (go in to the cell) (stay put no significant movement across the membrane).

A-4. When the cells are exposed to heat, you expect the cells to become (hyperpolarized) (depolarized)

(either way).

Explanation: In response to heat, the TRPQ channels will open. All three ions, Ca++,Na+, and Cl- are higher on

the outside of the cell. In addition, the inside of the cell is negative, relative to the outside. So clearly the positive

ions (cations) will move in if channels are opened for them, and the cell will become less negative, or depolarized.

What Cl- will do if channels open depends on the electrical gradient (and the specificity of the channels). In this

case, the TRPQ channels do not transport Cl-, so the ion wont move even if the electrical gradient is favorable.

(If there is enough heat, an AP will be generated, but the immediate response will be a receptor potential, not anAP.)

B-1. In this case the sensory neurons (have a depolarizing receptor potential).

B-2. In this case, the sensory neurons (have a depolarizing receptor potential) & (fire action potentials).

B-3. If you increase the concentration of capsaicin you will increase (the number of spikes or APs) &

(the absolute magnitude of the receptor potential).

B-4. Capsaicin changes (the distance to threshold)

Explanation: There is no EPSP in receptor cells; they are presynaptic only. There is a graded response in the

receptor cells that is proportional to the stimulus this is called the receptor potential, not an EPSP or IPSP. In

this case, the rec. pot. is depolarizing (see A). The more channels are opened, the bigger the receptor potential. If

it is big enough, it reaches threshold and the cells fire action potentials. Both capsaicin and heat open channelsand depolarize the cells. The receptor potential caused by capsaicin and the rec. pot. caused by heat are summed,

and if the total is over threshold, the cell fires an AP. The bigger the rec. pot., due to either heat or capsaicin, the

more APs. If capsaicin is present, the cell is already partially depolarized and the amount of heat needed to

depolarize to threshold is decreased. (Note this is a one cell system the receptor cells themselves fire APs.)

C-1. It is likely that the heat receptor proteins of the worm are (ionotropic).

C-2. The compounds in hot peppers probably (bind to TRPQ channel proteins)

Explanation: C-1. The only new protein made in the recombinant epithelial cells, one that is not made in normal

epithelial cells, is the TRPQ channel. This is sufficient to give a response to both heat and capsaicin. In the

detached patches, there is no possibility for generation of a second messenger, and there is still a response to

heat. So the obvious explanation is that the receptor is ionotropic that is, the receptor is part of the channel.

There is no separate receptor. (The only other possibility is that normal epithelial cells already have unusedreceptors for heat, and corresponding G proteins, and that the G protein or a membrane bound component can

open the added channels without a soluble second messenger.)

C-2. It does not say in the problem whether or not there was a response to capsaicin in the detached patches.

Given that the receptor is ionotropic, it makes sense that capsaicin should bind to the receptor/channel itself, and

open the channels, as heat does. There is no reason to invoke a separate receptor, because epithelial cells do not

normally respond to capsaicin or heat, and should not have any part of the signaling system except the added

TRPQ channels. (Note: In the actual experiments, the response was measured, and the detached patches do

respond to capsaicin.)

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C2006/F2402 12 Key to Review Questions for Exam #3

5. A. At the time of transplantation in this case, the mouse skin cells were (specified) (differentiated) (determined).

B. At the time of transplantation tin this case, the cells were (specified).

Explanation: As cells progress down the cell fate cascade they are specified (will change fate if transplanted) and

determined (wont change fate if transplanted) multiple times. Only at the final stage do they differentiate express the

proteins that confer the cell with the structure and function of that specific cell type. The cells always progress in thefollowing order: specification, determination and at the end stage differentiation. So a cell that has been determined to

mesoderm has already by definition been specified to be mesoderm.In A the cells are expressing the proteins that confer a skin phenotype and they do not change their cell fate after

transplantation so they are specified, determined and differentiated to become skin cells. In B the cells are specified to

become skin (under normal circumstances they would develop into skin cells) but are not determined or differentiated

since they will change their fate when transplanted.

6. A. In induced pluripotent (iPS) cells (mitochondria DNA and most of the nuclear DNA are derived from the

same cell)

B. In somatic nuclear transfer(mitochondrial DNA and nuclear DNA are derived from different cells)

Explanation: iPS cells are created by infecting differentiated cells with the DNA that encodes four transcription factors

a result the nuclear DNA and the mitochondrial DNA come from the same cell but the nucleus now also contains the DN

of the four transcription factors that was infected.

Somatic nuclear transfer (SNT) is a method where the nucleus of a differentiated cell is placed inside an egg cell which

has had its nuclear DNA removed. In SNT the mitochondrial DNA is derived from the egg and the nuclear DNA is deriv

from the nucleus of the somatic donor cell.

7. A. Methylation of H3 K9 (tightens chromatin)(decreases TF accessibility to DNA).B. Acetylation of H3 K9 (loosens chromatin) (increases TF accessibility to DNA).

Explanation: Decreased transcription, caused by H3 K9 methylation, is associated with tighter chromatin and decreas

TF accessibility to DNA.Increased transcription, as a result of H3 K9 acetylation, is associated with looser chromatin and increased TF

accessibility.

8. A. The sex chromosomes directly determine the fate of(gonads).

B. Which of these reproductive organs are bipotential? (external genitalia) (gonads).

C. If gonad is removed 6 weeks into development, the (Mullerian duct remains)

Explanations:

8A: The sex chromosomes only determines the fate of the somatic tissues of the gonads and then once this is determin

the somatic tissue of the gonad determines the fate of the germ cells, the ducts and the external genitalia.

8B: Unlike most of the organs in our body, the external genitalia and the gonads can become two different tissues. Th

Mullerian ducts and the Wollfian ducts can each only form specific organs.8C: This question asks what would be different from Terry in the case study if we removed the gonad before the gona

differentiated into testes. The testes secretes two molecules that are important for sex determination of the other sexorgans, testosterone and anti-Mullerian hormone (AMH). The only tissue that responds to AMH is the Mullerian duct. In

the case study, where the testes were present until birth, the AMH secreted by the testes would destroy the Mullerian du

If the gonads were removed before they differentiated into testes, AMH would not be produced and the Mullerian ducts

would remain. All of the organs that are normally affected by testosterone would respond in the same way, i.e. no

response. If you dont have the receptor for testosterone (the androgen receptor) testosterone will not be detected,

regardless of whether or not testosterone is produced.

C2006/F2402 12 Key to Review Questions for Exam #3

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9. Which of these proteins during normal development do not have their histones bivalently methylated at H3K4 and

H3K27 simultaneously? (Cdx2) & (Oct4).

Explanation for 9: Both Cdx2 and Oct4 are turned on at the 8 cell stage, that is very early, so they do not have a

period when they are poised to go on, but not yet expressed. Bivalent methylation is the state of master regulatory

genes before they are turned on when they are not expressed, but ready to go either way. They can be turned on or

off by removing one of the methyl groups.

10. A. Which of the following progresses in the opposite direction, becoming more potent?(creation of zygote using SCNT) & (iPS cell creation).

B. Which of the following changes increases potency and occur in the natural process of human development?

(none of these).

Explanation for 10: iPS cells and zygotes created by SCNT are pluripotent and totipotent respectively. The

differentiated cell whose nucleus was used as a donor for SCNT was less potent, as were the cells transformed by the

4 genes used to make iPS cells. Natural human development does not include SCNT or creation of iPS cells these

require laboratory manipulations. All the other changes listed occur during human development, but they all involve

increases in differentiation and decreases in potency.

11. As a child which of the following should be present? (clitoris) (testes) (labia major) (Wolffian duct derivatives).

Explanation for 11: The initial stages of differentiation should be normal for a male. The SRY gene on the Y should

turn on production of proteins needed to develop testes. Sertoli cells of the testes should secrete AMH, which should

cause degeneration of the Mullerian ducts; Leydig cells should secrete testosterone which sustains the Wolffian duct.

However, male external genitalia will not develop. The perineum is bipotential, and develops into male genitalia

(penis and scrotum) only if DHT is present; otherwise it develops into female structures (clitoris and labia major). Ifthere is no reductase, there will be no DHT because conversion of testosterone to DHT will be blocked, and female

genitalia will develop by default.

12. A. What could you expect regarding the fetal development of this individual? (male levels of testosterone)

(testes are present)

B. What could you determine regarding this individuals chromosomes and the expression of genes on those

chromosomes? (XY individual) (Sry expressed) (testosterone receptor expressed)

C. If this individual had both Mullerian and Wolffian ducts what could you conclude?

(Leydig cells are okay) (Sertoli cells arent functioning properly).

Explanation for 12:

A. If tissues derived from Wolffian ducts are present, the fetus must have developed testes that produced male

levels of testosterone (in the fetus) to sustain the Wolffian ducts.

B. You would expect this person to be XY. The SRY gene must be expressed (to trigger testes formation andtestosterone production) and the testosterone receptor must be expressed (for the Wolffian ducts to respond to

the testosterone).

C. If the person had both Mullerian ducts and Wolffian ducts, you assume the person had high testosterone

(from working Leydig cells) but not AMH (so Sertoli cells arent working properly). If there is testosterone,

granulosa and thecal cells wouldnt form.