Biochemistry, cell biology, and genetics question thread

The cell is considered to be haploid after the first round of division (i.e., after meiosis I). This is true even though there are still two copies of each chromosome from the sister chromatids in the daughter cells. However, the two homologues have been separated at this point, and it is because of this that we consider each of the daughter cells from meiosis I to be haploid, not diploid any more.

What do our taste receptors actually detect?

I mean chemically, i.e.:

sweet = OH groups?
salt = ions, but which ones?
sour = protons?
bitter = ?
umami (meatiness; whatever MSG adds) (assuming this "taste" exists) = ?

How much do I really need to know about the cellular-level energy cycles?

What do I really need to know about organelles? Which organelles?

Shrike said:

How much do I really need to know about the cellular-level energy cycles?

Click to expand...

For all of the pre-health tests (MCAT, PCAT, DAT, and OAT), you should be able to follow the generation of ATP in each step, and also the energy carrier reduction (NAD and FAD) in each stage. You do NOT need to memorize any enzymes or pathway intermediates; they will make you do that in your professional school biochem class. You should also know that oxygen is the final electron acceptor of the electron transport chain, and that anaerobic respiration is insufficient to sustain human life. In addition, fermentation produces lactic acid as a byproduct in humans, and ethanol in yeast. Finally, you should know where in the cell each stage of respiration occurs. Here is a list of the energy conversions for each stage and where in the cell they take place:

Glycolysis: (anaerobic, occurs in the cytoplasm)

• 2 net ATP (4 total made, but 2 needed to complete this stage)

• 2 NADH produced (making 4 ATP in ETC for eukaryotes and 6 ATP for prokaryotes)

Fermentation: (anaerobic, occurs in the cytoplasm)

• 0 ATP; its main purpose is to reoxidize the NADH produced in glycolysis

Pyruvate Decarboxylation: (aerobic, occurs in the cytoplasm for prokaryotes, mitochondrial matrix for eukaryotes)

• 0 ATP produced

• 2 NADH produced (making 6 ATP in ETC)

TCA Cycle: (aerobic, occurs in the cytoplasm for prokaryotes, mitochondrial matrix for eukaryotes)

• 2 ATP produced

• 6 NADH produced (making 18 ATP in ETC)

• 2 FADH2 produced (making 4 ATP in ETC)

Electron Transport Chain (ETC): aerobic, occurs across the inner cell membrane for prokaryotes, inner mitochondrial membrane for eukaryotes

• NADH oxidation back to NAD and FADH2 oxidation back to FAD occur along with ATP production, allowing the earlier stages to continue

Summary: 36 net ATP produced in eukaryotes, 38 net ATP produced in prokaryotes (because the electrons from the NADH produced from pyruvate decarboxylation do not have to be transported across the mitochondrial membrane in prokaryotes; doing this causes a net loss of two ATP in eukaryotes)

QofQ Thank you for your clear, well organized answer. I was wondering about your thoughts on texts, including TPR materials, that assume: that each NADH yields only 2.5 ATP and FADH2 yields only 1.5 ATP. The total would therefore, yield 30 ATP in euk. & 32 in prok. . Is this a significant discrepancy?
Also how significant is prok. anaoerbic respiration (as opposed to fermentation). My understanding is that in anaoerbic respiration, there is e- transport leading to the phosphorylation of ADP, but that some chemical other than O (often S) is the final electron acceptor.
Thanks

Lindyhopper said:

QofQ Thank you for your clear, well organized answer. I was wondering about your thoughts on texts, including TPR materials, that assume: that each NADH yields only 2.5 ATP and FADH2 yields only 1.5 ATP. The total would therefore, yield 30 ATP in euk. & 32 in prok. . Is this a significant discrepancy?
Also how significant is prok. anaoerbic respiration (as opposed to fermentation). My understanding is that in anaoerbic respiration, there is e- transport leading to the phosphorylation of ADP, but that some chemical other than O (often S) is the final electron acceptor.
Thanks

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I am not familiar with the TPR materials, so I am not sure how they came up with those numbers that each NADH yields 2.5 ATP instead of 3. Maybe Shrike or another TPR instructor can chime in on this one? But in the whole scheme of things, I would say it is less important to know how many ATP each NADH yields than it is to know how many NADH are formed at each stage. It is entirely plausible that both numbers are not perfectly accurate, because there are probably other factors that affect the efficiency of the entire respiration process. (Biological processes are always more complex than an introductory level textbook makes them out to be. ) But of course we are way over the level of knowledge you are expected to have for the MCAT here.

For your second question, I think you are asking about sulfur-reducing bacteria? (They make H2S rather than H2O.) If so, you do not need to know about this at all for the MCAT; if they asked you any questions about it, it would be explained to you in a passage.

Shrike said:

What do I really need to know about organelles? Which organelles?

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I will post further details about them at a later time, but for MCAT, DAT, OAT, and PCAT biology, you should know the structure and function of the following organelles:

• nucleus
• cell membrane
• cytosol (cytoplasm)
• cytoskeleton (microtubules, microfilaments, and intermediate filaments)
• endoplasmic reticulum
• Golgi apparatus
• vesicles
• vacuoles
• ribosomes
• lysosomes
• microbodies
• mitochondria
• chloroplasts
• cell wall
• centrioles

You should also know that prokaryotes do NOT have any membrane-bound organelles, such as a nucleus, mitochondria, or Golgi apparatus.

There has been recent debate as to whether or not NADH produces 3ATP and FADH2 yields 2 ATP. The recent texts seems to say that the actual ATP produced is slightly less then the previously accepted values. However, I was told that for the MCAT we should use the old numbers of 3 and 2.

Lindyhopper said:

QofQ Thank you for your clear, well organized answer. I was wondering about your thoughts on texts, including TPR materials, that assume: that each NADH yields only 2.5 ATP and FADH2 yields only 1.5 ATP. The total would therefore, yield 30 ATP in euk. & 32 in prok. . Is this a significant discrepancy?
Also how significant is prok. anaoerbic respiration (as opposed to fermentation). My understanding is that in anaoerbic respiration, there is e- transport leading to the phosphorylation of ADP, but that some chemical other than O (often S) is the final electron acceptor.
Thanks

Click to expand...

This really is not required for the MCAT, but if you are interested in knowing, the debate is still going on.

It all boils down to the P/O ratio. The P/O ratio refers to the number of molecules of ATP formed in oxidative phosphorylation for every two electrons flowing through a defined segment of the electron transport chain. There is still some debate regarding this value.

10H+ are transported out of the matrix for every two electrons that pass from NADH to O2

4 H+ are transported into the matrix for every ATP synthesized
P/O = (1 ATP/4H+)*(10H+/NADH)*(1 NADH/2e') = 2.5 ATP/2e'

This is more or less close to the 3 ATP's for each NADH that enters oxidative phosphorylation quoted previously

QofQuimica said:

For all of the pre-health tests (MCAT, PCAT, DAT, and OAT), you should be able to follow the generation of ATP in each step, and also the energy carrier reduction (NAD and FAD) in each stage. You do NOT need to memorize any enzymes or pathway intermediates; they will make you do that in your professional school biochem class. You should also know that oxygen is the final electron acceptor of the electron transport chain, and that anaerobic respiration is insufficient to sustain human life. In addition, fermentation produces lactic acid as a byproduct in humans, and ethanol in yeast. Finally, you should know where in the cell each stage of respiration occurs. Here is a list of the energy conversions for each stage and where in the cell they take place:

Glycolysis: (anaerobic, occurs in the cytoplasm)

• 2 net ATP (4 total made, but 2 needed to complete this stage)

• 2 NADH produced (making 4 ATP in ETC for eukaryotes and 6 ATP for prokaryotes)

Fermentation: (anaerobic, occurs in the cytoplasm)

• 0 ATP; its main purpose is to reoxidize the NADH produced in glycolysis

Pyruvate Decarboxylation: (aerobic, occurs in the cytoplasm for prokaryotes, mitochondrial matrix for eukaryotes)

• 0 ATP produced

• 2 NADH produced (making 6 ATP in ETC)

TCA Cycle: (aerobic, occurs in the cytoplasm for prokaryotes, mitochondrial matrix for eukaryotes)

• 2 ATP produced

• 6 NADH produced (making 18 ATP in ETC)

• 2 FADH2 produced (making 4 ATP in ETC)

Electron Transport Chain (ETC): aerobic, occurs across the inner cell membrane for prokaryotes, inner mitochondrial membrane for eukaryotes

• NADH oxidation back to NAD and FADH2 oxidation back to FAD occur along with ATP production, allowing the earlier stages to continue

Summary: 36 net ATP produced in eukaryotes, 38 net ATP produced in prokaryotes (because the electrons from the NADH produced from pyruvate decarboxylation do not have to be transported across the mitochondrial membrane in prokaryotes; doing this causes a net loss of two ATP in eukaryotes)

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I was thinking about making index cards to memorize. Should I just study as to what goes in and what it ends up making. Ex: Krebs starts with Pyruvate---> ATP, etc... or should I put more detail. I'm still struggling as to how to best tackle memorize these.

I often see stated as a fact that triaclglycerol is the body energy storage and that it supplies the most ATP per carbon. TPR has a straight forward question reinforcing this "fact". When I do the math, I can't make it add up.

Using the simple conversion of 3 ATP per mitoch. NADH & 2 ATP per FADH2. Each glucose molecule yields 36 ATP or 6 ATP per C.
For every 2 Cs in the fatty acid chain we get 1 turn of the Krebs cycle. Each turn of the Krebs cycle yield 3 NADH, 1 GTP, 1 FADH2 for a total of 12 ATP. Or 6 ATP per C. The glycerol will also be converted to PGAL and enter pathway at a point that seems to also generate 6 ATP.

I can see that the C of triaclglycerol with many bonds to H and few to O is more reduced than the C of glucose, But why doesn't my math confirm this by yielding more ATP per C? Can ATP possibly be generated by the beta- oxidation of fatty acids to acetyl-CoA?

in response to lindyhoppers question regarding anaerobic respiration in prokaryotes, many bacteria create energy which works somewhat similarily to aerobic respiration. you seem to have the correct idea. in aerobic respiration, bacteria utilize oxygen as a terminal electron acceptor. in anerobic respiration, they use things like nitrate, sulfate, etc. it is not as efficient as aerobic respiration, however it is far more energy efficient than fermentation. just know that aerobic respiration, anaerobic respiration and fermentation are all separate things (and very basically, why). hope this helps a bit.

Hi Q of Q, can you please go over the repressible and inducible operons and can you also please go over the fungi stuff we need to know? Thank you

Hi-

I can try to answer your question regarding operons, but the fungi would best be left to someone else.

operons in prokaryotes consist of groups of genes that typically work together and are all controlled by one promoter, which is in turn controlled by an operator region. this means that all of the genes in the operon are turned on and off together.

(i wish i had the means to draw pictures, if you'd like pictures, let me know and i'll try to make some with powerpoint)

As far as inducible operons go, the first example that comes to mind is the lac operon. the genes in the operon, lac Z, Y, and A, code for proteins that DEGRADE/utilize lactose. these genes are preceeded by an operator region. the operator region is the "control region" of the operon. with the lac operon, in the absence of lactose, an inhibitor (protein) binds to the operator region DNA, and the operon is effectively turned off because the inhibitor interferes with RNA polymerase binding. when lactose is around, it binds the inhibitor and releases it from the operator DNA. once the repressing inhibitor is removed, RNA polymerase can bind and transcribe the operon (the Z,Y and A genes), and the operator is considered "induced" or turned on. if you look at this from the standpoint of the cell, it makes sense. the cell only needs the proteins that utilize lactose when lactose is around, right? so it's set up that the genes that utilize lactose are only "turned on" when there is sufficient lactose in the cell (ie. the cell doesn't waste energy making proteins it doesn't need).

For "repressible" operons, the tryptophan operon comes to mind. in this operon, the genes for tryptophan (trp) SYNTHESIS are grouped together. these genes are also preceeded by an operator region that controls expression of the operon. however, in this operon, there is a repressor that is only active (ie it will only bind the operator) when it is also bound to trp. this means that there has to be a relatively high level of trp in the cell to bind the repressor which will then turn off the genes for trp production. this also makes sense from the standpoint of the cell... if you need more trp, the genes will be turned on until there is enough trp in the cell. when there is enough trp in the cell, the genes will be turned off.

i hope this makes sense, it really helps if you have a picture to look at, but let me know if you have any other questions...

Could somebody clarify what the P, A, E sites are in the translation.
I got the P site is when the methione codon(initialtion codon) hooks up with the ribosome(rRNA and proteins) in the mRNA. A site is the same for the tRNA(anticodon) which pairs with the mRNA and the E site is the next codon after the start codon, when the actual elongation starts and the actual sequence begins.

I PMed one of the thread moderators of this sub-forum, and when this mod didn't know the answer I was told to post this in the biochem thread. I saw the disclaimer at the top saying not to post embryology questions here, so I hope this isn't in the wrong place.

When a zygote divides in the early stages to produce identical twins, is it possible for the split to be uneven? Also, I've read that often the splitting is caused by problems with the genes, and the zygote is trying to dispense of the bad genes to produce a more viable embryo. If one twin dies in-utero, and then has a lot of health problems later in life, could this be because of the split of the zygote? Because there were problems with the embryo to begin with, and nature tried to remedy that, but it didn't work out all the way and the embryo actually survived and was born?

I realize this isn't really MCAT related, and it's impossible to identify the cause of different health problems. I know someone who has been told he has bad genes by his doctor, while his siblings have no health problems at all. I was just curious if someone could point me towards some research in this area, or answer my question.

Thanks!

The ribosome sites are as follows:

A site is the "acceptance" site where incoming tRNA's go before their amino acid gets tacked onto the growing chain (it's like the on deck circle at a baseball game)

the P site is the "peptide" site where the amino acid gets tacked onto the chain. the growing amino acid chain elongates and stays here.

the E site is the last site, the "empty" site where the empty tRNA's are released from the ribosome.

make sense??

LT2 said:

The ribosome sites are as follows:

A site is the "acceptance" site where incoming tRNA's go before their amino acid gets tacked onto the growing chain (it's like the on deck circle at a baseball game)

the P site is the "peptide" site where the amino acid gets tacked onto the chain. the growing amino acid chain elongates and stays here.

the E site is the last site, the "empty" site where the empty tRNA's are released from the ribosome.

make sense??

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Thanks a lot
I didn't find the EK to be very clear on this.
empty tRNA's are the portion of the tRNA that is done or "use up" right? after that portion is translated.

yes, "empty" tRNA's are tRNA's that have no amino acids attached to them, so i suppose they could be considered "used up".

i'll try to do a blurb on transcription and translation sometime soon...

LT2 said:

yes, "empty" tRNA's are tRNA's that have no amino acids attached to them, so i suppose they could be considered "used up".

i'll try to do a blurb on transcription and translation sometime soon...

Click to expand...

Thanks.
If somebody could do the same for the cellular respiration, because it doesn't seem like we have to know that much about the krebs cycle with the exception of Acetyl CoA, pyruvate, and NADH. Also the electron transport chain with the ATP synthase with the proton motive force.

LT2, thanks for that nice summary on "operators" control of the operon. I know it is hard to say ALWAYS in biology (Just too much damn diversity) but is it generally true that:
Activation of an inducible operon, like the lac operon, is induced by the release of the repressor from its OPERATOR, while activation of a repressible operon, like trp, is repressed by the binding of a repressor to its operator.

RE the lac operon, I would like to point out, that there is also an activator that binds, not at the operator site, but just upstream of the promoter at the CAP binding site. CAP the "catabolite activator protein" is activated by cAMP.
low glucose levels leading to high cAMP levels, and therefore, the activation & binding of CAP.
Therefore, if glu levels are high, & lac are low, the CAP activator does not bing and the repressor stays bound to the operator resulting in no transcription.
If Glu & lactose levels are both high, the CAP activator does not bind, but the repressor is freed from the operator resulting in low levels of transcription.
If Glu level are low but Lac is high - The low Glu levels result in high levels of cAMP resulting in the activation of CAP. The lactose will induce the release of the repressor from the OPERATOR resulting in high levels of transcription.

Can I get some help with the immune system? T cell mediated vs. B-cell mediated methods of fighting off pathogens. Primary vs. secondary action, and to what extent we have to know this stuff for the MCAT. It seems that T-cell is for extracellular pathogens, while B is for intracellular, while their methods are very similar. Just need some boning up on my immunology.

B cells make antibodies, not T cells!!

For Humoral Immunity:
Antibodies (also known as Immunoglobulins) neutralize toxins, bacteria, etc (in a matter of speaking). They have two heavy chains and two light chains. within these chains are constant regions and a variable regions. the constant region defines the "type" of antibody (IgM, IgA, IgD, IgG, and IgE), and the variable region is specific for epitopes of the toxin, bacteria, etc. B cells are derived from bone marrow and they produce antibodies when they encounter an antigen. this is the primary response (and is relatively slow 7-10 days). The original proliferation of B cells in response to the antigen become memory cells. when the body encounters the antigen again, these memory cells activate and produce antibodies quickly. this is the secondary response.

Cell-mediated Immunity:
This type of immunity is regulated by T cells (which also come from bone marrow). Cell mediated immunity is typically used in response to viral infections. There are a couple of types of T cells, cytotoxic T cells, suppressor T cells and helper T cells (these help activate the B cells mentioned above). T cells secrete interferons and cytokines that help deal with infection. T cells are also responsible for inflammatory response.

Hope this helps...

LT2 said:

B cells make antibodies, not T cells!!

For Humoral Immunity:
Antibodies (also known as Immunoglobulins) neutralize toxins, bacteria, etc (in a matter of speaking). They have two heavy chains and two light chains. within these chains are constant regions and a variable regions. the constant region defines the "type" of antibody (IgM, IgA, IgD, IgG, and IgE), and the variable region is specific for epitopes of the toxin, bacteria, etc. B cells are derived from bone marrow and they produce antibodies when they encounter an antigen. this is the primary response (and is relatively slow 7-10 days). The original proliferation of B cells in response to the antigen become memory cells. when the body encounters the antigen again, these memory cells activate and produce antibodies quickly. this is the secondary response.

Cell-mediated Immunity:
This type of immunity is regulated by T cells (which also come from bone marrow). Cell mediated immunity is typically used in response to viral infections. There are a couple of types of T cells, cytotoxic T cells, suppressor T cells and helper T cells (these help activate the B cells mentioned above). T cells secrete interferons and cytokines that help deal with infection. T cells are also responsible for inflammatory response.

Hope this helps...

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Is cell mediated immunity specific in its action? I know that humoral immunity is highly specific due to antibodies but it's confusing me if this level of specificity is involved with T cells.

Cell mediated immunity is non-specific in that the products secreted (interferons, etc) will kill what ever is in the vicinity. They are also somewhat involved with humoral immunity in that they activate B cells to start making antibodies. so they are technically non-specific, but they dabble in both modes of immunity.

hope that helps...

LT2 said:

Cell mediated immunity is non-specific in that the products secreted (interferons, etc) will kill what ever is in the vicinity. They are also somewhat involved with humoral immunity in that they activate B cells to start making antibodies. so they are technically non-specific, but they dabble in both modes of immunity.

hope that helps...

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But they have proteins on their membranes that have to match to the antigen don't they?

yes, the t cells recognize antigens (specifically) but the response (ie IFN's etc) are not specific.

I am confused as to how I should understand the ribosomes.These are the two ways that I am thinking of them just tell me which is correct.

1. The major components(large, small subunits and the rRNA) are made in the nucleolus which get "activated" in the cytosol by the substances contained in the cytosol and do their stuff eventually "depositing" the synthesized protein into the rough ER.

2. The rRNA are made within the nucleolus and move to the cytosol which combines with the subunits and get activated in the cytosol, and do the same as above. The rRNA acts as a key "turning on" the ribosome.

Which one is the closest representation of the Ribosomes?

BTW I am beginning to understand why membranes are key in Eukaryotes aside from being the distinguishing feature from Prokaryotes.

Just wondering if anyone had advice on who best covers molecular biology --Kaplan, TPR, EK -- because I know the MCAT is increasingly slanted that way and I didn't know if the courses had changed their material accordingly.

Thanks.

blankguy said:

I am confused as to how I should understand the ribosomes.These are the two ways that I am thinking of them just tell me which is correct.

1. The major components(large, small subunits and the rRNA) are made in the nucleolus which get "activated" in the cytosol by the substances contained in the cytosol and do their stuff eventually "depositing" the synthesized protein into the rough ER.

2. The rRNA are made within the nucleolus and move to the cytosol which combines with the subunits and get activated in the cytosol, and do the same as above. The rRNA acts as a key "turning on" the ribosome.

Which one is the closest representation of the Ribosomes?

BTW I am beginning to understand why membranes are key in Eukaryotes aside from being the distinguishing feature from Prokaryotes.

Click to expand...

The ribosomal subunits are synthesized in nucleolous and assembled when they attach to mRNA. They can be found both in cytosol and ER. The ribosomes in the cytosol make proteins to be used in the cell and the other to be exported. Hope that helped.

Nutmeg said:

Not to be a thread Nazi, but perhaps issues of blood vessels would be more appropriate for the organismal biology thread...

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I agree. I'm moving those posts to the organismal bio thread: http://forums.studentdoctor.net/showthread.php?t=207484 Posters, please post about systemic level biology in that thread, and only post about cellular/molecular level biology here.

myfavred said:

The ribosomal subunits are synthesized in nucleolous and assembled when they attach to mRNA. They can be found both in cytosol and ER. The ribosomes in the cytosol make proteins to be used in the cell and the other to be exported. . .

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I wanted to elaborate on "myfavred" nice concise statement. Translation begins on a free floating ribosome in the cytosol. If the nascent polypetide expresses a hydrophobic SIGNAL SEQUENCE in the first 16-30 amino acids, a signal recognition particle (SRP) will carry the entire complex to the ER. The growing protein will either be inserted into the ER lumen or threaded through the membrane of the ER.
What peptides have the SIGNAL SEQUENCE, & thus will be translated at the ER?
Proteins destined to be secreted, or end up in the "secretory pathway" will have the SIGNAL SEQUENCE. These include polypeptides that will eventually be in the lumen, or membranes of the ER, Golgi, lysomes, plasma membrane, as well as all secreted proteins.

What polypeptides will be translated in the cytosol because they lack the hydrophobic signal sequence?
Pretty much all the rest, including, proteins that will "live" in the nucleus, cytolsol, & mitochondria.

Nitya2284 said:

. . . I don't think you're responsible for the intricate details regarding the signal sequence and polypeptide recognition.

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I'm not sure just what level will ultimately be tested, but, this level of detail is taught by TPR & a similiar level is presented in EK's book.
I think understanding the start of the secretory pathway is a pretty fundamental step in cell bio.

Nitya2284 said:

. . . They can be either found free in the cytoplasm or attached to the ER which is then called the RER. . .

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A good point to keep clear is that translation begins on ribosomes in the cytoplasm. It is only if the nascent polypeptide is destined to enter the secretory pathway will the ribosome/mRNA complex be taken to and become attached to the ER.
Your summary was simple & good, but at TPR (where I teach MCAT bio) we stress the above point.

Please excuse me if the material is slightly long or beyond the scope of MCAT. But, I think, this should summarize most of eukaryotic transcription. For the purposes of this post, I'm only going to summarize RNA pol II (mRNA synthesis) mediated transcription. RNA pol I regulates rRNA synthesis and Pol III regulates 5s RNA and tRNA synthesis.

The mechanism of gene transcription by RNA pol II follows 3 general steps, initiation, elongation and termination. These 3 steps are followed by RNA processing.
Initiation
In eukaryotes initiation is regulated by the presence of regulatory regions, promoters and enhancers. Common promoter elements are the TATA and CCAAT boxes, found upstream of the transcription start site. Enhancers can be found upstream, downstream, or within the coding region
Promoters are recognized by basal transcription factors and are necessary for initiating transcription, while enhancers, as the name suggests are necessary for enhancing transcription and also for regulating and mediating cell and tissue specific transcription.
The basal transcription factors (TFs) such as TFIID, along with other TFs, recruit RNA pol II to the promoter element and initiate basal transcription.
Other enhancer elements and TFs mediate higher levels of transcription.
Elongation
During elongation, RNA pol II moves along the DNA, close to the bubble that represents separation of the two strands of DNA. As the enzyme moves forward along the bubble, RNA is synthesized in the 5’ to 3’ direction. DNA ahead of the bubble is unwound and behind it is rewound. Elontation continues until the enzyme reaches a termination point.
Termination
If I may, termination in eukaryotic genes is not very specific. Pol II continues to transcribe RNA for a few thousand (1000-2000) bases past the end of the mature mRNA. The exact end is determined during RNA processing.
Processing
RNA processing is characterized by capping at the 5’ end, polyadenylation at the 3’ end and intron splicing.
5' Capping
A methylated guanine nucleotide is added to the 5’ end of the mRNA in a 5’ to 5’ phosphodiester linkage. This capping is essential for mRNA recognition by ribosomes during translation.
3' Polyadenylation
A polyadenylation signal (AUAAA) is present in most of the mRNA transcripts and this signal is reconized by an enzyme that cleaves the transcript about 20 nucleotides downstream and adds a series of As (~200) to the 3’ end. These As are added without the need for a template and prevent the mRNA from degradation.
Splicing
Removal of the introns from the pre-mRNA to yield mature mRNA is called splicing. Splicing is carried out by spliceosomes that contain at least 5 known small nucleotide ribonucleoproteins (snRNPs). These snRNPs contain small nuclear RNAs (snRNAs) and together they detect intron/exon boundaries and cleave the RNA at those specific junctions. The spliced RNA is then joined together to form the mature mRNA transcript.

Salient points of eukaryotic mRNA transcription:
1) occurs in 5’ to 3’ direction
2) mRNA synthesis regulated by RNA pol II
3) mRNA synthesis involves initiation, elongation and termination followed by processing to make the mature transcript.
4) Initiation is mediated by promoters and enhancers, and elongation by the RNA pol II. Termination in eukaryotes occurs way downstream and is not very specific, unlike in prokaryotes.
5) mRNA processing to produce the mature trancript involves 5’ capping, 3’ polyadenylation and intron splicing.

Travelbug that's definately the post of the week!
I've always been a little cloudy on enhancers. I assume that enhancers are sequences of DNA. To enhance or lessen transcription, must a protein bind to the enhancer? Are these proteins called "enhancer elements"?
In the case of enhancers thousands of bp from the promoter what possible mechanism can enhance the binding of TFs? Does the chromatin bend to put the enhancer elements in physical contact with the basal TFs?
In the case of enhancers within the promoter- How are they different from sequence specific TFs?

Any guidance would be welcome.

Reply to Lindyhopper

I assume that enhancers are sequences of DNA. To enhance or lessen transcription, must a protein bind to the enhancer? Are these proteins called "enhancer elements"?

Yes, enhancers are sequences of DNA that enhancer binding proteins or enhancer elements bind to. In addition, it is possible (and more often the case) that these enhancer binding proteins also bind to TFs present in the promoter region, thereby creating a loop, like you said.

In the case of enhancers thousands of bp from the promoter what possible mechanism can enhance the binding of TFs? Does the chromatin bend to put the enhancer elements in physical contact with the basal TFs?

Look above

In the case of enhancers within the promoter- How are they different from sequence specific TFs?

Enhancers within or outside of promoter region are not TFs per se but require the binding of TFs. The role of enhancers is to stimulate transcription, often in differentiated cell or tissue types, thereby making it cell or tissue specific transcription.

I hope I have answered your questions, if not, please feel free to ask.

Thank you

Please feel free to add if I may have missed anything or correct if I'm even partially wrong.

tRNA molecule
• Single chain, contains 73-93 ribonucleotides
• Contains many unusual bases such as inosine, pseudouridine
• tRNA is L shaped
• 5’ end is phosphorylated
• 3’ end ends in CCA and contains the amino acid attachment, it is at one end of the L
• The other end of the L, far from the amino acid end, is the anticodon loop


The process of translation, like transcription, is also divided into three phases:
initiation, elongation and termination. These three phases are regulated by initiation, elongation and termination factors respectively.

Initiation

Initiator tRNA (tRNAi) that carries methionine is the only tRNA capable of initiating translation. An initiation complex called 43S, comprising methionine tRNAi, the small 40S ribosomal subunit, and initiation factors such as eIF2. The 43S complex is recruited to the 5’ end of the mRNA by eIF4E. This complex now scans the mRNA in the 5’ to 3’ direction to find the first 5’-AUG-3’. Scanning is an ATP dependent process. As soon as the met-tRNAi finds the first AUG, the larger ribosomal subunit is recruited and this recruitment is mediated by eIF5. Assembly of the large ribosomal subunit completes the initiation step. The large subunit has 3 binding sites, E, P and A and the first codon (AUG) is aligned at the P site.

Elongation

Elongation begins with the delivery of an amino-acyl tRNA (corresponding to the appropriate codon on the mRNA) to the A site on the ribosome by EF-Tu and this is followed by GTP hydrolysis. A peptide bond, catalyzed by peptidyl transferase, is formed between methionine and the aminoacyl tRNA by the transfer of methionine to the A site, leaving the deacylated tRNA at the P site. The next step of elongation is translocation, where, the deacylated tRNA moves to the E site, the dipeptidyl-tRNA (met + aminoacyl tRNA) moves to the P site and the mRNA moves forward by 3 bases, thereby aligning the next codon for the appropriate aminoacyl tRNA. Translocation is mediated by elongation factor G. A and E sites cannot be occupied at the same time, therefore, as soon as the A site is occupied, the E site containing the deacylated tRNA is emptied. Elongation proceeds in this fashion until a stop codon is encountered.

Termination

Normally, tRNAs do not have anticodons corresponding to the stop codons (UAA, UAG or UGA. At termination the polypeptide chain is at the P site and the stop codon is at the A site. Stop codons are recognized by proteins called release factors (RFs) or termination factors. Peptidyl transferase is activated when an RF binds to a termination codon at the A site. The activated peptidyl transferase hydrolyzes the bonds between the polypeptide and the tRNA at the P site. The released polypeptide chain, tRNA and mRNA leave the ribosome in that order. The ribosome dissociates into its subunits ready for another round of protein synthesis.

Summarizing eukaryotic protein translation

• mRNA is always translated in the 5’ to 3’ direction
• proteins are synthesized in the amino to carboxyl direction
• several ribosomes can simultaneously translate an mRNA molecule and such an mRNA molecule (with many ribosomes attached) is called a polysome or a polyribosome
• amino acids are added sequentially to the carboxyl end of a polypeptide chain
• aminoacyl tRNAs are the activated precursors in which the carboxyl group of an amino acid is attached to the 3’ hydroxyl group of a tRNA
• the above step is catalyzed by an aminoacyl tRNA synthetase and is driven by ATP
• initiator tRNA, met-tRNAi, occupies peptidyl (P) site, the next aminoacyl tRNA, added during elongation, occupies the aminoacyl (A) site
• peptide bond is formed between carboxyl group of met and aminoacyl tRNA
• dipeptidyl tRNA moves from A to P site
• deacylated tRNAi moves to E (exit) site and leaves ribosome
• a new aminoacyl tRNA occupies A site
• elongation proceeds until stop codon is encountered
• stop codon (UGA, UAA, or UAG) recognized by release factors that facilitate release of the completed polypeptide from the ribosome

Nondisjunction the failure of homologous chromsomes to successfully seperate results in diseases such as trisomy 21 (Down's Syndrome) & Turner Syndrome (XO).
When does this failure take place? There would seem to be two possible candidates. In Anaphase I it seems quite possible that the homologous chromosomes could get tangled, resulting in the imperfect seperation.
In Anaphase II when the sister chromatid are seperated, it seems possible that an inappropriate split at the centromere could result in both sister chromatids going to same gamete.

For the MCAT, is it necessary to know what intermediates the carbhohydrates, fats and proteins are converted into and where they enter the Krebs cycle?

Lindyhopper said:

Nondisjunction the failure of homologous chromsomes to successfully seperate results in diseases such as trisomy 21 (Down's Syndrome) & Turner Syndrome (XO).
When does this failure take place? There would seem to be two possible candidates. In Anaphase I it seems quite possible that the homologous chromosomes could get tangled, resulting in the imperfect seperation.
In Anaphase II when the sister chromatid are seperated, it seems possible that an inappropriate split at the centromere could result in both sister chromatids going to same gamete.

Click to expand...

Nondisjunction could occur during mitosis, meiosis I or meiosis II. However, monosomies and trisomies are more common because of nondisjunction during meiosis than mitosis and more so during meiosis I than meiosis II.

Hi,
In incomplete dominance, the dominant allele is only partially expressed. The classic example is the red rose being incompletely dominant over the white rose resulting in pink offspring.
In codominant alleles both alleles are expressed. One common example is the ABO blood types.
The other common example of codominance, fur color, is less clearly different than incomplete dominance. If one expresses both co-dominant allele for say brown & red fur the result is sort of a "blend". Does anyone know how on a physical level this "co-dominant" expression is different from the incomplete dominance seen in the "blended" pink rose. In the co-dominant fur, are alternate hairs brown & red? Or perhaps, are both pigments translated in each hair but our eyes see a blend? Or something else?

Hi, Could someone explain reciprocal cross? Also, explanations & tips with reading pedigrees would be really helpful! Thanks!

Hi, maybe someone can clarify this.
-A mountain climber living at sea level ascends to a very high altitude during the course of a day long climb. By the end of the day all of the following acclimazations will occur EXCEPT:

A- increased tidal volume
B- increased respiration rate
C- right shift of hemoglobin dissociation curve
D- increased concentration of erythropoietin in the blood.

The answer they give is C

However I picked D. At the end of the climb 2,3 DPG should be increased thus shifting the curve to the right. The purpose of 2,3 DPG is to allow for a better deliver of oxygen to the muscles until erythropoeitin can increase which takes a few days. It has been a while since i delt with biochem so if someone can help out this would be appreciated.

(Im hoping that Kaplan made a mistake so I get a 12 instead of an 11 haha)

http://us.commercial.lifefitness.com/content.cfm/exerciseataltitude
"Within 24 to 48 hours after exposure to altitude, the hormone erythropoietin is secreted from the kidneys. This hormone stimulates the bones to increase red blood cell production, thereby enabling the blood to carry more oxygen. At moderate altitudes of 7,000 feet, this acclimatization may take about two weeks. For higher altitudes, acclimatization will take several weeks."

D is wrong b/c the question states a one day hike which is a 12 hour max event.

As far as C goes, if the curve is shifted towards the right, this means that there would be less hemoglobin affinity for oxygen. This would make sense b/c it means that hemoglobin would give up O2 to the tissue at a greater rate.

gotgame83 said:

Hi, maybe someone can clarify this.
-A mountain climber living at sea level ascends to a very high altitude during the course of a day long climb. By the end of the day all of the following acclimazations will occur EXCEPT:

A- increased tidal volume
B- increased respiration rate
C- right shift of hemoglobin dissociation curve
D- increased concentration of erythropoietin in the blood.

The answer they give is C

However I picked D. At the end of the climb 2,3 DPG should be increased thus shifting the curve to the right. The purpose of 2,3 DPG is to allow for a better deliver of oxygen to the muscles until erythropoeitin can increase which takes a few days. It has been a while since i delt with biochem so if someone can help out this would be appreciated.

(Im hoping that Kaplan made a mistake so I get a 12 instead of an 11 haha)

Click to expand...

Thank you for responding but it says EXCEPT lol. Therefore a one day climb ISNT enough time to make D true... which is my reasoning.