C2006/F2402 '02 -- Answers to Review Questions for Exam #2  -- Last Update: 03/08/02 11:07 AM

For all questions, each choice = 2 pts; each explanation = 2 pts unless otherwise indicated. 

 1.  A. Both; either way. Both genes should have have enhancers. Enhancers are sequences of DNA and the DNA is the same in all cells. What changes is not the enhancers but the transcription factors that bind to them.  The enhancers are different from each other and bind different sets of transcription factors. The TF's for gene A are found in muscle cells; the TF's for gene B are not. (They are found in some other cell type.) Enhancers can be far away from the genes they control and can be either upstream or downstream from the start of transcription. 

    B. Gene A; less sensitive. Hypersensitive sites = regulatory regions that are currently binding transcription factors = regulatory regions of active genes. The TF's displace the histones and make the DNA more accessible to nucleases than DNA associated with histones (in nucleosomes). The DNA in the hypersensitive site should be more sensitive than DNA in nucleosomes but less sensitive than naked DNA, which has no protein at all to protect it. 

    C. Both should be about the same; 10. Transcribed regions and untranscribed regions both contain DNA associated with nucleosomes; only the regulatory regions lack significant numbers of nucleosomes. The nucleosomes of the transcribed regions may lack H1, but they contain all the histones of the inner octamer -- 2 molecules each of H2A, H2B, H3 and H4 per nucleosome. So there should be the same amount of H2A in the transcribed regions of both genes (whether they are being transcribed or not). There should be 2 H2A molecules per 200 BP of DNA = 1 per 100 or 10 per 1000.


2. A. Nucleolus. That's where rRNA is made and ribosomal subunits assembled.

    B. All of these. (3 pts) None of these should enter the endomembrane system or be made in mitochondria. (Only proteins encoded in mitochondria are made there.)

For part C, first answer 2 pts, same or different 1 pt; explanation 2 pts. 

    C-1. NLS; same. Protein F functions in the nucleus. Therefore it should be made in the cytoplasm with an NLS to allow it to be transported through nuclear pores into the nucleus. The protein will remain in the nucleus during interphase but it will be released when the nuclear membrane is disassembled during mitosis, and it must be returned to the nucleus when the nuclear membrane reforms after mitosis.  Therefore it likely that the NLS will not be removed when the protein first enters the nucleus -- the NLS will remain so the protein can re-enter the nucleus after division.  

    C-2. NLS; different. Protein G is made and functions in the cytoplasm, but it must travel into the nucleus (and back out) before it can function. Ribosomal proteins are made in the cytoplasm but the proteins fold up with rRNA and assemble into subunits within the nucleus. So the protein must be made in the cytoplasm, go to the nucleus, meet up with an rRNA and form a subunit, and return to the cytoplasm. Once the subunit reaches the cytoplasm it must stay there, as there are no functioning ribosomes in the nucleus. Protein G needs an NLS to enter the nucleus and find its rRNA partner, but once the protein returns to the cytoplasm, it should have its NLS removed or masked so it can remain permanently in the cytoplasm. 

    C-3. Transit peptide; different. Protein H needs a TP to cross both the outer and inner membranes at a contact point (where the two membranes come together) and reach the matrix. Once it enters the matrix, the TP is generally removed (to prevent exit?). The protein does not need an SP since it remains in the matrix and does not enter or cross the inner mito membrane. (Mito. ribosomes are assembled in the matrix from rRNA encoded and made in mito DNA and proteins encoded in nuclear DNA and made in the cytoplasm.)

3. A. Binding of ATP; Alpha. Adenylyl cyclase (AC) catalyzes ATP  à    cAMP.  Of the choices given, the only way to block this reaction would be to prevent binding of the ATP. When a trimeric G protein is activated, it separates into alpha and beta + gamma. Either half can bind to and affect the next protein in the signaling chain. So when G1 is activated, either alpha or beta + gamma must bind to AC and prevent ATP binding. The alpha subunit is the only one that is different in the 2 G proteins, so it must be responsible for the action on AC. 

    B. Binding of GTP; hydrolysis of GTP.  All G proteins are activated by GTP/GDP exchange = release of GDP and binding of GTP in its place. The proteins are then deactivated after a short period by hydrolysis of the GTP to GDP. G1 hydrolyzes the GTP more slowly, so its effects last longer. The ability to bind GTP sets the sensitivity of system -- it determines how much signal is required to activate the G protein. The rate of hydrolysis sets the timing of the system -- it determines how long the G protein will stay activated in response to one dose of signal.


4. A. Activity of individual PKA and # of cAMP molecules. Ligand L binds to receptor forming a complexà    complex binds to and activates G2 à    activated G2 binds to and activates AC à    AC catalyzes conversion of ATP to  cAMP (so you get more molecules of cAMP) à    cAMP  binds to PKA and activates it (no phosphorylation or change in number of molecules) à    activated PKA phosphorylates enzyme Z à    lower enzyme Z activity. (Note that AC itself, and PKA, are not phosphorylated. Binding of activated G protein and cAMP respectively are what activate AC and PKA, not phosphorylation. Once PKA is activated, then the phosphorylations start.)

    B. (answers 1 pt each; 4 pts for explanation.) Higher; lower. Enzyme Z activity will be higher; PKA activity will be lower. Phosphodiesterase (PDE) degrades cAMP; Ca++ activates PDE. Therefore higher Ca++  à     higher PDE activity à   hydrolysis of cAMP to AMP (not ATP) à   lower levels of cAMP à   PKA is less active [PKA-cAMP complex (active) <-->  cAMP + PKA (inactive) shifts to right]  à   phosphorylation of enzyme Z is reduced and activity of enzyme Z is increased.


5. A. sweating and weight loss (1 pt).

    B. After eating the thyroid gland, it would be broken down in the digestive tract.  Thyroglobulin (1), a protein, is broken down by digestive enzymes in the intestines to the individual amino acids.  This separates the thyroxine (T3/T4), which is really iodinated tyrosine at this point, from the thyroglobulin molecule. (1)  Thyroxine is lipid-soluble(1) so would be absorbed directly through the apical membrane of the intestinal epithelial cells, across the basolateral membrane, to the blood.(1)

    C. No.(1) TSH is a peptide, so would be broken down by intestinal enzymes before absorption, and couldn't diffuse directly through intestinal wall into blood.(3)

    D. (2 points for any two of these explanations) 

Negative feedback is not relevant here, since this high level of T4 is not being secreted from the thyroid gland, so a decrease in TRH/TSH would not affect it.  The high T4 comes from ingested hormone.

Thyroxine has negative feedback effects (2) on the hypothalamus, to decrease TRH secretion (which could lead to decreased TSH)    (1)  and on the anterior pituitary gland, to decrease TSH secretion (1)