homework - What is the main general difference between Mitosis and Meiosis?

I found such a clause:

The general principle is that mitosis creates somatic cells and
meiosis creates germ cells.

However, I cannot agree. Each gametogonium needs to go through mitosis before it can enter meiosis I. So in that case mitosis is happening with germ cells so the clause is false.

I would rephrase the sentence to be

The general principle is that meiosis creates only germ cells with the possibility of a decrease in chromosome number, while mitosis can create both somatic and germ cells while the ploidy stays constant.

Ok, not perfect.

How would you say the main general difference between mitosis and meiosis?

immunology - How does herpes (HSV) infection suppress HIV?

Alright, having read the citation linked, and doing a little poking of my own, here's my approach at an answer:

Some human herpes virus infections may compete with HIV infection. Essentially, some strains (not the ones you normally think of) infect CD4 cells - the same cells targeted by HIV. These strains down regulate transcription in CD4 cells, which in turn interferes with the HIV infection process. This pertains, it appears most notably, to HHV-7.

However the actual impact on HIV disease isn't clear. Strain competition triggers some fascinating evolutionary pressures, but HIV is notoriously prone to mutation, and competition for CD4 cells might not impact HIV infection on a clinical - rather than microbiological - scale.

Additionally, the two most commonly thought of forms of herpesvirus infection, HSV-1 and HSV-2 are associated with increased acquisition of HIV infection. The clearest reasons for this are genital lesions and inflammation at the site of HIV infection. There's also some interesting dynamics in play for active coinfection, such as the impact of acyclovir treatment for HSV impacting HIV, or HAART treatment for HIV impacting HSV.

genetics - Pedigree Probability of Autosomal Recessive Trait

Starting with the left hand side of the diagram:

• III:2 is definitely a carrier (Tt) as one parent (II:2) is affected (tt).


• III:1 is also definitely a carrier (Tt) as when mating with III:2 they produce an affected (tt) offspring (IV:1)


• This means that we can work out the possibilities for IV:4 as we know the parent genotypes. It follows the standard arrangement for two carrier parents giving the options of:
  
  • TT (1/4)
  
  
  • Tt (2/4 = 1/2)
  
  
  • tt (Normally 1/4 but in this case 0 as individual not marked as affected).
  
  


• Therefore for this scenario, the probabilities for IV:4 are :
  
  • TT (1/3)
  
  
  • Tt (2/3)
  
  

Now if we look at the right hand side of the diagram.

• IV:5 is definitely a carrier (Tt) as one of their parents (III:5) is affected.

This gives two possible Punnett squares to be examined:

|-------------------------------------------------------------------------------|
|                                        ♂ (IV:5)                               |
|                          T                                   t                |
|            -------------------------------------------------------------------|
|           |                                |                                  |
|         T |              TT                |                Tt                |
|           |                                |                                  |
| (IV:4)    |-------------------------------------------------------------------|
|    ♀      |                                |                                  |
|         T |              TT                |                Tt                |
|           |                                |                                  |
|-----------|-------------------------------------------------------------------|

This gives nil affected offspring so we can disregard this option for your question (as we are ONLY looking for scenarios which produce affected individuals).

Therefore the alternative is:

|-------------------------------------------------------------------------------|
|                                        ♂ (IV:5)                               |
|                          T                                   t                |
|            -------------------------------------------------------------------|
|           |                                |                                  |
|         T |              TT                |                Tt                |
|           |                                |                                  |
| (IV:4)    |-------------------------------------------------------------------|
|    ♀      |                                |                                  |
|         t |              TT                |                tt                |
|           |                                |                                  |
|-----------|-------------------------------------------------------------------|

Giving 1/4 affected offspring.

As mentioned above, in order to have affected offspring then IV:4 must be Tt. There is a 2/3 chance of this being the case. If this is the case, then there is a 1/4 chance of the child being tt.

Both conditions need to be true for this to happen so we multiply the fractions:

2/3 * 1/4 = 1/6

molecular biology - PDB Mining: Why Do I Find Atoms Less than 1 Angstrom Apart?

I am attempting to find potential Hydrogen bonds between Hydrogen donors and aromatic ring acceptors. I do this by predicting the location of Hydrogens on residues and then calculating how far these Hydrogens are from aromatic rings. If a certain Hydrogen is <7.0 Angstroms from a certain aromatic ring, then I take it under consideration: I form the N-H vector, which is the vector created by the Hydrogen under question and the Nitrogen in the backbone of the residue that the Hydrogen belongs to. I test that this N-H vector is pointing toward the plane of the aromatic ring, and I also test that the point of intersection between the plane of the aromatic and the N-H vector is within 6 Angstroms of the center of the aromatic ring.

If all of these conditions are met, then I consider it a Hydrogen bond between the Hydrogen and the aromatic ring. However, my data must be incorrect, because I am seeing situations where a Hydrogen is < 1.0 Angstrom from the plane of the aromatic. Atoms should not be getting that close to each other.

I thoroughly tested my method by hand using an example situation where my code identified one of the sidechain Hydrogens on an ASN is 0.3 Angstroms from the plane of the aromatic of a TRP. Unfortunately, I could not find any bugs. You can find a PDF of this verification here.

Any suggestions on how my method might be flawed would be greatly appreciated.

terminology - Calculating Protein Concentration from Kilo Units (KU)

The figure of 350 - 600 Units per mg refers to the specific activity of the enzyme.

The Unit is International Unit or IU and is usually defined as that amount of enzyme that will catalyze the transformation of 1 micromole of substrate (or product) per min, under defined assay conditions (such as pH, temperature, substrate concentration, presence of Mg⁺⁺, etc). It is thus a measure of activity.

When the enzyme is pure (no other extraeneous proteins present), the specific activity provides important information about the catalytic capacity of the enzyme.

It is usually calculated by measuring

• the activity of the enzyme preparation under defined assay conditions


• the protein concentration of the same enzyme preparation (using, say, the Lowry or Biuret method for protein estimation).

  Alternatively, if the E(1%, 280) is known (see below) and the enzyme is pure, measurement of the absorbance at 280 nm gives a very good estimate of protein content (and the enzyme may be recovered 'unharmed' at the end of the measurement).

Thus, taking a figure of 450 Units/mg for the specific activity of pyruvate kinase,
25 KU (25 Kilo-Units, I presume) contains 500/9 mg (~55 mg) protein.

I notice that the Sigma product sheet provides a figure for E(0.1%, 280) = 0.54.

This means that a 1 mg/ml solution of the protein will have an absorbance at 280 nm of 0.54

E(0.1%, 280) can be used as a very convenient measure of the protein content provided that the enzyme preparation supplied by Sigma is pure.

A 'rule of thumb', useful when the E(0.1%, 280) is unknown, is that a 1mg/ml protein solution has an A₂₈₀ of 1.

Thus if, say, the A₂₈₀ (absorbance at 280 nm) of the resuspended lyophilized powder is 1.08 and you have 5 ml of this, the protein concentration is 2mg/ml and you have 10 mg of protein in total. You may wish to assay the enzyme yourself to determine an accurate specific activity.

The EC (Enzyme Commission) Number may also be of interest. For pyruvate kinase (EC 2.7.1.40) see here.

For a great ref on PK (pdf may be downloaded) see here (Ainsworth et al.)

---

For your second question, I do not have access to that paper from home.

However, if calmodulin has a specific activity of 40 000 Units/mg,

• 25 000 Units is equivalent to 0.625 mg; this is in a volume of
  0.5 ml. Therefore, the calmodulin concentration is 1.25mg/ml.



• Taking the molecular weight of calmodulin to be 16 000,
  then 16 000 mg /ml (theoretical) would be a 1 Molar solution.
  Thus a 1.25 mg/ml solution is about 78 micromolar.

immunology - Harmless virus? - Biology

It is possible for viruses to live in mutualistic relationships with their hosts, these associations are often overlooked due to the devastating effect that many viruses can have.

To give an example in humans, when HIV-1-infected patients are also infected with hepatitis G virus, progression to AIDS is slowed significantly (Heringlake et al., 1998; Tillmann et al., 2001). Also hepatitis A infection can surpress hepatitis C infection (Deterding et al., 2006).

There are many other notable examples within plants, fungi, insects, and other animals, reviewed by Shen (2009), and Roossinck (2011), in two excellent papers.

The table below, summarises some beneficial viruses across all organisms, and is taken from Roossinck (2011).

References

• Deterding, K. et al., 2006. Hepatitis A virus infection suppresses hepatitis C virus replication and may lead to clearance of HCV. Journal of Hepatology, 45(6), pp.770-778.


• Heringlake, S. et al., 1998. GB Virus C/Hepatitis G Virus Infection: A Favorable Prognostic Factor in Human Immunodeficiency Virus-Infected Patients? Journal of Infectious Diseases, 177(6), pp.1723 -1726.


• Roossinck, M.J., 2011. The good viruses: viral mutualistic symbioses. Nature Reviews Microbiology, 9(2), pp.99–108.


• Shen, H.-H., 2009. The challenge of discovering beneficial viruses. Journal of Medical Microbiology, 58(4), pp.531 -532.


• Tillmann, H.L. et al., 2001. Infection with GB Virus C and Reduced Mortality among HIV-Infected Patients. New England Journal of Medicine, 345(10), pp.715-724.

genetics - T7 promoter leakiness

Can a gene be expressed under the T7 promoter in an E. coli strain (e.g. DH5 alpha), which does not have the T7 polymerase gene encoded in its genome? In other words, is T7 promoter leaky?

To be more specific, how is it possible that a regular E. coli strain, which does not encode for the T7 polymerase, can grow on kan selective media if it was transformed with a plasmid that has the kanR gene under T7 promoter?