Chem 451

## Biochemical Toxicology

Spring 2010

Lecture Notes:: 16 February

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## Chronic Toxicity

Chronic Toxicity can be quantified in similar ways and can reveal evidence of accumulation and therefore estimates of body t1/2. For chronic toxicity measure TD50 or LD50 for a specific time period, such as 90 days, and compare to the same total dose delivered in a single dose. Determine the chronicity factor.

Chronicity factor: (LD50)1 dose/(LD50)90 doses with units of mg/kg vs. mg/kg/day.

• Note in each case we have the same total dose top and bottom.
• If CF = 90 then the toxin is absolutely cumulative. (e.g. for a compound with LD50 = 90 mg/kg, if 90 1 mg/kg doses also gave LD50, then CF = 90/1 = 90)
• If CF > 2 then relatively cumulative.
• If CF < 2 then relatively non-cumulative.

Accumulation can be due to accumulation of the compound in vivo or an accumulation of effect!

Example: tri-o-cresyl phosphate (TOCP), a cholinesterase inhibitor and neurotoxin. This toxin has the same effect in chickens in one 30 mg/kg dose or in 30 1 mg/kg dose delivered daily for 30 days. Here either situation could hold, because the t1/2 of replacement for cholinesterase is > 30 days.

# An Aside on Interpreting Statistics

[Examples modified from: Ulrich Hoffrage, Samuel Lindsey, Ralph Hertwig, and Gerd Gizerenzer. (22 Dec. 2000) "Communicating Statistical information." Science 290, pp 2261-2.]

A common situation seen in news reports, and even the literature is the interpretation of how likely a particular event is given a prevalence, and for example, the rate of false positives.

For instance. let's assume a drug is used by 0.1% of the students at HSU, and in the standard test for this drug has a false positive rate of 2%. What is the likelyhood that a student who tests positive in a random drug test is actually a user? Think about this for a few minutes and write down your answer, assuming there are 7,000 students at HSU.

Now let's solve this problem.

• First we need to determine how many users there are: Users = (0.001) (7000) = 7
• Next, we determine the number of false positives: False +'s = non-users x rate = (7000 - 7) (0.02) = 139.9
• The total number of positive tests will then be; 140 + 7 = 147
• And the probabity of a positive test indicating actual use then is: users / positives = 7/147 = 0.048, or 5%!

As another example let's consider the situation for HIV testing. If we assume a false positive rate of 0.01% and a sensitivity of 99.9%, what is the probability that you have AIDS if you test positive, and you do not belong in a known risk group (assume a prevalence of 0.01% HIV in the general, non-risk group population).

Solving:

• First, the prevelance is 1 in 10,000
• Assuming a population of 10,000 for ease of calculation, positives = 1 + 1 false = 2
• probability of the positive test indicating the disease state = 1/2, or 50%

Similar reasoning can be used to analyse the probability that an individual's disease state or symptoms are due to a particular exposure.

Another important consideration is the recognition that "hot spots" occur spontaneously as a result of probabilities and natural background rates. That random events don't occur evenly over a population, there is an expected "clumping" or noise. (Over a long sampling time the noise will even out, but over short spans, such as a human lifetime for rare phenomena, noise spikes will be common.)

# Absorption & Distribution

Transport across membranes: [overhead Fig 9.10, M&vH] (Timbrell fig. 3.2)

• Membrane structure:
• Lipid bilayer provides barrier for transport of polar molecules, but will allow passage of non-polar molecules. (However, non-polar molecules have low solubility in aqueous solution at membrane surface, so exposure to non-polar is restricted. Overall then, between the aqueous environment and the lipid bilayer barrier, we have a very effective isolation from the outside world.)
• Protein channels and specific binding/transport provide transport of polar molecules across the membrane, both non-specific and specific.
• Note, have similar membranes enclosing various compartments within the cell as well.
• Mechanisms of transport:
• Filtration.
• Passive diffusion.
• Facilitated diffusion: This is a catalyzed diffusion dependent on specific binding to a carrier molecule, commonly a protein. It is generally very specific, only catalyzing the transport of molecules which bind to the specific carrier active site. An example is the glucose carrier, which catalyzes the rapid equilibrium of glucose concentrations across membranes. Note that transport is driven by concentration gradients, there is no direct involvement of energy.
• Active transport: This is very similar to facilitated diffusion in the sense that a specific carrier is involved. however in the case of active diffusion, particles are "pumped" against a concentration gradient with the expenditure of energy (e.g. ATP hydrolysis).
• Pinocytosis: This involves the invagination of membrane to form a vesicle inside the cell. Both particles bound to the cell surface, and particles trapped from the bulk fluid during the invagination process are included in the vesicle. The bulk particles are obviously transported without specificity.

# Factors affecting transport:

### ## Molecular Properties

1. size/shape
2. lipid solubility/hydrophobicity
3. structural similarity to endogenous molecules
4. charge/polarity

## Filtration

Filtration involves the flow of substances along a concentration gradient through pores. The only specificity will then be size - can the substance fit through the pore. The flow is filtered because large molecules or particles cannot pass through the pore.

## Passive diffusion

2. Lipid solubility
3. Ionization.

Fick's Law: Rate of diffusion = k(A(C1 - C2)/d

A = area, d = thickness, C1 = [ ] outside, C2 = [ ] inside, k = constant.

Generally will be first order with k = 0.693/t1/2, where t is the time of penetration. However, can get deviations with multiphasic processes.

## Effects of Ionization on Distribution

Generally only non-ionized form passively absorbed across membrane: pH partition theory.

The proportion ionized depends on [H+] and Ka

HA H+ + A-

Ka = [H+] [A-] / [HA]

and A + H2O HA+ + OH-

Ka = [HA+] / [A][H+]; Kb = [HA+][OH-] / [A]

It is often more convenient to use log transformation.

For Acid: logKa = log[H+] + log [A-]/[HA]

rearrange - log[H+] = - logKa + log [A-]/[HA] to get the

Henderson-Hasselbalch Equation: pH = pKa + log [A-]/[HA]

For base: pH = pKb + log [A]/[HA+] (derive using Kb= Kw/Ka etc.)

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