Examined Life: 2005

Friday, October 28, 2005

Plasmodium versus the Sickle Cell Mutant

Malaria is a disease caused by infection with one of the few species of the parasite, Plasmodium, often Plasmodium falciparum (the others being vivax, ovale, and malariae ). Plasmodium is transmited by the bite of a female blood-sucking mosquito, each species of plasmodium has its own species of mosquito that acts as a vector. The parasite then goes thru a series of physical changes and stages as its life cycle progresses, residing in Red Blood Cells in one stage. They reproduce inside a person's blood stream and body, making a person host to a thriving population of them. This causes, as might be imagined, all sorts of problems, which is what the disease malaria is. Eventually some of them will leave the human body by being picked up by other blood-sucking mosquitos, thru which they spread to other hosts.
Sickle Cell Anemia is a disease of the blood, wherein, because of a point mutation, the hemogloblin molecules in the Red Blood Cells (RBCs) aggregate; they tend to stick to each other, forming long chains. This deforms the RBCs, making them shaped like little sickles. This effects the oxygen carrying capacity of RBCs negatively, and also makes it difficult for them to move thru the smallest of the bodies capillaries; they tend to jam up inside them and prevent bloodflow.
The deformed RBCs are not proper homes for the malaria parasite, it cannot live in them. Without being able to do this, the parasitic infection is greatly reduced and the disease can be prevented from developing in an infected person all together. ( sickle-cell haemoglobin )

Physical Mechanism of Resistance
In 1978 Martin Friedman (1) of Rockefeller University, was investigating the mechanism of this known resistance. He was able to do this because of new advances in the ability to culture RBCs infected with falciparum. sdgfsdfs He grew the parasite in these new cell cultures, with RBCs from normal, heterzygous for sickle-cell, and homozygous for sickle cell, individuals, both under normal Oxygen concentrations and low oxygen concentrations, with sickle celling being induced by low concentrations. He found that the parasite was not effected by cells that hadn't gone sickle. Under the low oxygen concentrations, the parasites in normal cells were relatively normal, the parasites in heterozygous sickle-cell cells were inhibited, and the parasites in homozygous sickle-cells were destroyed. The mechanism of destruction in homozygous cells was found to be that the chains of deformed hemoglobin would function like needles, actually rupturing and puncturing the bodies of the parasite. In heterozygous sickle-cells however, a different mechanism was at work. Freidman couldn't identify the mechanism, however he noted that these parasites were very similar to parasites grown under 'poor' conditions, indicating that there was something that was inhibiting their metabolism at work. A 1979 (2) study by Friedman also examined the physical destruction of the parasite, finding that disintegration of the membranes and cytoplasm of the parasite following their puncturing. In heterozygous cells Freidman observed vacuolization of the parasites, again indicating that a metabolic mechanism was inhibiting their growth.
Kodjo Ayi et al (3) completed a study that found that RBCs that were infected by Plasmodium and that had the sickle cell trait, or indeed any of a number of blood hemoglobin diseases such as thalessima, "homozygous hemoglobin C (Hb-C), and glucose-6-phosphate dehydrogenase (G6PD) deficiency", were preferentially phagocytosised by monocytes, particularly in the signet-ring stage of the parasite.
Shear et al (4) created transenic mice that expressed fetal hemoglobin in their related study of the disease, finding that the parasite can't properly digest fetal hemoglobin, and reasoning that this is also a protective mechanism in sickle-cell anemia.
Becker et al (5), in a recent study, found that the parasites are "highly susceptible to alterations in the redox equilibrium", that is, on the balances of reactive oxygen species, charged ions and compounds of oxygen. They note that the immune response to infestation with Plasmodium is to destroy the RBCS that are infected and also to create nitric oxide and oxygen radicals. The parasite, for its own part, metabolizes haemoglobin into free haeme and hydrogen peroxide, which damages the cell and might also set up a finely-tunned redox equilibrium, which permits the parasite to function normally. The production of radicals by the immune system along with the inability to properly metabolise the chained, malformed sickle-cell haemoglobin, throws off that balance.
Williams et al (6) concluded that sickle-cell affords protection also by enhancing the innate and acquired immunity to the parasite itself. They reasoned that if protection from malaria varies with age in sickle-cell individuals, then that might indicate that there is an immune response at work. Their study found that protection in sickle-cell individuals increased throughout the first ten years of life, and returned to a 'baseline' after wards, and noted that this is 'most likely explained' by accelerated immune acquisition. Citing Ayi's study above, they reasoned that the increased phagocytosis of RBCs containing the parasite, involving the marking of the infected RBC's membrane, is responsible for this accelerated response. They also suggested that, alternatively, since heterozygotes for sickle cell anemia have a greater number of different types of falciparum strains when they are infected, and a longer exposure to these different strains (tho they still have a reduction in symptoms), that this permits the innate immune system to 'familiarize' itself with the different types of parasites and thus be better able to deal with the repeating cycles of parasite proliferation.

Digestion of normal haemoglobin
The malarial parasite, as it resides in the RBC, metabolizes and digests haemoglobin into consituents, one of which is Ferriprotoporphyrin IX (aka FP). (7) This product is detoxified via biomineralization by to a compound called hemozoin. Chloroquine, an antimalarial agent, prevents this detoxification, FP accumulates in the membranes and kills the parasite. The chloroquine does this by comepetively interfereing with glutathione.

Quinine, however, is the historically traditional drug used to treat malaria, comming from the bark of the chinchona tree. In the Imperial Age, Great Britain had outposts, forts, and colonies all over the world, including regions infested with malarial mosquitos. The British peoples, however, did not have any historical exposure to malaria, and thus any time the sickle-cell anemia popped up in their population, it represented a great loss in fitness for the individual and was subsequently suppressed and wiped out via natural selection. The British, therefore, had to develop a method of protecting their citizens and soldiers from the disease that they were being exposed to. Initially quinine was given in pill form, however the drug is hideously distasteful. Thus it was mixed with water, to dilute it and make it more palapable. This is called tonic water. Unforntunately, it too is far too distasteful to easily encourage people to drink it in the amounts needed to prevent malarial infection. Finally tonic water was mixed with gin, and thus the drug was delivered to the patients effectively. Indeed, the patients would deliver themselves to the drug.

More on malaria and Sickle-Cell anemia

If one looks at a map of the world and compares the incidences of malaria infection and the incidence of sickle cell anemia, one finds that they overlap. This is because the pressence of high transmission of malaria creates a selection pressure upon humans where resistance to malaria is favoured. Sickle Cell anemia comes about by a single point mutation, a change of a single nucleotide in a gene that codes for part of human hemoglobin. This mutation of the gene is an allele, an allele being merely any alternate form of a gene. A person can inherite this mutant allele. If they inherite one copy from their father, and one copy from their mother, then they have two copies of the mutant allele, and will have the horrendous disease known as sickle cell anemia, their body cannot produce normal hemoglobin, their RBCs sickle, and they die rather young, and painfully at that.
If they have only one copy of the mutant allele from one parent, and a normal copy of the gene from the other parent, then they are producing some hemoglobin that is normal, and some hemoglobin that is abnormal. Their RBCs will show a variation, from fully normal to fully sickled and everything in between, but they will tend to not die young from sickle cell anemia.
They will also be resistant to malaria, because their RBCs are inhospitable to the malaria parasite. This means that while others around them are dying from malaria, or, at least not producing all that many children because of having to deal with malaria, they are producing children, and, indeed, those children will have a chance to also inherit that parent's copy of the sickle cell gene, and thus their fitness will be greater than that of those around them. This means that sickle cell anemia, normally a destructive and harmful disease that would decrease an individual's fitness, is actually of great benefit in an environment where malaria is present.

The Plasmodium parasite goes thru an incredible array of body-forms during its development, not unlike how a frow hatches from an eggs as a tadpole and then later matures into an adult frog, only with more forms. Sporozoites are released from the salivary glands of the host mosquito. They enter the bloodstream and work their way into the cells of the liver. Therein they replicate as merozoites. They break out of the liver cell, bursting it, and it is now that the merozoites enter the RBCs. Once there, each merozoite divides and reamin together as a form called the Schizont. The parasite stays within the RBCs for a set period of time, different lengths for different species, anywhere from two days to three days. While invested within the RBCs, they can appear in a typcical 'signet ring' shape. After that, the schizonts rupture and the RBC itself is destroyed, wherein the merozoites move on to invade other RBCs. This rupturing causes the clinical symptoms of malaria itself, the "bone breaking" chills/fevers. This process also explains why the symptoms occur in cyclic episodes with malaria, the merozoites are destroying large batches of RBCs every few days, all at once. Mature forms of the parasite are also called trophozoites.
The parasite can also enter a different cycle, a sexually reproductive cycle, once entering the RBCs. Instead of dividing into schizonts, the merozoites can form into gametocytes, which are taken up by blood-sucking mosquitos. Once in the guy of the mosquito, each male gametocyte divides into 8 microgametes, which can fertilize the undivided female macrogamete. The fertilized form is called the ookinete, the moving egg. This egg burrows thru the guy wall and encysts on the outside of the gut wall (iow still inside the mosquito itself). The cyst eventually ruptures, spilling sporozoites into the mosquito body, which migrate and end up residing in the salivary glands, to begin the cycle renewed.
This intricate and complex cycle becomes even more complex when one considers that different species of Plasmodium are transmitted by different species of Anopheles (the genus of the mosquito vector for human malaria).

a few of the relevant stages

Cited Articles:

Saturday, October 01, 2005

Purification of the product of the 2nd PCR reaction, Digestion set up

A small sample of the 2nd PCR reaction will be run on gels, to ensure that the product was acheived. The rest of the PCR will be purified for later steps.

PCR Reaction Procedure

Pipet a small 1 microliter dot of DNA loading buffer onto a peice of parafilm
Remove 10 microliters from the PCR tube and mix with that dot via pipet
Pick up the mixed smample and load it into on fo teh wells on teh gel, using a gel with small wells. A DNA ladder is loaded into the first lane
Run the gel at 120V for 1 hour.
Stain in ethidium bromide for apporximately 15 minutes and visualize.

As the gel runs, perform the purification.

Purification of 2nd PCR product

Add 5 volumes of Buffer PB to 1 volume of the PCR sample and mix. Since 40 microliters of sample used, 200 microliters of buffer is used.
Place a QIAquick spin column in a provided 2ml collection tube
Apply the sample to the QIAquick colum and centrifuge for 1 minute to bind the DNA
Discard the flow-through. Place the QIAquick column back into the same tube.
Add 0.75 ml Buffer PA to teh QIAquick column and centrifuge for 1 minute, to was the sample.
Discard flow thru and place the QIAquick column back into the same tube and centrifuge for an additional minute.
Place the QIAquick column in a clean 1.5ml microcentrifuge tube
To elute the DNA into the microcentrifuge tube, add 30 microliters of water to teh center of the QIAquick membrane, allow it to sit for 1 minute to increase the DNA concentration, then centrifuge for 1 minute.

Now the product is run thru digestion.

Digestion

Add 4 microliters of 10x Buffer#2 to the eluted DNA
Add 4 microliters of BSA to the tube
Add 2 microliters of the premade enyxme mix to the tube (composed of HindIII/XhoI)
Mix the contents ths far by setting the pipet to 25 microliters and gently pipetting up and down 2-3 times

The primers have the sites for the restriction enxyme, and will cut the sample there. Later, it will be mixed with similarly restricted bacterial plasmids. Because both the plasmid and the sample DNA has been cut in the same with, they will be able to 'stick' to one antoher, and thus the mutant DNA will recombine into the bacterial plasmid, thus priducing a recombinant, transgenic, mutant genome.

DNA extraction from Gel Slices and the 2nd PCR Reaction

The DNA bands in the gel will be extracted using a QIAquick gel extraction kit, and then a 2nd PCR reaction will be done using the purified DNA as teh template.

QIAquick Gel Extraction Protocol

Weight the gel slice in a tube, making note to subtract the weight of the tube itself.
Add 3 volumes of Buffer QG to 1 volume of gel. For this experiment it will be 630 microliters and 480 microliters for the A and B gels respectively.
Incubate at 50 degrees Celsiu for 15 minutes.
After the gel discolces completely, (noting that the mixture is a yellow colour), add 1 gel volume of isopropanol to each sample and mix.
Place the QIAquick spin colum in a 2ml collection tube and put sample in.
Centrifuge for 1 minute to bind the DNA
Discard flow-through and place QIAquick column back in teh same collection tube.
Add 0.75 ml of Buffer PE to QIAquick column and centrifuge for 1 minute.
Discard the flow-through and centrifuge thet QIAquick column for an additional minute, at 13,000 rpm.
Place the QIAquick column into a clean 1.5 ml microcentrifuge tube.
Elute the DNA by adding 40 microliters of elution buffer to the center of the QIAquick column membrane. allow it to stand for 1 minute, then centrifuge for 1 minute.

2nd PCR Reaction Protocol

Into a single eppendorf tube, add via pipet:

40.5 microliters of H2O
1 microliter of Primer 1 ((the forward start primer)
1 microliter of Primer 2 (the reverse primer with teh stop codon)
1 microliter of dNTPs
1 microliter of the 10X PCR Buffer
1 microliter of Vent Polymerase
1 microliter of the template DNA from PCR A
1 microliter of the template DNA from PCR B
Mix by pipeting 40 microliter volumes 2-3 times.

The toal volume should now be 51 microliters. Place this mixture into the PCR machine, set at 95 degrees Celsius. Then have one cycle of 5 minutes of denaturation at 95 degrees Celsius. Then have 30 repitions of this set:

Denaturation at 95 degrees Celsius for 1 minute
Annealing at 50 degress Celsisus for 1 minute
Extension at 72 degress Celsisus for 1 minute

Followed by a single 10 minute extension at 72 degress Celsisus.

Gel Electrophoresis of PCR Products and DNA Purification

Continuing with the mutation of the VHL tumor suppressing gene experiment, the previous PCR products will ge run on agrose gels, and the PCR DNA will be purified for later use.

40 microliters of the PCR product will be run on teh gel via the following method.

Pipet 2 small dots containing 4 microliters of DNA loading buffer onto a peice of parafilm.
For PCR A, remove 40 microliters from the PCR tube by pipet and mix with the 4 microliters of DNA loading buffer by pipetting up and down a few times on the parafilm
Pick up the mixed sample and load it inot one of the large wells on the gel, using the 10-100 microliter pipette, releasing slowly
Repeat steps 2+3 for the PCR B. Leave an empty well between A and B PCR samples. There will be other samples from other groups sharing the same gel. A DNA ladder will be put into the first lane.
Gels will run at 200 Volts for 20-30 minutes.
The gel is then stained in ethidium bromide for approximately 20-30 minutes and then visualized under ultraviolet light.
The glowing band of DNA will be cut out with a scalpel, taking care to get as little of the gel and as much of the DNA as possible.
The two cut out bands will be placed into seperate eppendorf tubes, labeled, and will be used to extract the DNA itself later.

Unfortunately, upon following this procedure, it was found that no DNA from teh original PCR was present. There was no banding seen on the agarose gel, for any of the group samples, whereas the ladder did show banding. The staining was permited to continue for approximately another 20 minutes, with new concentrated ethidium bromide added to the staining solution, still the PCR bands failed to show up, stronly indicating that there simply was no PCR DNA.

Upon review, it was discovered that the orignal stage of the experiment was missing a critical component, the dNTPs, from which the PCR DNA is built out of.

The orignal stages were repeated with the correct procedure. After repeating the steps above, the bands did appear, and those bands were cut out and set aside in eppendorf tubes for later stages.

While the above gel was running for the second time, the remainder of the DNA (approx 40microliters) was purified using columns and buffers from the QIAquick PCR purification kit, using the following procedure.

Add 5 volumes of Buffer PB to 1 volume of the PCR sample and mix. Since 40 microliters of sample are left, 200 microliters of buffer is used.
Place a QIAquick spin column in a provided 2ml collection tube
Apply the sample to the QIAquick colum and centrifuge fro 1 minute to bind the DNA
Discard the flow-through. Place teh QIAquick column back into teh same tube.
Add 0.75 ml Buffer PA to teh QIAquick column and centrifuge for 1 minute, to was the sample.
Discard flow thru and place the QIAquick column back into the same tube and centrifuge for an additional minute.
Place the QIAquick column in a clean 1.5ml microcentrifuge tube
To elute the DNA into the microcentrifuge tube, add 30 microliters of water to teh center of the QIAquick membrane, allow it to sit for o1 minute to increase the DNA concentration, then centrifuge for 1 minute.

Now a restriction enzyme digest of this purified DNA will be set up, using the following protocol:

Add 4 microliters of 10x Buffer#2 to the eluted DNA
Add 4 microliters of BSA to the tube
Add 2 microliters of the premade enyxme mix to teh tube (composed of HindIII/XhoI)
Mix the contents ths far by setting the pipet to 25 microliters and gently pipetting up and down 2-3 times

What is now had is a tube containing a total volume of 40 microliters (the 309 microliter water+DNA sample, and the combined 10 microliters above). It is important to not hold the base of the microcentrifuge tube and thus warm it, because the restriction enzyme are refrigerated and are very heat sensitive.

This tube is now incubated at 37 degrees Celsius overnight, stored, and then used in later steps.

By this time the gels have run and have been stained and are ready for viewing (with UV protective face shields) via UV illumination.

The ethidium bromide is UV reactive and also binds with DNA, thus when illuminated with UV light, the locations of the DNA bands, which have moved during electrophoresis, show up as glowing bands, similar to the gel phot shown below, however in this experiment there is only a single band (excluding the bands in the ladder DNA row), because only one fragment has been prepated (per row, the PCR row A, and PCR row B). It is these bands that are now cut out and placed into eppendorf tubes for later procedures.

Monday, September 19, 2005

Human Reproductive Ratios

Jacobson et al in the journal Human Reproduction conducted a study in 1999 that showed that the sex ratios of children born in a study group had a schewed sex ratio. Normally a 1 to 1 ratio of boys to girls is expected, since there is a 50 percent chance that a male parent will pass on either one of the sex determining chromosomes. When this ratio is schewed, explanations are needed. Jacobson showed that the sex ratio decreased with increased number of chiren per plural birth and with paternal age, for example. This meant that there were more girls born than expected.
Sex allocation theory hypothesises that parents will increase their fitness by controlling the sex ratio of their offspring, schewing it torwards producing more of the rarer sex. Lumma et al are an adaptive feature and can change as the adaptive environment requires. Lummaa et al showed that this occured in a set of pre-industrial humans living between 1175-1850 in Finland. Where males were rare, more sons were produced. Mackenzie et al studied a population of Canadian aboriginees and showed that their sex ratio had declined from a stable ratio to a very low one, and hypothesised that this was due largely to the influx of industrial pollutants in the river that the aboriginees lived in close association with.
When considering what affects the human sex ratio, there are pre-fertilization factors to consider, and post-fertilization factors to consider.
Rex-Kiss in Act.Biol.Hung. found that fetal-maternal blood group incompatibility will lead to a higher newborn sex ratio, iow more males than expected. That researcher felt that the blood group incompatibility has an effect on the X chromosome, and that whatever this effect is the elimination rate of the zygotes fertilized by Y chromosome-carrying sperm will decrease .
Rex-Kiss also found that incompability within the Rh-factor system also increased the sex ratio.

Andersson and Bergstrom foundd that short maternal stature and obesity in African populations of the C.A.R. were related to a lower sex ratio at birth, meaning that more females were produced than males. Experiments in animals indicated that this was a result of maternal malnutrition on the male fetuses. TM Allan in the journal Reproduction studied, along with many others, the effects of blood type upon the at birth sex ratio. Allan found that AB mothers tend to have boys, and that if a baby is type A, it will tend to be a girl. More specifically, he found that the ratio is low for AB babies of AB mothers, and also for A babies of A mothers. The ratio was high for O babies of O mothers B babies of B mothers. He hypothesised that the ratio schewing is caused by "sex-differential mortality caused by interaction of the ABO genes, and some o the sex-determining genes, with oestrogen and progesterone".
Environmental factors such as pollutants can affect the sex ratio by affecting the already formed zygote, or they can also affect the parents in such a way, pre-fertilization, as to change the ratio. Moracelli et al in Lancet in 2000 studied the paternal concentrations of the highly dangerous industrial poison Dioxin on this. The 2000 study was a continuation of a previous study examing a population exposed to a dioxin release in Italy in 1976. They showed that there was an increasing probability of producing female offspring with increasing concentrations of dioxin in the fathers. Furthermore, fathers exposed when they were less than 19 years old produced a sgnificantly greater number of girls than boys (with a ratio of 0.38, as opposed to 0.5). They also quantized their study and found that the median concentration of dioxin that produced this effect was similar to doses that induced epididymal impairments in rats. Many people have hypothesized that this might explain the sex ratio schewing that is occuring in industrialized nations, however they found that the media dose in their study was twenty times the average estimated concentration in people in industrialized nations.
Another important pre-fertilization factor to consider is meiotic drive. This is where whole chromosomes will have one effect or another and cause themselves to be represented in the gametes in higher than normally expected amounts. This can occur with autosomal chromosomes, but also with sex chromosomes. Jaenike in Evolution in 1999, for example, noted that a particular species of Drosophila is polymorphic for X-chromosome meiotic drive, and that matings with males who have a 'sex-ratio' X chromosome (designated XSR) result in the production of strongly female-biased ratios, and also that there was variation amoung the species for suppression of this drive. In humans, Jaenike in Annual Review of Ecology and Systematics showed in 2001 the presence of driving sex chromosomes can reduce fitness within a population, and even bring about intragenomic conflict between the X, Y, and autosome chromosomes.
A theoretical study conducted by Norberg in 2004 and published under the NBER Working Paper Series found that sex determination can actually be affected by the pressence or absence of two-parent care. Norberg found that, from a sample population, 51.5% of respondants who were living with a partner before the child's conception or birth reported male births. 49.9% of those who were not reported male births.
And finally Grech et al in BMJ reported on "unexplained differences in sex ratios at birth in Europe and North America", noting that mammals in general have more males than females born, and that in humans the actual ratio is expected to be 0.515. In their study it also was found that statistically more boys were born in soutern countries than in central europe or the nordic countries. A low ratio was found in mexico, higher in teh US, and even higher in Canada. In a reply however, Voracek and Fisher reanalysed the data and found that the ratio was varying with latitude, and by virtue of that varying seasonal changes in the climate and varying photoperiods. Sheilds et al also responded to the Grech study, noting that maternal infection with the cytomegalovirus also influenced the sex ratio torwards more males, and that this infection was also associated with "social deprivation and unmarried status"
So studies have shown that the sex ratio is offset from the expected 50:50, in many animals, by adaptive and non-adaptive traits, and by pre and post zygotic factors, ranging from biochemical incompatibility to environmental pollutants and climate.

Saturday, September 17, 2005

Mendelian Genetic Laws

The Law of Independant Assortment

This simply states that during the formation of gametes that pairs of alleles will segregate independantly of one another. This means that when considering two genes (with 2 alleles each, therefore 4 alleles all together), all combinations of gametes genotypes will occur.

The Product Rule
This permits one to calculate the probability of two independent events both occuring. Simply multiply the probability of one event by the other. The probability of an individual having a genotype of yy is therefore
1/2 X 1/2 = 1/4

However, the probability of an individual having a genotype of Yy is not as simple, as these are two mutually exclusive events; because one parent must give Y and another must give y and not anything else.

The Sum Rule
To calculate the probability of two mutually exclusive events both occuring, you add the individual probabilities.. So the Probability of having Yy is
1/2 + 1/2 = 1/4

Multiple traits can be analyzed via these methods. Consider two individuals heterozygous for 3 traits being crossed

AaBbCc X AaBbCc

The Probability that the offspring will have a genotype of aaBBcc is

1/4 x 1/4 x 1/4 = 1/64

Because the events are independent and thus only the Product Rule applies.

The probability of having the phentoype ABc can also be calculated

3/4 X 3/4 X 1/4 = 9/64

Chi-square Analysis
This type of analysis allows one to determine how close to the theoretical results one's experimental results are, and if something other than statistical chance is creating unexpected results. The formulae is

Where E is the expected results and O is what is actually observed. The answer to the equation is then compared to a chi-squared table, using the degrees of freedom and the above value to determine the 'p' or proability for the experiment. For genetic analyses, if there are two phenotypic classes then there is one degree of freedom. The p values inform you as to theprobability that the variations were due to chance, a

nd usually p values lower than 0.05 are considered a cut off. That p value would mean that there is a less than 0.05 percent chance that the variations were statistically insignificant.

Patterns of Inheritance and Pedigrees

A pedigree is a diagram showing the inheritance patterns of traits. A horizontal pattern of inheritance indicates that a trait is rae and recessive. It occurs in several members of any particular generation, but doesn't tend to occur generation to generation. A vertical pattern of inheritance is where the trait is in every generation. Often these are dominant traits, however a very common recessive allel can give the same pattern. Approximately half of the family will get the trait when its within the family, and everyone that is affected will have an affected parent.

Incomplete Dominance

The appearance of the intermedaite pink trait is a result of gene dosage, gene A produces a red pgiment, gene 'a' produces a white pigment. Homozygotes are either red or white, but heterozygotes are pink, because they have both a red and a white pigment. This is called Incomplete Dominance and is sometimes called 'blending inheritance'

Co-Dominance

Human blood types illustrate the concept of co-dominance. There are three alleles, Ib, Ia, and i. Regardless of what other alleles are part of teh genotype, so long as allele Ib is present, a particular sugar is added to the coats of red blood cells. If Ia is present, then a different sugar is added. So when both are present, both are added, and when neither is present, and the genotype is ii, no sugars are added.

A child with type A blood born to a mother with type B blood could not have come from a father with types O or B blood. This is why blood typing is sometimes used in paternity testing.

Some genes are not simply dominant to each other, but infact occur ina dominance series, such that A >ax>af>a

Other alleles can infact be leathal when present in the homozygous condition.

In general, by examining the ratios of crosses, one can determin ehow the genes are acting. A 12:3:1 ration indicates that Dominant Epistasis is occuring. The 9:7 ration indicates that complementation between multiple genes is occuring.

Chromosomes

In the somatic cells of an individual, there is a fixed number of chromosomes, humans have 46 and flies have 8 for example. The chromosomes are present in pairs, this condition is called the diploid condition or the 2N condition. Gametes, however, are haploid, they posses just the 1N number of chromosomes. In humans the gametes possess 23 chromosomes.

Mitosis is the process by which new cells are produced, the end result is two daughter cells with identical sets of chromosomes.

Chromosomes duplicate, with copies remaining attached to each other, these attached copies are called 'sister chromatids'
The Sister chromatids line up in the center of the cell
They seperate to opposite poles of the cell
The cellular cytoplasm divides, resulting in two new cells

In meiosis, there are two rounds of division and a duplication. Meiosis is often called 'reductive division'. A diploid cell that completes meiosis will yeild haploid cells.

The chromosomes duplicate into Sister Chromatids
Homologous duplicated chromostomes pair (aka synapsis).
Homologous chromosomes(but not sister chromatids) will exhange material (aka crossing over)
Paired homologous duplicated chromosomes (aka tetrads) line up independently
Homologous duplicated chromosomes seperate from their homolog, forming daughter cells. These cells are not identical. This is what makes up the basis of mendellian Independent Assortment)
Duplicated chromosomes divide (iow, sister chromatids seperate), forming 4 daughter cells overall, each with half the original chromosomal content (iow, they are now haploid).

The Chromosomal Theory of Inheritance

When chromosomes were first observed in teh stains of cells, their function was not known and their behaviour wasn't understood. Over time, their complicated and bizzare movements noted above were observed and in 1902, Sutton hypothesized that these chromosomal bodies infact carry genes. He based this hypothesis largely the fact that Mendels theoretical Laws could be explained by the movements of the chromosomal bodies.

Chromosomes:

Occur in pairs
Seperate in gametogenesis
But reconcile in fertilization
And they assort independantly

The linkage between sex determination and the chromosomes strengthened this hypothesis.

Tuesday, September 06, 2005

Simple Genetic Analysis

Gregor Mendel is a great person to start off with when considering the analysis of genetic traits. This is becuase not only was he the first person to systematically do this, but also because he used surprisingly 'modern' seeming methods. He was extremely thorough in choosing his paths of investigation, and also extremely detailed and systematic in recording his results.
His investigations were so successful because he used an excellent organism to study, the Pea plant. It has a short generation time, which is vital when studying the distribution of characteristics over generations. They're also easy to control in terms of breeding, he could use a small brush or tweezers to control their breeding by physically manipulating their pollen bearing and receiving structures. They also produce a large number of offspring in each generation, this means that there are large datasets on which statistical analyses can be performed. Mendel used whats called "pure breeding lines" to begin his experiments, these are lines that are the result of a long series of breedings and that 'breed true' for a characteristic or trait, such as pod shape or pea colour. He'd then mate different pure breeding lines to create hybrids. The specific traits he picked to study were clearly identifiable. In systematic-phylogenetic terms he used 'bivariate' traits, ie traits that were either one of two extremes (green coloured or yellow coloured), rather than multivariate traits where there are a number of states that can be occupied or even a smeared out spectrum (ie, red, blue, yellow, or a continuum from blue to green to yellow).
Mendel performed reciprocral crosses for his study, mating, for example, purple flowering female to white flowering male, and then white flowering female to purple flowering male.

Some of the traits Mendel studied

Monohybrid Cross

The simplest type of cross Mendel would do is called a monohybrid cross, this means that the result is a generation hybridized for a single trait, rather than multiple traits. An individual from a pure breeding line for one trait is mated with an individual from a pure breeding line with the opposite trait.

Now the whole of the new generation, called the F1 generation, is only showing one of these clear traits, such as green colour or smooth pea shape, and the other traits, yellow colour or wrinkled shape, are lost. Next, the F2 generation is obtained by self-fertilization of F1 plants. Mysteriously (at the time), the lost trait reappeared, in proportion with the other trait also. The two types show up in a 3:1 portion, the lost trait reappears as a third of the whole population. Mendel concluded that there is latent information that is not expressed in the plant. That the lost trait was preserved somehow.

The trait that appears in all of the F1 generation is called 'dominant', the trait that is lost but will reappear is called 'recessive'. Mendel reasoned that each plant contains two discrete units of inheritance. Today we call the unit a gene and the discrete and different forms alleles.

Individuals in this case are diploid, they have a 2N genotype, iow they have two alleles for each trait. In gametes however, there is only a 1N genotype, the haploid condition. During gametogenesis only one allele is packaged into the gamete. This packaging process is called Segregation. From these genotypes result phenotypes, the actual displaying of characteristics and traits. Individuals with both alleles in the dominant form or both in the recessive form are said to be homozygous, whereas individuals with a mix of the dominant and recessive alleles are heterzygous.

In order to predict the percentages of different phenotypes and genotypes in a new generation, a Punnett Square can be used.

The outer edges of the square essentially represent individual and distinct gametes from the crossed individuals (the parental generation). The boxes within the square represent the genotypes of the individuals in the offspring generation, and from this the phenotype can also be determined. In a self-fertilization of the F1 generation, there are two phenotypes and three genotypes. Half of that generation are heterozygotes. A quarter are homozygous recessive and another quarter are homozygous dominant. This is a genotypic ratio of 1:2:1 , and a phenotypic ratio of 3:1

The Punnett Square follows two simple probability rules, the Sum and Products rules. The Product Rules states that the probability of two independant events both occuring is the probability of the first event times the probability of the second event. In a monohybrid cross (Yy X Yy), the probability of a gamete having a y allele is 1/2. The probability of a gamete having a Y allele is 1/2. So the probability of an individual having a genotype of YY or even yy is 1/4. The probability of getting the heterozygous condition is not independant. The Sum Rule states that the probability of two mutually exclusive events occuring is the sum of their individual probabilities. So the probability of one parent giving the y allele and then the other giving the Y allele is 1/2 x 1/2 = 1/4. The probability of the reverse, say the female parent giving y and the male giving Y, is also 1/4. The sum of these two events is the probability of an individual being heterozygous, Yy, which is 1/2.

Mendel then crossed his F2 generation. When he crossed F2 green peas he got all green offspring, and with F2 yellow peas he got all yellow offspring. Of this, 1/3 ended up being pure breeding. 2/3 gave rise to yellow and green peas in a ratio of 3:1 (iow they were hybrids).
To distinguish between pure breeding individual and a hybrid you perform a test cross, wherein the unknown is crossed with a known homozygous recessive. The ratios and types of offspring that result will allow one to work backwards to the parental genotypes.

<--phenotypes are all one type

phenotypes are half and half --->

For example, in cattle, the polled phenotype (which is hornless) is dominant over the horned condition. If a polled bull is crossed with

Cow A, horned, and yields a horned calf
Cow B, polled, and yields a horned calf
Cow C, horned, and yeilds a polled calf

Then what are the genotypes of all those involved?

To determine, you'd consider a test cross with a homozygous recessive individual. A homozygous recessive individual is anyone who displays the recessive trait. A horned cow, such as Cow C, is therefore homozygous recessive. In order for a cross with Cow C to result in polled individuals, there must be at least a single polled allele in the bull. Thus we now know part of the polled Bull's genotype.

The cross with Cow A however results in a horned calf. Since that calf is horned, its also homozygous recessive, and that means that it has two recessive alleles, only one of which can have come from the Cow. So the other must have come from the bull, and therefore the bull has at least one recessive allele. These results taken together mean that the polled bull under consideration must be a heterozygote.

This means that in the cross with the polled Cow B, which results in a horned calf (which, again, must be homozygous recessive ), that Cow B must be heterozygous, in order for it to produce a gamete with the recessive allele.

The phenotypic ratios of offsprings from each crossing set can also be determined. Because crossing A and C is a homozygote and a heterozygote, this is the punnett square.

This results in half the offspring being polled and half being horned. A ratio of 1:1

The crossing with Cow B, which is another heterozygote, results in this Punnett Squar. Here the offspring phenotypes are in a 3:1 ratio.

Dihybrid Cross

Mendel also considered dihybrid crosses, wherein the individuals to be crossed have opposite characteristics for multiple traits. For example, in a cross of Pure breeding yellow-round peas with pure breeding green-wrinkled peas, (YYRR x yyrr), the F1 generation individuals were all yellow and round (all hybrids). In the F2 generation, he received the following:

Yellow Round....315 (a parental type)
Yellow Wrinkled....101 (recombinant type, new combination of characters)
Green Round....108 (recombinant type, new combination of characters)
Green Wrinkled...32 (the other parental type)

Notice that individual traits, such as Yellow, still appear in total in a 3:1 ratio.

In the dihybrid cross, the different pairs of alleles segregate independantly into the gametes. Thus in a YyRr individual, the possible gametes are YR, Yr, yR, yr.

The genotypes will become present in their specific ratios, as shown.

Given a mulitple cross of two heterzygous individuals for three traits (AbBbCc x AbBbCc) , what is the probability that offspring of aaBBcc genotype occuring?

1/4 x 1/4 x 1/4 = 1/64

Tuesday, August 30, 2005

The VHL tumor suppressing gene and a Mutation of it

The von Hippel-Lindau tumor supressor gene is located on human chromosome 3p25, a group A chromosome. VHL disease is caused by germ line mutations of this gene, that means that the mutation is passed on to offspring. Inheritance of the gene results in an increased likelyhood for having certain types of tumours. The type of tumours that develop depend upon the type of mutation. Truncation mutations, wherein the long sequence of genes are shortened, put the indiviual at increased risk for renal cell carcinomas and hemangioblastomas. There is also a slight increase in risk of pheochromocytoma; adrenal tumours. If the mutation is a Missense type, there is an increased likelyhood for developing these tumors also, however some more so than others, depending upon the location of the missense mutation. Missense mutations for what will become the Alpha-domain of the protein product of the gene result in increased likelyhoods across the board.
The protein product of the VHL gene, pVHL, enables the degradation of another chemical called "HIF-alpha".

This degradation ability (actually it helps marks HIF-alpha for degradation) is thought to be connected to its ability to bind to complexes containing elongin B and C, an cullin 2 (Cul2). Elongin C and Cul2 are homologous to Skp1 and Cdc53 (respectively), and because of this are suspected to target certain proteins for covalent modification with Ubiquitin, and thus target them for degradation.
So the loss of this marking (with Ubiquitin) function results HIF-alpha not being degraded in the cells lysosomes. Marking a normally produced protein with another molecule to signal that it should be degraded and destroyed by an entirely different set of chemicals (in an entirely different cellular organelle, the lysosome) is an off-handed and perhaps seemingly needlessly complex way to control things. But thats just so typical of evolutionary processes anyway, working in subtle and diverse manners to have dramatic affects.

a schematic action of ubiquitin, the mutation interupts this process

Misssense mutation RC 161/2 QW effects the Alpha-domain. I will be making this mutation in a genetic sample and analysing it. The numbers in the name 161/2 indicate amino-acid positions along the primary structure chain of the protein. RC is what is normally there, Argenine and Cysteine. QW is what they are replaced by in the mutation, Glutamine and Tryptophan. This results in the wrong amino acids being in that position, as the coding is off.

The universal genetic code

Specifically the Elongin C binding region will be mutated, as it occurs over positions 157-172.
Proteins affect other proteins and materials (in which cases they are called substrates) by having binding regions that have very specific shapes, sort of like a lock and a key. What also comes into play are the charges and chemical properties of the amino acids that make up these keyed regions. This can have an effect on the protein's ability to bind, or can also effect changes in the proteins tertiary structure.

The interesting thing about pVHl is that one domain, the alpha-domain, starts by binding with Elongin C, and from that a series of other proteins. The Beta-domain of the protein, meanwhile, has bound to HIF-alpha.

HIF-alpha is a transcription factor, its responsbile for the expression of some growth factors, some of which cause new blood vessel growth, aka angiogenesis. This is why VHL disease tumors are so highly vascularized, because the HIF-alpha builds up and this results in the genes that bring about angiogenesis being amplified.

HIF-alpha will be bound to ubitquitin by the complex attached to the alpha-domain, this 'marks' it for digestion/destruction by proteosomes, which apparently is triggered by an increase in the oxygen concentration of the local environment.

pVHL and the specific action underconsideration in schematic

With the RC 161/2 QW missense mutation, [b]the ability to bind Elongin C is severly comprimised[/b], and thus the chain of events leading to ubiquitinization and ultimate degradation is broken. The mutation [b]does not necessarily affect the ability of pVHL to bind to HIF-alpha[/b] itself. Because its a missense mutation, one can expect [b]the Disease Phenotype that results from it to be of the type 2B[/b]. This is a disease phenotype that results in very great risk of developing any of the three types of tumors.

Manufacturing the mutation for analysis

The specific mutation will be created via PCR. PCR will also clone the mutated genetic segement so as to permit further analysis.

In normal PCR, a primer is mixed with DNA and DNA polymerase enzymes cause the replication of the desired segement. Two primers will be used, a forward and a reverse, so as to yeild opposing strands of DNA. Three stages are required to create the missense mutation via PCR.

a general example of PCR with primers

In Stage A, a forward primer with the "START" codon, ATG, is run up to the point where the mutation will be. A reverse primer with the mutation sequence is permited to run in the oppositte direction. This results in everything downstream (or upstream, its relative) from the mutation being copied.

In Stage B, the same is done, but for everything in the opposite direction along the genetic sequence. A forward primer with the desired mutation is used, and the reverse primer has a "STOP" codon (TAA, TAG, or TGA)This yeilds two halves, with the mutation at opposing ends.

In Stage C, the strands of DNA are denatured/unzipped. Opposite strands are then re-annealed, resulting in one strand of what would be a double helix. PCR is then repeated, and this fills in the opposing strand to make up a full gene sequence.

Following is the sequence I'll use for stages A & B

First 41.5 micro-liters of cold water are pipetted into a 0.5 micro-liter PCR.
Then 1 micro-liter of the first primer, the forward start primer is pipetted.
Then 1 micro-liter of the second primer, the reverse mutation primer, is pipetted.
Then 5 microliters of a buffer, 10x PCR Buffer, is pipetted.
0.5 micro-liters of Vent Polymerase is then added.

Now another 0.5 ml PCR tube is prepared for Stage B. The above process is repeated, with the exception that different primers are used; again the forward primer has the mutation, whereas the reverse primer now has the "STOP" codon. While preparing this Stage B tube, the Stage A tube is closed and placed on ice. The seemingly insiginificant heat of the room temperature and handling of the tube is enough to start the process prematurely.

When both tubes are ready, only now is the final step taken.

1 micro-liter of Template DNA is added to tube A, and 1 micro-liter is added to tube B.

The tubes are now sealed, labeled, and placed into the PCR device, set at 95 degrees Celsius. The mixture will be heated and denature over the course of 5 minutes. After which, there will be 30 cycles consisting of:

Denaturation for 1 minute at 95 degrees Celsius
Annealing for 1 minute at 50 degrees Celsius
Extension for 1 minute at 72 degrees Celsius

After that, the process is completed with a 10 minute extension at 72 degrees celsius. The products of the PCR will now be ready for gel electrophoresis and puryfication.