Family DNA

Abstract:

The experiment presented deals with genetic analysis, in order to determine what genotype a sample has, at a specific loci. We have two family trees with samples of DNA from 24 persons of each family. We are looking for one trait in each family, one called Sloth and the other called Jerk.

We use a method called electrophoresis to do so. The procedure is briefly as follows:

First we cut the sample of DNA with a restriction enzyme. Then we mix it with a loading dye (bromophenol blue) and incubate it (37°C, 30-45 minutes). As we are waiting we prepare the tank with Agarose gel and we create small wells using a panel with “teeth”. We then pour a buffer (TBE) so that the gel does not desiccate. A Microsyringe is used to pour the sample into one of the wells. Electrodes (carbon fiber tissue) are then installed in the tank and a power source is connected to the system. Two hours later you have a tank with DNA bands and from there you can start interpreting your results.

The result could be one strand, two strands or three. One strand means Homozygous dominant, two meaning homozygous recessive and three meaning heterozygous.

Combining all the results we got in a family tree we can, after analyzing, guess what type of trait it is (if it's autosomal, sexlinked, recessive, dominant and so on.)

Introduction:

Out of a sample, we can derive the genotype of the specified locus of the DNA. We are trying to find whether the genotype is homozygous recessive, dominant or heterozygous. We want to see if the pedigree is consistent and how the disease is transmitting.

Methods:

A restriction enzyme is an enzyme which chops off the DNA at specific sequences. The sequences are different depending on the enzyme concerned. A dominant allele has no restriction site (place where DNA will be chopped) and can therefore not be cut. The recessive one, on the other hand, has a restriction site and is therefore chopped in two pieces, when the restriction enzyme is introduced to the DNA. The uncut DNA sample will have 6500 base pairs and is therefore the largest one. I'll come back to why this is something we should have in mind. The DNA that is cut is divided in two pieces; one of the parts will contain 4000 base pairs. The other one will have 2500 base pairs, making it the smallest one.

The method we use for deriving the genotypes is called gel electrophoresis and it allows us to separate the different DNA fragments as mentioned above. The fragments are negatively charged, so they will move from the wells to the positive anode. Since the sizes of the DNA fragments differ, they will have different speed when loaded. The smallest will of course be travelling the fastest through the gel, since it's the most flexible, and the largest will be travelling the slowest. As the different fragments travel at dissimilar speed, the smallest fragment is going to get to the end of the tank as the second one is going to be somewhere in the middle while the largest one will be found closer to where they all started. Each strand represents one of the fragments. If we have one strand, we know that the two alleles must be the same, and since they're only consisting of one fragment, they have to both be dominant. If we have two strands we know the number of fragments is two and therefore it's the homozygous recessive. If we have three strands, we know that the two which have been travelling furthest are the recessive allele and the slowest one is the dominant; we know that it is heterozygous.

Since we have different combinations of the alleles, the number of bands differs from person to person depending on which genotype the person has.

Results:

I and my partner were given two samples to analyze. The results we got were that daughter 7 has 6500+6500 base pairs and daughter 21 has 6500+2500+4000 base pairs.

The pattern of the Sloth-tree is consistent at all points except for person number 11. The daughter should have an affected X chromosome from her father.

For Jerk, number 14 is the person who cuts the pattern. A heterozygote can't mate with a homozygote dominant and create homozygote recessive offspring.

Discussion:

There are a few things I need to point out about the results we got.

In the Sloth-pedigree, sons 22 and 24 don't fit in with the results of their parents (if we consider this being an autosomal trait). One parent is heterozygous and one is homozygous dominant. This cannot result in a homozygous recessive offspring since there is no recessive allele in the genotype of the father. Same thing goes for son 16 and son 19. This tells us that the disease must be transferred from the mother to the sons - it is, in other words probably either mitochondrial or x-linked since the boys get their mitochondria and their x-chromosome from the mother. Let's dig deeper into it!

All the daughters of affected fathers in the pedigree are carriers except for number 11. This could be owing to one of many reasons; she is not their biological daughter, or there could have been some kind of mutation (to know if it this is likely to be the case, we need to know more about the disease. If it's controlled by one single allele and so on). Another reason could be that the mother 3 slept with a man who was either a heterozygote or a homozygote dominant to this trait. If none of these statements are true then there must have been an error made in the lab, like someone getting in physical contact with the sample or someone mixing the sample up, or maybe even someone made a mistake interpreting the results. (Another reason could be that the company who created the kit sent us the wrong sample for number 11, but I think it is very unlikely for this to happen.)

Another thing I noticed was that no daughters were affected; many were carriers.

Now, if we ignore daughter 11 for one moment, we see that all the daughters of affected fathers (and homozygous dominant mothers) have an allele which is of the disease. This excludes the fact that it is mitochondrial, since the fathers cannot transfer mitochondria. The sloth trait is therefore an X-linked recessive disease.

For Jerk, we can exclude the option of it being sex linked, since males and females are both carriers and affected. It can't be mitochondrial since father 9 transmits it to all his children. So we know it's autosomal. Now, the question is if it is dominant or recessive? This is pretty hard to tell, but part of the answer is in the children of 12 and 13. 50 percent are homozygous recessive and 50 percent are heterozygote.that means that a fraction less than 1/3 should be homozygous dominant

If we use the numbers from the pedigree then generally this is the results we get:

1+1 can produce 1.

2+2 can produce 2.

3+3 can produce 1, 2 and 3.

1+2 can produce 3.

1+3 can produce 1 and 3.

2+3 can produce 2 and 3.

Changes might occur when its sex chromosome linked but these are the general inheritance patterns.

As for the 12th and 13th parent, they have a few options;

- Have the child.

- Not have the child.

- Have the child with special nutrition plan. (If there is something you can eat to weaken the phenotype).

- Have the child with Genetic Engineering, also known as GMO (genetic modification).

- Have a child, but with a donated egg (goes for the sloth pedigree).

- Have a child with a donated sperm (goes for jerk).

- Adopting a child.

I would advise them to really think through the decision because there is a big risk of the child being affected. They would have to think about the disease. Is it going to be a crucial disease? How serious is it? Is the life of their child going to depend on it being healthy? Can they consider one of the options stated above?

2/4 (a)

2/4 (A)

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