S4: The Use of Linked Markers in Genetic Disease

Question 1

What are homologous chromosomes?

Answer 1

• All our chromosomes exist as pairs that are similar in length, gene position, and centromere location. The position of the genes on each homologous chromosomesis the same, however, the genes may contain different alleles. Homozygous and heterozygous refer to the alleles.
• One homologue is inherited from mother and one from dad.
• Even our sex chromosomes pair up to a degree. Pseudo-autosomal region on x and y chromosome which allow them to pair up.

Question 2

What is locus?

Answer 2

Locus is used to refer to any position on the chromosome.

Question 3

What is genotype and haplotype?

Answer 3

• Combination of alleles on one locus is known as genotype.
• Combination of alleles one a single chromosome (homologue) at several loci is a haplotype (also a group of alleles in an organism inherited from a single parent).

Question 4

What is gene mapping?

Answer 4

• Gene mapping is using an observed locus (marker, we know where it is and we can work out the genotype) to try find an unobserved locus (the disease causing gene, we don’t know where it is, we can just see the phenotype).
• We can map these unobserved loci to the markers.
• Linkage analysis is a family based design (from a few large families to many small nuclear sibpairs).

Question 5

What causes sister chromatids to be different in meiosis I?

Answer 5

In meiosis I, the chromosomes are present as identical sister chromatids joined by a centromere. They line up with their homologous chromosomes and under crossing over/ recombination between non-sister chromatids.
This causes each sister chromatid to now be different to each other which is why each daughter cell is different.

Question 6

What is linkage?

Answer 6

Phenomenon of two loci being close together and therefore alleles on the same chromosome being likely to be inherited together.

Question 7

Describe the principles of genetic linkage

Answer 7

• The tendency of alleles at neighbouring loci to segregate together at meiosis the the phenomenon of genetic linkage. Therefore to be linked, loci must lie close together.
• Alleles at linked loci is called a haplotype. Haplotypes mark chromosomal segments which can be tracked through pedigrees and populations.
• The further apart two alleles are on a chromosome, the more likely it is that a chiasma will form between them and that cross over is more likely to occur. The closer two alleles are on a chromosome, the less likely a chiasma will form between them, therefore the two alleles will stay together. Therefore cross over are more likely to occur between loci separated by some distance than those close together.

Question 8

Example of genetic linkage and recombination where A1 + B1 are one parental haplotype (on same chromosome), A2 and B2 other parental haplotype.

Answer 8

• On the left, the two alleles on the chromosomes are far apart. Therefore during meiosis, a chiasma can easily form between the homologues as there is so much space, as a result, they get separated as they cross-over. So the alleles that were on the same chromosome have been separated and there is generation of recombinants with regards to these two loci (A1 and B1, A2 and B2). A1 now with B2, A2 now with B1. Recombinants indicates no linkage as loci are not close together. Recombinants are also non parental haplotypes.
• In contrast on the right, the two alleles are close together, therefore it is unlikely a chiasma can form between them, rather it will form above or below. Therefore these alleles will travel together and not be separated. There has been no recombination with regards to these loci.
• Therefore to be linked, two loci must be close together.

Question 9

Describe recombination frequency

Answer 9

• The proportion of recombinants seen in a generation and this is the recombination fraction/recombination frequency.
• Recombination frequency = theta = RF.
• The recombination frequency is simply the number of recombinants over the total number of individuals we’re looking at.
• RF was introduced to describe the proportion of recombinants in the total number of offspring. Therefore RF is how frequently recombination occurs in a family.

Question 10

What is the RF for the example above?

Answer 10

In the example above, there are 2 recombinants and 4 non recombinants + recombinants. This means the RF = 50% (these are in the gametes in the example above). We often refer to the RF with the value theta.

Question 11

Describe linkage equilibrium for the example above

Answer 11

• If loci are far apart, there is independent assortment. There is no particular pattern of getting A1B1 or A2B2 or A1B2 or B1A2, you are equally likely to get any of these mixes in the gametes (equal proportions of all possible haplotypes).
• So when loci are far apart and you are equally likely to get any of the outcomes, the RF will tend towards 0.5, because you are equally likely to get recombinants or non-recombinants.
• Therefore there is no linkage between these two loci A and B, we are seeing linkage equilibrium which refers to independent assortment there is equal probability of recombinants or non-recombinants.

Question 12

Describe linkage disequilibrium for the example above

Answer 12

• However when two loci are close together, you are less likely to see recombination of the parental haplotypes. So the probability of no recombination is more than ¼ and the probability of having recombination is less than ¼.
• In this situation, the proportion of recombinants we see in the population will tend towards 0, so RF/theta will tend towards 0.
• This is linkage disequilibrium, when we see two alleles at different loci being co-inherited more often than would be expected by chance. This (low RF, little recombination) indicates the two loci are close together (relationship between two alleles being coinherited together).

Question 13

Describe genetic distance

Answer 13

• The frequency of recombination events is related to distance measured in centimorgans (cM).
• 1 cM = distance between 2 points where 1% of the products of meiosis are recombinant.
• 1cM ≈ 1Mb (1,000,000 base pairs). This enables us to work out how close together two loci are.
• E.g. If we see 10 percent recombination fraction =10 cM = 10 mB which aids us in gene mapping (working out where genes are in respect to each other).
• 2 or 3 recombination events per chromosome.
• The recombination fraction gives an estimation of genetic distance which is the physical distance.

Question 14

Why genetic distance and recombination fraction is useful?

Answer 14

• Can work out how close to one another loci are for genetic mapping.
• Can determine where in the genome a disease gene is (disease gene mapping): today’s example = autozygosity/homozygosity mapping.
• Predictive testing for genetic diseases, today’s example: Huntington’s Disease.

Question 15

Describe Huntington’s

Answer 15

• Gene HTT on chromosome 4p16.3.
• Encodes Huntingtin = a transcription factor.
• Autosomal dominant.
• Neurodegenerative, usually 10 year survival from symptom onset.
• Symptoms: Chorea, dystonia, cognitive decline, behaviour changes, incoordination.
• Progressive neural cell loss and atrophy.
• 3-7 per 100,000 people.

Question 16

Genetic cause of Huntington’s

Answer 16

HTT gene is on chromosome 4. There is a trinucleotide repeat in this microsatellite and this is present on everybody but with variable repeats. If there is a certain number of repeats, there is an issue as the repeat CAG codes of amino acid (glutamine residue) which creates a protein with lots of glutamine. If there is more than 36 repeats in the gene, it causes polyglutamine disorders.

• Normal range of repeat: 11-26.
• Normal but can expand: 27-35 repeats.
• May or may not have disease: 36-69 repeats.
• Affected range: >39 repeats.

Question 17

What are the two types of molecular genetic testing for Huntington’s?

Answer 17

1. Direct mutation analysis with PCR which determine size of repeat.
2. Exclusion testing which uses genetic markers to determine risk of disease.

Question 18

What are genetic markers?

Answer 18

A specific DNA sequence with a known location of a chromosome. It is typically not a gene.

Question 19

Features of ideal genetic markers

Answer 19

• Polymorphic.
• Randomly distributed across genome.
• Known location in genome.
• Frequent in genome.
• Easy to assay (to determine what the genotype is)

Question 20

Examples of genetic markers

Answer 20

• Microsatellites: Short repeats of <10bp, multi-allelic, random, frequent.
• Minisatellites: Large repeats of >500bp, multi-allelic, non-random. Typically used for DNA fingerprinting.
• Single Nucleotide Polymorphisms (SNPs):
  Single base, bi-allelic, random, frequent.

Question 21

Describe how markers and disease genes are used for genetic mapping

Answer 21

• Markers are in known position due to human genome project etc. and they mapped based on recombination fractions and evidence for and against linkage, how close together they are with respect to each other.
• Map of markers (e.g. Microsatellites and SNPs) and disease causing genes are at various positions (close to some markers and distant from other markers).
  The markers close to disease genes are often linked to it.
• So there will be patterns of inheritance of healthy and disease alleles at these markers i.e. particular alleles at this marker you will always get disease causing mutation whereas if you get different alleles at this marker you get healthy allele.

Question 22

Describe microsatellites as genetic markers

Answer 22

• Microsatellites are an example of a good genetic marker.
• As a reminder of microsatellites are, they are repetitive (di,tri,tetra) nucleotide units, the number of repeats varies between individuals. But in the same individual it will be constant (i.e. on the same chromosome). Although they may be heterozygotic, i.e. different no. of repeats for each allele.
• Each microsatellite is surrounded by a unique flanking sequence, unique to that microsatellite and this is identical between individuals. This is really important to allow us to work out the genotype of these microsatellites, we do this with PCR

Question 23

Describe how microsatellites are detected with PCR

Answer 23

1. Isolate DNA from individuals to be studied.
2. Then we design primers specific to the flanking sequences (will be on both chromosomes, remember as we are diploid!).
3. We then do PCR amplification of microsatellite region and this will give us just that part of the genome we are interested in, bound by the primers.
4. We can then do gel electrophoresis of the products (both alleles/microsatellites). The fragment length will be based on the number of repeats. Gel electrophoresis allows us to genotype the individual for that microsatellite, the shorter fragments will travel further (less repeats) and the longer fragments will travel shorter distance.
   - If we see a single band, this indicates all the DNA fragments are the same length and this shows both microsatellites on our chromosomes are the same, i.e. we are homozygous.
   - If we see two bands, it means both microsatellites are different length as they have a different number of repeats. Therefore this shows we are heterozygous for that microsatellite.
   - Multiple markers can be tested to produce haplotypes

Question 24

What two tests for genetic disease for foetus?

Answer 24

1. Direct mutation analysis.
   - Parent affected or known gene carrier.
   - Determines gene status of foetus.
2. Foetal exclusion test.
   - Gene status unknown in ‘at-risk’ parent where they don’t want to know their own gene status (if they actually have disease, just know risk).
   - Test determines if foetus has same risk as parent i.e. 50% risk.
   - Allows risk assessment to foetus without diagnosing at-risk parent who doesn’t want to know if they have the condition.

Question 25

Describe Huntington’s disease exclusion test (type of foetal exclusion)

Answer 25

• Risk of HD on chromosome 4 either on yellow or orange (individual with both is affected).
• 50% risk individual has one gene low risk one gene high risk but doesn’t want to know if they have the disease.
• Children of 50% risk can see risk using multiple microsatellite markers (using linked markers to health gene or HTT gene). So outcome is either foetus with 50 percent risk or foetus with 0 percent risk.
• Any couple that decide to have this test may consider termination in foetus with 50 percent risk of HD.

Question 26

Factors to think about when taking HD exclusion test

Answer 26

• Takes time to set up test.
• If does not carry HD gene having test unnecessarily.
• If high-risk face the decision of TOP (termination of pregnancy).
• Possibility that couple have ended an unaffected pregnancy if later HD predictive test returns negative.
• Ethical implications as could terminate healthy pregnancy as only 50 percent risk of HD.

Question 27

What is homozygosity mapping?

Answer 27

• Establish trait follows autosomal recessive inheritance only and in a consanguineous family.
• Isolate DNA and perform marker analysis (based of microsatellite, SNPs etc) to analyse data to identify shared region(s) of homozygosity in affected individuals.
• The disease causing homozygous mutation is likely to be one of those regions.

Question 28

Describe inheritance pattern for homozygosity

Answer 28

• Double line in pedigree is consanguineous marriage (related ancestors).
• A recessive allele is often seen and is passed down generation until two consanguineous individuals with recessive gene meet and their children are affected (they are homozygous for that mutation).
• Disease gene in this case is the genotype with respect to mutation that is homozygous recessive.
• Any linked markers to disease gene will also be homozygous with respect to alleles that are present. That is because they are similar bits of chromosome

Question 29

Describe identical by descent

Answer 29

• This is when homozygous mutation and homozygous marker genotypes which are identical by descent (most due to consanguineous family).
• Red bar mutation and light grey haplotype is broken up because of recombination events. Affected individuals are homozygous for red bar or homozygous for grey haplotype.
• That means that all the bases in the area are identical (region of homozygosity) –> it is just 2 copies of specific chromosome being passed down from ancestors that is identical.
• All genotype in any position in region of homozygosity is homozygous and identical haplotype to parents transmitting it. This is called: being identical by descent.

Question 30

Why is identical by descent useful clinically?

Answer 30

• Minimise need for sequencing multiple genes in genetically heterogeneous conditions.
• Uncover previously unsuspected mutations.
• Remove need for specialised biochemical testing.

Question 31

Describe retinitis pigmentosa screening (with markers)

Answer 31

• Genetically heterogeneous. There are loads of different mutated genes that cause this (not just one mutated gene), Mutations in >60 genes can cause RP.
• It is therefore difficult to sequence all genes in someone with RP to determine to mutation. It is more efficient is to sequence genome-wide genotyping of thousands of markers (SNPs). This is easy, quick and cheap with modern technology. There are the homozygous regions in respect to SNPs so just screen RP genes in homozygous regions in specific family. As long as consanguineous family.

Question 32

Describe Bardet-biedl syndrome screening (with markers)

Answer 32

• Patient negative for mutations in known genes (only exons screened).
• Genome-wide homozygosity mapping highlighted region on chromosome 11 as only ROH (region of homozygosity). Found that it contains BBS1: mutations in BBS1 most common cause of BBS.
• Then deep-sequenced gene (including introns) and found novel intronic mutation which affects splicing.
• So this disease found using linked markers.

Question 33

List for neuronal ceroid lipduscinosis , the battery of clinical tests that patient had to have which didn’t really diagnose

Answer 33

• Fundus of the eye examination (RP-like).
• Electroretinograph (absent).
• Visual evoked potential test (absent).
• Conjuctival biopsy (negative).
• Serum lactate levels (normal).
• Plasma very long fatty acid levels (normal).
• Brain MRI (atrophy).
• EEG (focal slowing in parasagittal region).
• Enzyme assay for PPT1 and TPP1 (normal).

Question 34

What was used to diagnose neuronal ceroid lipduscinos?

Answer 34

• Homozygosity mapping with SNP markers.

- Genes in homozygous region likely to share mutation and MFSD8 was found.

Question 35

Why is homozygosity mapping useful clinically?

Answer 35

• This is all based on linkage of markers to disease genes.
• Minimise need for sequencing multiple genes in genetically heterogeneous conditions.
• Uncover previously unsuspected mutations.
• Remove need for lots of testing.