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SouthCity High School: Student BC -  25th Arpril 2pm

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General Discussion

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Challenge 1

Image:Challenge title2.jpg

(a) Explain what a Barr Body is and why they are only found in females.

(b) How could the presence of Barr Bodies affect the phenotype of a female who is:
Heterozygous for an x-linked gene
Homozygous for an x-linked gene.

(c) All the females in the family who are affected are heterozygous for the mutation. Within the females in the whanau, there is variation in the severity of the vision impairment. What biological explanation could account for this variation in the heterozygous female phenotype?.

Campbell Heron – Horowhenua College Firstly, Barr bodies Are present in some males, e.g. those with Klinefelter’s syndrome (XXY or XXXY), however the overwhelming majority of Barr bodies are found in females as they have two X-chromosomes and one in each cell is deactivated early in development. Which X-chromosome happens to be deactivated is purely by chance and cell groups descended from a cell with a specific X-chromosome deactivated will also have the same X-chromosome deactivated. This causes females to be, to at least some extent, Genetic mosaics.

For female homozygotes, the presence of Barr bodies would have little effect on the phenotype of the X-linked condition, they will probably be similar in phenotype to males, however female heterozygotes will be genetic mosaics of this condition and so only some cells will be affected.

The altered or mutant gene was identified to be located on the X-chromosome. As all the affected females are heterozygous for the mutant gene, Barr bodies may explain the variable severity of the vision impairment. Since some cell groups in the eyes will be descended from a cell whose mutant gene-carrying X-chromosome was deactivated, this will cause those cell groups to be unaffected, while other cell groups will be affected if the normal X-chromosome was deactivated. As the eyes are relatively small organs and X deactivation occurs early in embryonic development, the proportion of affected eyes is highly variable and thus the level of effect on the eyes is also highly variable.

Jacquie Bay 15th September
Cambell - some excellent points in your answer. Yes you are correct to note that in males with extra X-chromosomes such as Klinefelter's Syndrome males (XXY), X inactivation occurs in the same way that it occurs in a normal female, although these individuals are male because of the presence of the Y chromosome. Similarly in Turner's Syndrome (XXX), X-inactivation occurs in two of the X-chromosomes in each cell. Importantly you have picked up on the concept that it is possible that the heterozygous females could be genetic mosaics.
I will leave my comments there for now and look forward to more contributions from other students during the week.

Campbell Heron - Horowhenuas College Thanks for ther helpful comments! :)

Pat and Mike Horowhenua College 10.15pm 15/sep/2009

Only one X-chromosome in each cell is fully active while the other X-chromosomes are randomly, ‘almost deactivate’ as some genes on the chromosome are still active. The X-chromosome(s) that are deactivated are condensed and we call them Barr bodies. Usually humans have either XX or XY but there are cases where this is not so, Turner’s syndrome and Klinefelter’s syndrome. Therefore males can contain Barr bodies. Also in some examples more than one X chromosome is barred so that only one X-chromosome is fully active at a time.

Do Barr bodies exist in birds? They have ZZ sex chromosomes.

What causes the X chromosomes to become inactive and how is the process regulated? this is since only the X-chromosomes are targeted.

After cell replication daughter cells will have the same barred X-chromosome as the parent.

Heterozygous- if the x-linked gene is recessive and a Barr bodied x-chromosome containing the dominant allele becomes deactivated then the mutant gene is expressed as there is no dominant allele to mask the effect. Some cells will be infected while other will not, depending on which X-chromosome is Barred.

However if the dominant gene is not deactivated in a Barr body and is allowed to remain active then the phenotype will be the dominant gene’s trait as it will mask the effect of the infected gene. Thus the phenotype is kept the same as none of the cells are affected.

Jacquie Bay 17th September
Pat and Mike- some interesting and good questions that you have raised.
I recommend that you read an article on Scitable :Sex Chromosomes in Mammals X-Inactivation. This article explains some of the history of the discovery of X-inactivation and looks at how Barr Bodies are formed.
You are correct that there are a small number of genes that are still expressed in the inactive x-chromsome. For the purpose of this question. let us assume that the gene in question is not one of these. Once an x-chromosome has been inactivated in a cell, that inactive state will be passed on in mitosis. Because X-inactivation occurs early in development, when an individual is heterozygous for an x-linked gene there will be some areas of the developing embryo that have the dominant allele in the active x-chromosome and some where the active x chromosome has the recessive allele. leading to a mosaic form of phenotype. A good example of this is the tabby cat coat pattern. A gene for coat colour in cats that controls for Black and Orange occurs on the X chromosome. In the heterozygous individual, there will be patches of black and patches of orange fur dependent on the allele present in the active x-chromosome in that section of the skin.
You have explained the situation for expression in a heterozygous individual. Can you take this one step further and link your answer to the context of the CACF1 gene in question. Can you use the information that you have presented to explain why there is significant variation in the phenotype of females whose genotype is known to be heterozygous for the mutant allele?

Pat and Mike Horowhenua College 17/sep 3:46pm

A simple answer is that since X-chromosomes are randomly barred early in the embryonic stage so that only one X-chromosome is fully active in each cell. The daughter cells will have the same x-chromosome barred as the parent cell. If the female is heterozygous for the mutant gene then the barring of a X-chromosome will change the severity of the phenotype of the female. If the Barr body contained the mutant gene then the gene in that cell will not be expressed and the normal gene will be expressed. Therefore by chance depending on how many cells in the eye have the ‘infected’ X-chromosome Barred will determine the severity of the condition, lots means less severe not many means more severe.

Now using the pre-resource sheet page 5 notably there is some major variation in the phenotype of the female, most have poor eye vision and many are very short sighted but all have involuntary eye movement, therefore the mutant gene has pleiotrophy effects. Depending on the amount of infected genes barred in eye cells will determine the amount of properly functioning protein, mutant gene being the gene that causes protein to become not properly functional. ‘Protein being the ion channel’.

In conclusion the more infected X-chromosome not barred the more severe the condition and visa versa if it is barred.

If I am right with my ideas then shouldn’t it be possible to methyl tag the mutant gene, if found’, on the X-chromosome? By doing this all daughter cells will also have the gene disabled. In effect the infected protein will in time ‘die off’ and since only the ‘non infected’ cells are able to produce protein will replace the dead ‘protein’ with fully functional protein. Therefore the ion channel will act normal.

Marion Maw Geneticist  17th September

You raise an interesting point about whether cells that lack the ion channel protein are more likely to die than cells that contain the ion channel protein (or whether cells that contain the mutant channel protein are more likely to die than cells that contain the healthy channel protein). This is an issue because the ion channel protein is retained within the cell in which it was made. (Some proteins are secreted from cells into the surrounding environment and this is important when understanding the phenotype of female heterozygotes in some X-linked disorders, such as hemophilia which impairs blood clotting in affected males.) 

If in a particular cell the unaffected (not “uninfected”) CACNA1F gene is on the inactivated X-chromosome, and it were possible to also inactivate the affected (not “infected) CACNA1F gene, then neither copy of the gene would be active in that cell. This would mean that the cell would completely lack the ion channel protein. Mutations in the CACNA1F gene that result in a lack of ion channel protein cause the inherited retinal condition called X-linked congenital stationary night blindness 2 (CSNB2). In this condition, heterozygote females are carriers ie their vision is not impaired. In effect, you might convert affected I745T CACNA1F heterozygote females into carrier CSNB2 heterozygote females.  

Tangaroa College: Bonnie and Karen 23 September 2009
1. a) A Barr Body is a small, dense body formed by condensation and inactivation of an X chromosome. This occurs in the early development of the embryo, and the number of Barr Bodies is always one less than the number of X chromosomes found in the individual, which can be expressed in the formula:
B = n - 1
Where B is the number of Barr Bodies, and n is the number of X chromosomes in the individual.
In normal females, there is only one Barr Body, as there are two X chromosomes present.
b = n - 1
   = 2 - 1
   = 1
In normal males, Barr Bodies are not present, as there is only one X chromosome present.
b = n - 1
   = 1 - 1
   = 0

 b) i. Once an X chromosome has been inactivated, it gives rise to a clone of cells, all of which have the same inactivated X chromosome and produce the same phenotype. The phenotype depends on which X chromosome was inactivated, as the only gene expressed is that which is present on the activated X chromosome. As X chromosome inactivation is completely random, the phenotype of the female is varied.
 e.g. Tortoiseshell cat hair pattern.

  ii. The phenotype for the X linked gene will not change, as there is no difference in alleles on the two X chromosomes, and therefore, if one is inactivated, the gene will still be expressed as if two alleles were present.

 c) The inactivation of the X chromosome when forming a Barr body is completely random, and therefore different phenotypes can be produced in different individuals due to the variation in the genotype of the activated X chromosomes in the cells of the body. When these variations occur in the eye, they can affect the severity of vision impairment from the whanau.

Jacquie Bay 24th September
Bonnie and Karen - good to see an answer from you. I like your definition in (a) - very clear.
In (b-i) I would like to have seen you add that the x-inactivation is random and happens very early in development - then is set. The Tortoiseshell cat hair inheritance pattern is a good example - I know I have outlined that above - in an examination when you use an example like that please make sure you outline the example.
In (b-ii) can you re-look at your wording. Rather than state it will not change (which sounds as if it will start as one phenotype and move to another later in life) you would be better starting the phenotype will be consistent within each individual and within a group of female individuals who have the homozygous X-linked genotype.
In (c) I would like to see you state that the inactivation of each cell is set randomly in early development. You could improve your answer by explaining how this can lead to variation in the individuals with this genotype dependent on the inactivation pattern set in the tissue where the gene is going to be expressed. ''


Challenge 2


What evidence from the family tree suggests that this is an x-linked condition?


Campbell Heron – Horowhenua College Males can only pass their one X-chromosome onto female offspring, and the family pedigree chart shows that all the female children of affected males were affected by the condition while males were not, this suggests that the condition is X-linked. For example, individual 15 had four daughters all of whom had the visual impairment and one son without the condition. Also individual 12 whose three sons have not inherited the condition but his daughter has. Assuming that the mutant gene is rare in the general public, and thus it is unlikely that these children’s mothers were carriers of the mutant gene, this provides a lot of evidence that it is X-linked and not purely due to chance that the daughters were affected and not the sons.

Jacquie Bay 15th September

Cambell - you have used examples well to illustrate your reason.  In an examination question I would advise you to make a clearer statement before your examples. Your first sentence starts well but before going into the family pedigree (2nd half of that sentence) you need to fully explain the reason why. Something along the lines of therefore if a gene is X-linked it will only be .......... Then launch into the specific situation in this pedigree. 

Campbell Heron - Horowhenua College Thanks, good point. Would this be better? Males have one X-chromosome and one Y-chromosome, so daughters always inherit one of their X-chromosomes from their father. While sons cannot inherit an X-chromosome from their father (because they only inherit one sex chromosome from each parent, and they inherit a Y-chromosome from their father). Therefore if the condition is X-linked fathers will always pass on the mutant gene to their daughter and not to their sons. Continue on to examples...

Jacquie Bay 17th September

Campbell - yes that you give you a clear explanation linked to the example. 

Pat and Mike Horowhenua College 10.15pm 15/sep/2009

All Female’s gametes contain a X-chromosome (with some exceptions, non disjunction occurring), therefore an infected mother’s offspring, doesn’t matter if they are male or female, may inherit the X-linked chromosome or not thus we have no solid evidence to determine whether this condition is autosomal or X linked. Therefore we will need to use the father’s gametes and offspring to help determine this condition.

Males can only pass a X chromosome to his daughters and Y chromosome to his sons. Therefore an infected father theoretically will have half of his gametes containing the X-linked chromosome and the other half containing a Y chromosome. All female offspring will inherit the x-linked chromosome from the father and all male offspring will inherit the Y chromosome.

By looking at the pedigree chart, out of all the infected male’s offsping only female’s offspring were infected while males were not infected. This can’t be a autosomal condition as the chances of offspring from over 5 families having these kind of offspring, this is highly unlikely and not a coincidence. This is shown with parent 8, 12, A left down one, 30, 15.

This was with the assumption that the infected male’s partner was not infected. This is because if she was some males would also be infected as they do not have another X-chromosome to mask the effect .At this point we know this is a sex-linked condition.

To determine that this is x-linked we look at the infected female’s offspring, we see some of the males are infected some are not which would be because the infected mother was heterozygous for an X-linked gene thus some male offspring inherited an X-linked chromosomes and others didn’t. Since males only have one X chromosome and one Y chromosome an infected male will contain a X-linked chromosome which will be expressed, no chromosome to mask the effect. In conclusion the condition is X-linked.

Jacquie Bay 17th September 

Pat and Mike - you have correctly explained the reasons why we can be sure that this is an x-linked inheritance pattern.  Well done.  A note about terms.  You refer to an affected individual as infected.  Biologically we speak of individuals affected by a mutation or allele that causes a disorder such as this as affected rather than infected.  Infected is used to describe situations where an individual is showing symptoms of a condition that results from a pathogen infecting the body.  Other than that, well answered.

2. The affected males in the family only pass on the gene for vision impairment to their daughters, and not to their sons.
e.g. Male 12 has three sons and one daughter, and only the daughter has the affected gene.
This shows us that it is an x-linked condition, as males only have one X chromosome that they can pass onto their daughters, and as all daughters are heterozygous for the condition, the allele for the vision impairment must be dominant.

Challenge 3


Compare person 16 in generation VII (affected) with person 17 in generation VI (unaffected). What does the difference between the microsatellite pattern for these two individuals tells us about where on the chromosome the mutation is? Look at the sequence “2 2 6” . Refer to your diagram in the handout

Alfriston College - Casey,Shaniya and Vanessa 16th September 4pm

The difference in the microsatellite pattern for the 2 individuals occurs in the numbers above the '226' sequence.  We can tell the mutation is in these numbers because the affected and unaffected chromosomes both have the '226' sequence.  The mutation does not occur on the '226' sequence because otherwise VI 17 would be affected.  Chromosome VI 17 could have taken part in crossing over and this is how it obtained the '226' sequence but not have the rest of the chromosome containing the mutation. 

After looking at this question some more and question 4 we were wondering how you ruled out the first 2 on the '226' sequence as in question four you say the mutation is in the sequence '1132' how can this be if the 2 was ruled out.   Our theory was that the mutation is more likely to have occurred in the '113'  sequence.

Jacquie Bay September 17th

Casey, Shaniya and Vanessa - you are on the right track but I think you can demonstrate to the examiner more of your understanding of concepts around crossing over and recombination in your answer. You state correctly that the mutation is in the upper region of the chromosome. In order for the individual VI17 to inherit the pattern of markers represented by 314311226, this individual must be a recombinant for the X-chromosome. Could you please re-write your answer, this time showing me that you understand what role crossing over the recombination plays in (a) the formation of the altered sequence of markers on the chromosome and (b) our understanding that the mutation is not in the section with the marker pattern 226. You will of course need to go back in your answer to looking at the chromosomes in the parents. Looking forward to reading your next posting..

Alfriston College - Nick and John 16th September 11pm

The only difference between VI 17 and VII 16 is the first 6 digits in the sequence. In VII this is 214113, thus making him an affected male. Because one is affected and one is not, the gene for the blindness must be in this sequence. Since 226 is common to both, it must not affect the condition, so it can be ignored. The reason 226 is common is because that sequence was crossed over. VI 17 has had crossing over on the bottom of his chromosome (the 226 end) from his mothers affected chromosome. Despite the fact this sequence has originated from an affected X chromosome, the individual is not affected.

Jacquie Bay September 17th

Nick and John - Be very specific in your description of the difference - and in doing so you can demonstrate a higher level of understanding of the biological concepts underpinning this argument.  Try starting with a statement that shows that you understand that the numbers represent microsatellite marker allele patterns of the x-chromosome.  Rather than proving that the gene for the condition was in the sequence 214113, the comparison of these two individuals demonstrates that the gene is not in 226.  That may sound pedantic - but while it proves it is somewhere in the 214113 region, it proves more specifically where it is not. I would also like to see an explanation in your answer of how the recombinant chromosome has formed. Use the biological terms that you know such as recombinant - simply by using this term correctly in the context of your answer you are demonstrating your understanding. Look forward to reading a re-post of this from you. 

Horowhenua College Pat and Mike 16 sep 2009 10:19pm

Looking at V 6 offspring males would have inherited from the mother either the X-chromosome with 314311315 pattern (does not contain mutation) or 214113226 pattern (contains mutation), considering that crossing over does not occur.VI 17 shows crossing over has occurred as the parent ‘V6’ X-chromosomes 315 segment of 314311315 has exchanged with the 226 segment of 214113226 forming the recombinant pattern 314311226 that VI 17 has inherited. The VI 17 carrying the 314311226 Pattern is not infected thus the mutation is not in the 226 pattern else it would be expressed as the male doesn’t have another X chromosome to mask the effect.

Now comparing the Males VI 17 314311226(not infected) with ‘VII 16’ 214113226 (infected). The difference between the sequences is the 226 pattern which we know is not where the mutation is. Therefore the mutation is within the pattern 214113.

Jacquie Bay September 17th 

Pat and Mike - I can see you understand, however your argument needs to be a bit tighter. Have a think about the following points:

  1. I like the fact that you have linked back to the affected parent in generation V - however you need to explain why you can be sure that crossing over has not occurred here. Explain why you know that.
  2. Rather than stating that VI 17 shows crossing over has occurred, state that VI 17 is a recombinant chromosome for the X-chromosome and explain (a) why you know this and then (b) how a recombinant is formed. Then go on to explain how this evidence suggests where the mutation

Pat and Mike Yee Horowhenua college 10:43 pm 18 sep

Thank you for you comment Jacquie, i have rewritten my answer below.

Offspring males receive one sex chromosome from each parent thus an X-chromosome from the mother and Y-chromosome from the father due to male being XY. Females inherit an X-chromosome from each parent as females are XX. V 5,6,15 has the microsatellite DNA pattern 214113226 on one of their X-chromosome(s).

Knowing this the chance that 214113226 being recombinant DNA is highly unlikely as crossing over would have occurred at the same point on the sex chromosomes three times. In addition parent’s gametes would either contain X-chromosome with the parent DNA, recombinant DNA or other recombinant DNA. Finally out of the parent and other recombinant DNA V 5, 6 and 15 by chance have all inherited this 214113226 X-chromosome during fertilisation, very unlikely.

We can now say that the microsatellite pattern 214113226 in not recombinant DNA.

Being a male V15 has only one X-chromosome and therefore cannot mask the expression of the mutant gene (affected male). Similarly V5 and 6, female, also contain the X-chromosome and is also affected despite having their other X-chromosome being different. In addition the mutation is dominant and thus is always expressed unless barred; we can say this as male in IV is affected and the X-chromosome containing the mutation in this example has to be passed from the father and inherited by the daughters V5 and 6. Daughters V5 and 6 also have a different X-chromosome due to different mothers yet just having the 214113226 satellite DNA sequence on the X-chromosome will cause the individual to be affected. The mutant gene is within the X-chromosome with the satellite DNA pattern 214113226.

(Ignoring other chromosomes except sex chromosomes) VI 17 X-chromosome contains recombinant DNA which has occurred in the female during meiosis as the homologous X chromosomes have undergone crossing over, 315 of 314311315 swapped with 226 with 214113226 and the gametes would either contain the parent DNA or recombinant DNA. Gametes formed have X-chromosome with either satellite DNA sequence 314311315,214113226, 314311226 or 214113315. The mother’s male offspring has inherited the 314311226 chromosome from the mother during fertilisation. Knowing that males are unable to mask the effect of the mutant gene VI 17 is not affected despite having the satellite DNA sequence 226 from the 214113226 pattern on his X-chromosome, therefore the 226 pattern seems to not contain the mutant gene.

To clarify this we will compare VII 16 with VI 17 although we could compare VI 17 with any other affected male except VII 48 which has undergone crossing over. VII is an affected male and contains the X-chromosome with the mutant gene, it has not undergone crossing over as its microsatellite DNA pattern 214113226 is the same as the affected male’s microsatellite DNA pattern in VI and V. The difference between VII 16 and 17 microsatellite DNA pattern is the 214113 of VII 16 (affected) and 314311 of VII 17 (not affected).

Therefore we may say that the mutant gene is within the pattern 214113.

Jacquie Bay 23 September

Pat and Mike - thank you.  You have in this answer offered clear explanation of your anser and in doing so have demonstrated knowledge and understanding of the underlying biological principles invovled.  Take care with use of language - from the analysis of these individuals we can conclude that the mutant gene must be in the region of the chromosome in which the microsatellite pattern 214113 is found. In preparation for your scholarship examination, this type of exercise is very valuable.  

3. Both person 16 and person 17 have the microsatellite pattern 226 at the end of the X chromosome. As person 16 is affected and person 17 is not affected, the mutation must not be in this part of the chromosome or both individuals would be affected. Therefore the mutation must be in the upper part of the chromosome

Challenge 4

Compare person 45 in generation VII (unaffected) with person 48 in generation VII (affected). The difference between these two individuals confirmed for the scientists that the mutation must be in the middle section of the chromosome where the satellite pattern “1 1 3 2” was found. What evidence did they use to come to this conclusion?

Alfriston College - Casey and Vanessa 16th September 4:25pm

They used the evidence that the only difference between the unaffected and affected chromsome was the middle sequence with the satellite pattern '1132'.  The rest of the chromosome sequence was the same.  The chromsome with the sequence  '1132' is the one with the mutation as this sequence is only shown in affected people. 

Jacquie Bay September 17th

Casey,and Vanessa  - yes - but have you compared and explained the evidence? I don't think you have shown me that you understand why there is a difference in these chromosomes and why that difference indicates that the 1132 section is the target. 

Alfriston College - Nick and John 16th September 11pm

Individual 23 is a heterozygous female, so has one X chromosome with the gene for the condition and one without. Individual VII 48 (a descendant of Indiviual 23) shows segments from both of his mother's X chromosomes (affected in blue, non-affected in black). He has had crossing over at two points of his X chromosome which resulted in him having roughly half of one of her chromosomes and half of the other chromosome still resulting in the mutation/condition. Therefore the half he inherited from the affected chromosome must include the gene for the condition. This half is '1132'.
When combining this knowledge with challenge question 3, we can infer the gene for the condition must be in the 113 sequence. We can assume this because VI 17 has undergone crossing over of '226' and is still not affected.

Jacquie Bay September 17th

Nick and John - well reasoned argument. A few points to think about.

  1. The crossing over occurred in the mother's calls during meiosis - formation of the gametes. So take care in stating that he has had crossing over. A more correct way to say this would be that the pattern of microsatellite markers on the X-chromosome in individual 48 indicates that a double cross over event occurred during meiosis when the gamete that contributed to this individual was formed. Then put in your explanation (which is good) of that evidence.
  2. Take care in talking about genes vs alleles. The mutant form of the gene CACNA1F is arguably a mutant allele of this gene, rather than a completely different gene. So it would be better to state that the mutant allele of the CACNA1F gene that causes the condition must be in ........

Marion Maw 18th September

Nick and John - Suppose that you were trying to locate a person living in the South Island, along the route Picton-Kaikoura-Christchurch-Ashburton-Timaru-Oamaru-Dunedin-Invercargill. Now imagine that you had excluded the area Picton-Kaikoura and the area Timaru-Dunedin. The area that you need to keep looking in is between Picton-Timaru, rather than Christchurch-Ashburton. Hence in this family, individual VI-17 shows the mutated gene allele maps above (“north of”) the bottom three markers and individual V-48 shows the gene maps below (“south of’) the top three markers. Thus it maps between 4-1-1-3-2, rather than between 1-1-3.

Kevin Horowhenua College 20 sept 10:30

In generation VI person 23 is female herterzygous for this mutant sex linked gene, this person has one X chromosome which is affected and the other unaffected. We are shown on the family tree that generation VII both persons 45 and 46 obtained the unaffected X chromosome from there mother giving the satellite pattern 122223331 (from being genotyped). Persons 47 and 48  are affected, obtaining the mutation squence for the affected X chromosome, person 47 has the original pattern 214113226 from previous generation. we see that person 48 obtained a recombinant allele of the X chromosome because there is a mixture of the two homologues this is due to crossing over in meiosis, the crossing of segments at two chiasmata points - the top and the bottom of the chromtid leaving the middle section of the original mutate pattern 1132. Since the person has the vision impairment mutation, so the satellite pattern is still in that X chromosome, the only pattern of the original mutate chromosome is 1132 we can thus prove that this is the section of the chromsome which the mutation occured where the satellite pattern was indicated.

Jacquie Bay September 23rd 

Kevin - thank you.  Your answer directly addresses the question and presents a logical argument for your conclusion.  

4. These individuals both possess very similar microsatellite sequences, but one is affected and one is not. Their first three microsatellite patterns are identical, as are their last two sequences. The only difference in their patterns is in the middle sequence, "1132", which is present in every individual affected with the eye condition. Therefore the mutation must be in the middle part of the chromosome, near this microsatellite pattern.