Lecture 14: Linkage and Mapping II

Question 1

explain Tetrad analysis in Neurospora

Answer 1

1. In most eukaryotes recombination analysis cannot be used to map centromeres.
   Few markers that display heterozygosity near centromere.

How can we define the position of the centromere genetically?
2. Fungi and algae produce either unordered or linear (ordered) tetrads. Neurospora is haploid and produces linear tetrads.

Question 2

explain The linear (ordered) meiosis of Neurospora = 4

Answer 2

1. Haploid fungi unique in that for any given meiocyte the spores are products of meiosis and held together in membranous sac called ascus.

2 Four products of meiosis lie in straight row.

3. Four products of a single meiosis undergo mitosis to produce octad.
4. Position of spores within ascus reflects chromatid segregation during meiosis

Question 3

First division (MI) segregation pattern (no crossing over) = 4

Answer 3

1 * A and a segregate into separate nuclei at the 1st meiotic division

2 * A/a meiocyte undergoes meiosis, then mitosis, resulting in equal numbers of A and a products.

3 * Two adjacent blocks of four spores:
AAAAaaaa

4 * Indicative of no crossing over between centromere and A at prophase I

Question 4

The second division (MII) segregation pattern = 4

Answer 4

1 * A and a segregate into separate nuclei at the 2nd meiotic division

2 * A/a meiocyte undergoes meiosis then mitosis

3 * Alternate pairs of spores: AAaaAAaa

4 * Indicative of crossing over between A and centromere

Question 5

Four second-division (MII) segregation patterns are equally possible - explain how? = 3

Answer 5

1. Homologous chromosomes attach randomly to the spindle pole, so sister chromatids segregate at random to give four different patterns
2. Crossovers only happen between non-sister chromatids on homologous chromosome arms (they cannot span a centromere)
3. Frequency of second-division segregation patterns is proportional to distance between A and its centromere, which always segregates in a predictable pattern.

Question 6

Centromere mapping: the steps = 3

Answer 6

1. Identify the two classes of divisions
   MI patterns, two types, alleles in two groups of 4:
   AAAAaaaa or aaaaAAAA
   MII patterns, four types, alleles in four groups of 2:
   AAaaAAaa, aaAAaaAA, AAaaaaAA, aaAAAAaa
2. Sum up MII octads and divide by the total, to work out the frequency
   of meioses that underwent recombination
3. Divide this frequency by 2 to get map units from locus to centromere. (map units are defined by the number of recombinant chromosomes from meiosis – 50% of chromosomes from these meioses will be recombinant)

Question 7

Tetrad analysis is useful because: 3

Answer 7

1. We can see the products of individual meioses and crossing over – visual confirmation (rather than inference) of equal segregation of chromatids
2. We can map centromeres genetically, very hard in most eukaryotes (and therefore the order of loci relative to the telomeres and the centromere)
3. Centromere mapping allows the integration of genetic, cytological and physical maps

Question 8

What gene is responsible for a given disease (phenotype)? = 3

Answer 8

– Cannot perform controlled crosses

– Small numbers of progeny (sampling error
becomes high)

– Testcross equivalents (e.g. A/a x a/a) are very rare

Question 9

• Remember the basis of mapping: looking for genetic linkage between traits = 2

Answer 9

– Nail-patella syndrome is autosomal dominant (Nn) – abnormal fingernails and kneecaps. N is rare allele

– N is linked to the ABO blood-type locus (I), which has multiple alleles: IA, IB, and i (null, group O). IA and IB are co-dominant

Question 10

Identifying underlying gene for a disease =4

Answer 10

1 * Two loci are linked if they appear nearby on the same chromosome.

2 * The task of linkage analysis is to find molecular markers that are TIGHTLY linked to the hypothetical disease locus (loci)

3 * When markers segregate very tightly with the disease phenotype, the underlying gene is likely to be nearby

4 * Integration of Genetic & Physical Maps:
– DNA polymorphisms provide
markers for the genetic trait
– DNA markers on a chromosome can help to locate a disease or other gene – by linkage and physical location

Question 11

Linkage mapping: Is a marker linked to the disease gene? = 3

Answer 11

1 * Collect families with affected individuals (more the merrier, but 1000+ families)

2 * Genome scan – test everyone for markers evenly spaced across the genome approx. every 10 cM, 400 markers (determine marker genotype)
– rough map to large region, can fine map later to improve resolution
– Calculate LOD score (“log of the odds”): a statistic that describes the strength of evidence for linkage

3. “Given the family data, how likely is it that there is linkage between my gene of interest and a given marker?”

Question 12

Determining LOD score = 6.

Answer 12

1 * The LOD score is an example of a maximum likelihood procedure (estimate the value of a parameter that cannot be directly observed, i.e. θ)

2 * The parameter value that gives the maximum likelihood is taken as the best estimate of the parameter.

3 * Calculate two different probabilities for obtaining the given set of recombinants observed in a family:
1. Assuming independent assortment (θ = 0.5)
2. Assuming a specific degree of linkage (θ = x)

4 * Calculate the ratio (odds) of these two probabilities

5 * Compute log10 of ratio – this is the LOD score

6 * Repeat calculation for range of different values of x, and compare: what value of θ gives highest LOD score?

Question 13

Determining LOD score; equation

Answer 13

Test to estimate whether the likelihood that two loci are linked is greater than the likelihood that two loci are unlinked.

Z = log10 (LIKELIHOOD OF LINKAGE (θ < 0.5) [A]/ LIKELIHOOD LOCI ARE UNLINKED (θ = 0.5) [B])

EXAMPLE:
A and B are probabilities between 0 and 1
If A = B, then Z = 0.
If A > B, Z becomes positive (linkage more likely) If A < B, Z becomes negative (linkage less likely) Z = 1 means 10:1, 2 = 100:1, 3 = 1000:1 etc

Question 14

Interpreting LOD scores

Answer 14

Z > +3 means
Conclusive evidence for linkage (higher is better)

+2 < Z < +3
Suggestive evidence for linkage, but relatively weak (inadequate)

Z < -2
Can confidently exclude linkage

-2 < Z < +2
Uninformative linkage analysis (not enough data)

Question 15

Fill in the blanks
1. During meiosis, …. keeps genes together, but crossing over breaks genes apart.

2. The proportion of recombinant gametes produced from a single crossover during meiosis is…
3. The maximum recombination frequency between any two loci is … and the minimum frequency is…
4. Recombination frequency is half the frequency of …
5. Completely linked genes show a recombination frequency of…
6. Two loci …. apart exhibit a recombination frequency of 25 %.

Answer 15

1. During meiosis, ***linkage keeps genes together, but crossing over breaks genes apart.
2. The proportion of recombinant gametes produced from a single crossover during meiosis is ***50 %.
3. The maximum recombination frequency between any two loci is 50%, and the minimum frequency is 0%
4. Recombination frequency is half the frequency of ***crossovers
5. Completely linked genes show a recombination frequency of ***0%
6. Two loci ***25cM apart exhibit a recombination frequency of 25 %.

Question 16

The genes Q, R and S are situated on the same chromosome. Q is 2 cM away from R. R is 10 cM away from S. Gene R is in the middle. Assume there is no interference between these three loci. In a QRs/qrS trihybrid, what proportion of gametes would you expect to be QRS?

Answer 16

1. Require single crossover between R and S, which have RF of 0.1
2. BUT remember that there could also be a crossover between Q and R, which would give us unwanted qR and Qr recombinants. So, we must consider this outcome as well.

So:

3 * R and S recombine at frequency of 0.1

4 * But we have to EXCLUDE rare gametes that have a crossover between Q and R as well (qRS and Qrs - these are the double recombinants)

5 * These occur at frequency 0.02*0.1 = 0.002

6 * Subtract 0.002 from 0.1 = 0.098

7 * Divide by 2 because we only want QRS (not qrs) = 0.049

8 * Therefore the frequency of QRS gamete = 4.9%

Question 17

You just bought two black guinea pigs, one male and one female, of the same genotype from the pet store that are known to be heterozygous (Bb). You also know that:

– black fur (B) is dominant over white fur (b)
– a lethal recessive allele k is located on the same chromosome and only 2 cM away from the recessive b allele
– both your animals are heterozygous for k as well.

What is the probability of finding a white individual among the progeny if you cross these two animals?

Answer 17

1. Let lethal allele be k (WT allele K). Then cross is: BK/bk x BK/bk
2. Each parent will produce four gametes:

BK 0.49 (non-recombinant)
bk 0.49 (non-recombinant)
Bk 0.01 (recombinant)
bK 0.01 (recombinant)

3. There are three ways of obtaining a white-furred guinea pig:
   Male gamete bk x female gamete bK = 0.49 * 0.01 = 0.0049
   Female gamete bk x male gamete bK = 0.49 * 0.01 = 0.0049
   Female gamete bK x male gamete bK = 0.01 * 0.01 = 0.0001
   Gamete bk x gamete bk is lethal, so will not produce white progeny.
4. Sum of all three probabilities = 0.0099 or ~ 1%.