Genetic Mapping

Question 1

What is genetic linkage?

Answer 1

• When 2 alleles are close together on the same chromosome they often get inherited together.
• Results in non-mendelian ratios of gametes and offspring types
• There is NO independent assortment

Question 2

Why are recombinant gametes rare in genetic linkage?

Answer 2

o Because crossover events between 2 genes that are very close together is not very common.

o Crossing over happens at random, so the frequency of crossovers between 2 genes depends on the distance between them.

Question 3

How do you find recombination frequency?

Answer 3

1. We need a heterozygote, in which we know which genes are together on the same chromosome.
   o Cross 2 homozygotes together (1 dominant homozygote, 1 recessive homozygote) -> heterozygote with 2 dom alleles on 1 chromosome, and 2 recessive alleles on the other.
2. Cross double heterozygote (from above) with a ‘tester’ (double homozygous recessive) & record phenotypic frequencies in F2 gen.
3. Calculate Recombination Frequency (RF):
   Recombination Frequency (RF)= Recombinants/(Total offspring) x 100%
   * RF ‘maxes out’ at 50%

1 Crossing Over event = 50% RF

Question 4

What is a linkage map?

Answer 4

• Chromosomal maps based on recombination frequencies.
• Shows relative positioning of genes on chromosome.
• Larger RF means the genes are further apart.
• Often use centimorgans or map units instead of RF:
  o 1% RF = 1 centimorgan or 1 map unit.
• Map distance isn’t always directly equal to the recombination frequency, because double crossovers can lead to underestimating the actual number of recombination events.
  o Double crossovers: ‘invisible’ if only monitoring 2 genes. They put the original 2 genes back on the same chromosome, but with a swapped-out section in between them.

Question 5

What is a 3 point test cross?

Answer 5

• Maps 3 genes at a time

- Cross 3-point heterozygote (heterozygous for 3 genes) with homozygous recessive individual.

Question 6

How do you find the order of genes and their distance apart on the chromosome?

Answer 6

Ll Gg Ss x ll gg ss

Progeny:

• L - G – S = 286
• L – G – s = 4
• L – g – s = 40
• L – g – S = 59
• l – G – S = 33
• l – G – s = 44
• l – g – S = 2
• l – g – s = 272

1. Identify the parental type gametes
   a. These tell us which alleles were on which chromosome in our heterozygote and are always the MOST abundant (because the most likely thing to happen in meiosis is for NO crossing over to occur)
   b. E.g. ( L – G – S: 286) and (l – g – s: 272) – means that all dominant alleles are on one chromosome (from heterozygote parent) and all the recessive on the other (from homozygous parent)
2. Identify the double-crossover gametes
   a. Least abundant (because least likely thing to happen in meiosis is double crossing over)… allow us to work out the order of the three genes on the chromosome.
   b. E.g. (L – G – s: 4) & (l – g – S: 2)
   c. Comparing the parental and DCO gametes allows us to work out which gene is in the middle
   i. The gene that has ‘flipped allele’ between the parentals and the DCO gametes.

                    Parentals: L G S / l g s
                    DCO:	      L G s / l g S
   
                    Therefore gene S is in the middle of the 3

3. Work out distances between the genes
   a. As usual our map distances are based on RF between pairs of genes
   b. Work out the RF between L&S and S&G in order to map them
   c. Do this by identifying gametes that result from crossovers between the 2 genes you wish to map.
   E.g. SCO between L&S gives:
   L S G L s g
   x gives
   l s g l S G

• When calculating distance between L & S we also need to include our DCO as they also represent a CO event between our 2 genes of interest.

Distance L & S = ((SCO L&S+DCO))/Total x100

	= ((40+33+4+2))/740 x100
	= 10.7 map units

Distance S & G = ((SCO S&G+DCO))/Total x100
= ((59+44+4+2))/740 x100
= 14.7 map units

Question 7

Why are 3 point test crosses better than 2 point test crosses?

Answer 7

1. We can map three genes at a time from a single experimental cross
2. We get more accurate distances as we can “see” some DCO event.

Question 8

What is interference?

Answer 8

• The inhibitory effect that one crossover event has on the probability of a second cross over event in the same region of the chromosome
• In most organisms: interference = 1 (over distances <20 cM)

Question 9

Why does interference occur?

Answer 9

o Not 100% sure.
o It is thought that crossing over requires the formation of a large protein complex. The formation of one complex prevents the formation of a second complex in the adjacent region

Question 10

What is the range of interference?

Answer 10

• From 0 (No inhibitory effect of 1 CO on a second CO) to 1 (Complete inhibition of 2nd CO; i.e. DCO cannot occur)

Question 11

Formula for coefficient of coincidence

Answer 11

(Observed DCO) / (Expected DCO)

Expected DCO calculated from Genetic maps

Question 12

Formula for interference

Answer 12

= 1 - coefficient of coincidence

Question 13

What is tetrad analysis?

Answer 13

• Special mapping technique that can be used to map the genes of haploid eukaryotic organisms in which the products of a single meiosis- the meiotic tetrad – are contained within a single structure.
• Occurs in fungi or single-celled algae – which are predominantly haploid
• First-division segregation: when the 2 types of centromeres (first-division segregation of centromeres) or the allele pairs (first-division segregation of alleles) segregate to different nuclear areas after the first meiotic division.
• To determine map distance between a gene and its centromere we measure the crossover frequency between the 2 chromosomal sites.

Question 14

The yeast life cycle

Answer 14

• 2 mating types:
  o MATa & MAT∝
• Haploid cells reproduce mitotically (each new cell arises from parental cell by budding)
• Fusion of MATa & MAT∝ cells produces a stable diploid cell that also reproduces by budding.
• Diploid MATa/MAT∝ cells sporulate (go through meiosis)
• 4 haploid meiotic products of a diploid cell (tetrad) called acospores.
• Acospores contained within a roughly spherical ascus.
• 2 of the acospores are of mating type MATa and 2 are MAT∝
• When ascus is ripe, acospores are released, they germinate to produce haploid cells.

Question 15

What are the 3 types of yeast tetrad?

Answer 15

• Parental ditype - (PD) tetrad that contains only 2 types of meiotic products, both of which are parental
• Non-parental ditype –(NPD) tetrad with 2 types of meiotic products, both of which aren’t parental.
• Tetratype – (TT) tetrad with 2 parental products and 2 non-parental products ( 1 of all 4 possible genotypes)
• The 3 types of tetrads arise for different reasons depending on whether the 2 genes are linked or not

Question 16

How do you determine if 2 genes are linked?

Answer 16

• We would expect PD = NPD if 2 genes are on different chromosomes.
• If 2 genes are linked, then PD&raquo_space;> NPD.

Question 17

What is the equation for calculating map distance in yeast?

Answer 17

ONLY IF THEY ARE LINKED

Map distance = [(0.5T + NPD)/Total ] x 100

Question 18

What are ordered tetrads?

Answer 18

• Physical constraints on movement of gametes once they are formed
  o i.e. they cannot move around within the tetrad
• Allows us to say something about when different spores within the tetrad divided away from each other
  o i.e. spores next to each other would have only divided away from each other in mitosis.
• Allows us to determine distance between genes by looking at segregation patterns
• M1 segregation = alleles separate during meiosis I (No CO occurs)
  o Will have one type of allele on each side of the tetrad
• M2 segregation = alleles separate during meiosis II (CO occurs meaning that daughter cells still have both alleles).
  o Will have both alleles on each side of the tetrads

Question 19

Formula for mapping a gene relative to its centromere

Answer 19

RF = [ (0.5M2 tetrads)/Total] x 100

Question 20

How to make 2 genes relative to each other and to the centromere

Answer 20

1. To determine whether genes are linked: assign tetrads as PD, NPD, TT
2. If they are linked calculate map distance as with yeast:
   MD = [(0.5TT + NPD)/Total] x 100
3. To determine distance between each gene and the centromere, determine which tetrads show M1 and M2 segregation of alleles for that gene.

Question 21

What are the 3 limitations of genetic mapping?

Answer 21

1. Breakdown relationship between RD and MD:
   o RF cannot exceed 50% no matter how far away 2 genes are
2. Genetic vs Physical maps:
   o Physical maps use DNA base pairs as their unit of distance
   o Genetic maps use recombination frequency as a proxy for distance
   o Both methods give same gene order
   o Anything that affects recombination will necessarily affect our estimate of the distance between 2 genes
   o Males have LOWER rates of crossing over than females (i.e. if male in test cross is heterozygote crossed with female tester, you would get less CO and so more parental type gametes, and would estimate distance to be less.)
   o In Drosophila, males have NO CO in meiosis
3. Heterochromatin represses crossing over

Question 22

If two genes are on the same chromosome in yeast, what kind of crossing-over events are responsible for generating the PD, TT and NPD tetrads?

Answer 22

PD = no cross-overs + 2 strand DCO
TT = SCO and 3 strand DCO 
NPD = 4 strand DCO