Review Question
If two genes are 13 map units apart on a linkage map, what proportion of recombinant offspring will be seen in a testcross?
What proportion of meioses experienced crossing over between to the two genes?
Two X-linked genes
Female
w P: m w+
— —————————
Male
m+
— —————————
• — ————————— •
w m Egg (A B) w m
—•————————
• — ————/ •
Gametes:
Sperm (a b) w+ m+
—•————————
—•———/
w+ F1:
m+
w
m
— —————————
• — ————————— •
w m w+
— —————————
X
• — ————/ •
w
m+
—•—————————
Par
m
—•—————————
w
m
—•—————————
Par
— ————/
•
w+
m
—•—————————
Rec
w
m+
—•—————————
Rec
Offspring (male or female) 314 Wild type 314 white, miniature 186 miniature 186 white
___% non-recombinant offspring, ____% recombinant offspring
Genes are how far apart? ___ map units
If two genes are 50 map units apart, what proportion of recombinant offspring would you observe?
What would you conclude about these two genes?
Multiple crossovers are useful for mapping many genes at a time
Three-point mapping
A B C —•—————————
A B C a b c —•————————— a b c
A B C —•—————————
A b C a B c —•————————— a b c
3 recessive phenotypes in maize (corn), coded by three linked genes
l l lazy or prostrate growth g g glossy leaves s s sugary endosperm
To map the genes, mate a triple heterozygote to triple recessive homozygote
Ll Gg Ss x ll gg ss
Gene order is not known, so the order shown here is arbitrary. Linkage phase is not known
How many different kinds of gametes can you get from triple heterozygote?
Ll Gg Ss L or l 2 G or g * 2 S or s * 2
Wildtpe for all
Recomb.
lazy, gloss, sugary
L G S // l g s
x
l g s // l g s
Number 286 33 59 4 2 44 40 272 740
Progeny Genotypes of offspring Phenotype L G S // l g s wildtype l G S // l g s lazy L g S // l g s glossy L G s // l g s sugary l g S // l g s lazy,glossy lazy,sugary l G s // l g s glossy,sugary L g s // l g s lazy,glossy,sugary l g s // l g s Total
Where to begin?
Parental types will constitute ≥ 50% of all progeny, so…
Rule 1
Two most-frequent gametes types are the __________ types Tells us the ___________________ of heterozygous parent: LGS l g s or LgS l Gs or l g S or LG s L g s l G S
L G S // l g s x
Progeny Phenotype wildtype lazy glossy sugary lazy,glossy lazy,sugary glossy,sugary lazy,glossy,sugary Total
l g s // l g s
Progeny Genotypes L G S // l g s l G S // l g s L g S // l g s L G s // l g s l g S // l g s l G s // l g s L g s // l g s l g s // l g s
Number 286 33 59 4 2 44 40 272 740
Linkage phase in heterozygous parent?
LGS l g s or LgS l Gs or l g S LG s or L g s l G S
Rule 2
The double-recombinant gametes will be the two least frequent types
A a B b C c
L G S/ l g s
Progeny Phenotype wildtype lazy glossy sugary lazy,glossy lazy,sugary glossy,sugary lazy,glossy,sugary Total
x
l g s/ l g s
Progeny Genotypes L G S/l g s l G S/l g s L g S/l g s L G s/l g s l g S/l g s l G s/l g s L g s/l g s l g s/l g s
Number 286 33 59 4 2 44 40 272 740
Rule 3
Effect of double crossovers is to interchange the members of the middle pair of alleles between the chromosomes
A a B b C c A b C
a
B
c
Parental types:
LGS and lgs
Double-crossover types:
LGs and lgS
Which gene is in the middle?
L S G
L s G
l
s
g
l
S
g
Now you know linkage phase of heterozygous parent and gene order
L S G l s g
How far apart are the genes?
Count the crossovers between adjacent genes
L S G l s g In parents, L allele on same homolog as S and l on same homolog as s. So if these get broken up ---> recombination between L and S loci In parents, S on same homolog as G and s on same homolog as g. If these get broken up --> recombination between S and G loci
L S G l s g
Progeny Genotype wildtype LGS / lgs lazy l GS / lgs glossy LgS / lgs sugary LGs / lgs lazy,glossy lgS / lgs lazy,sugary lGs / lgs glossy,sugary Lgs / lgs lazy,glossy,sugary l g s / l g s
Total
Progeny Phenotype
#
286 33 59 4 2 44 40 272 740
Crossover or NonCrossover?
Parental (NCO) single CO between L and S single CO between S and G double CO double CO single CO between S and G single CO between L and S Parental (NCO)
Rule 4: Reciprocal
products expected to occur in approximately equal numbers
Progeny Phenotype wildtype lazy glossy sugary lazy,glossy lazy,sugary glossy,sugary lazy,glossy,sugary Total
LGS ≈ lgs 272) LgS ≈ lGs Lgs ≈ lGS LGs ≈ lgS
Progeny Genotype LGS / lgs l GS / lgs LgS / lgs LGs / lgs lgS / lgs lGs / lgs Lgs / lgs lgs / lgs
(286 ≈ (59 ≈ 44) (40 ≈ 33) (4 ≈ 2)
# 286 33 59 4 2 44 40 272 740
Rule 5
Don't forget to include the double recombinants when calculating recombination frequency!
Progeny Phenotype
Progeny Genotype wildtype LGS / lgs lazy l GS / lgs glossy LgS / lgs sugary LGs / lgs lazy,glossy lgS / lgs lazy,sugary lGs / lgs glossy,sugary Lgs / lgs lazy,glossy,sugary l g s / l g s
Total
#
286 33 59 4 2 44 40 272 740
Crossover or NonCrossover?
Parental (NCO) single CO between L and S single CO between S and G double CO double CO single CO between S and G single CO between L and S Parental (NCO)
Rec Freq L-S l L L l g G S g s G s S 33 40 4 2 79
Rec Freq S-G L g S 59 l G s 44 L G s 4 l g S 109
2
Rec Freq L-S l L L l g G S g s G s S 33 40 4 2 79 Rec Freq S-G L g S 59 l G s 44 L G s 4 l g S 109
79/740 or 10.7% of gametes recombinant between L & S. So, map distance between L & S = ___ map units
109/740 or 14.8 % of gametes recombinant between S & G. So, map distance between 2 S & G=____ map units
Genetic Map
10.7 mu 14.8 mu _____________________________ L S G
Interference
Assuming independence, expected probability of double crossovers is the probability of recombination in one region times the probability of recombination in other (__________).
Maize example
Probability of recombination between L and S is 10.7% Probability of recombination between S and G is 14.8% If crossovers independent, probability of double crossover should then be
0.107
*
0.148 =
0.0158
In 740 events, the double crossover class should occur
Expected DCO = 12 Observed DCO = 6 Typical Result: O < E
L G s l g S
4 6
Conclusion: Crossing over in one region reduces probability of crossing over in adjacent regions
This is Interference
Why?
Physical constraints that prevent two chiasmata in close proximity during meiosis
Quantifying Interference
Coefficient of coincidence = Obs DCO Exp DCO cc = _____
Interference = (1 - cc) =
Recombination is not independent at small distances
If distance between genes is small (<10 map units in Drosophila) no double crossovers occur (interference is complete, I=1)
At large distances (> 45 map units, Interference disappears, Obs = Exp and I=0
In Drosophila, the allele b gives black body (wild type is tan); at a separate gene, the allele wx gives waxy wings (nonwaxy is wild type); and at a third gene, the allele cn gives cinnabar eyes (red is wild type). A female that is heterozygous for these three genes is testcrossed, and 980 progeny are classified as follows for body color, wing phenotype, and eye color:
Genotype of Gam ete fr om Heter ozygous Parent +++ b wx cn + wx cn b ++ + + cn b wx + + wx + b + cn
Phe notype brow n, nonw axy , red black, w axy, cinnabar brow n, w axy, cinnabar black, nonw axy , red brow n, nonw axy , cinnabar black, w axy, red brow n, w axy, red black, nonw axy , cinnabar
CO/NCO # DCO 6 DCO 6 SCO cn & w x 79 SCO cn & w x 77 Parental 369 Parental 369 SCO b & cn 38 SCO b & cn 36
Phenot ype brow n, nonw axy, red black, w axy, cinna bar brow n, w axy, cinna bar black, nonw axy, red brow n, nonw axy, cinna bar black, w axy, red brow n, w axy, red black, nonw axy, cinna bar
Haploid Genot ype + ++ b w x cn + wx cn b ++ + +cn b wx + + wx + b +cn
Number 6 6 79 77 36 9 36 9 38 36
a) What is the linkage phase of the heterozygous female parent? b) What is the order of the three genes?
c) Construct a linkage map with the genes in their correct order and indicate the map distances between the genes.
d) Calculate the Interference.
Threelocus mapping