1. 64% of males fell into above classes, but 36% were either wild type
white-eyed, normal-winged female x red-eyed, miniature winged male (wild type)
w m+
w m
w m+
w m+
white-eyed, normal-winged males x
w m+
w m
wild type females
for male progeny, EXPECT:
½ white-eyed, normal-winged
½ red-eyed, miniature winged
w m+
w m
64% of males fell into above classes, but 36% were either wild type
Or doubly mutant !!!!!!!

2. genetic recombination = chromosomal crossing over
36% of chromosomes in meiosis I:
w m+
white-eyed, normal-winged males x
wild type females
w m+
w m
36% of males are either doubly mutant or wild type :
w m+
w m

9. Chiasmata visible in
Locusta migratoria
spermatogenesis
A synaptonemal complex

10. vermillion (v+ = red eyes, v = vermillion eyes)
crossveinless (cv+ = normal wing veins, cv = missing crossveins)
cut (c+ = normal wing margins, c = “snipped” wing margins)
v+. cv . ct x v . cv+ . ct+
v+/v . cv/cv+ . ct/ct+ x v/v . cv/cv . ct/ct (three point testcross)
phenotype # of progeny % of progeny recombinant
v ..cv+ . ct % NR
v+ . cv . ct % NR
v .. cv . ct % v,cv ; cv,ct
v+ . cv+ . ct 40 3% v,cv ; cv,ct
v .. cv . ct 89 6% v,cv ; v,ct
v+ . cv+ . ct % v,cv ; v,ct
v ..cv+ . ct % v,ct ; cv,ct
v+ . cv . ct % v,ct ; cv,ct

11. v,cv recombinants: 45 + 40 + 89 + 94 = 268 = 18.5%
Phenotype # of progeny(T=1448) % of progeny recombinant
v ..cv+ . ct ~40% NR
v+ . cv . ct 592 ~41% NR
v .. cv . ct+ 45 ~3% v,cv ; cv,ct
v+ . cv+ . ct 40 ~3% v,cv ; cv,ct
v .. cv . ct 89 ~6% v,cv ; v,ct
v+ . cv+ . ct+ 94 ~6% v,cv ; v,ct
v ..cv+ . ct 3 ~0.2% v,ct ; cv,ct
v+ . cv . ct+ 5 ~0.3% v,ct ; cv,ct
v,cv recombinants: = 268 = 18.5%
v,ct recombinants: = 191 = 13.2%
ct,cv recombinants: = 93 = 6.4%
Aha! The genes must all be on the same chromosome! (RF’s < 50%)
v ct cv
13.2 m.u.
6.4 m.u.
Hmmm…why is the measured distance between v,cv (18.5m.u.) less than the sum of the measured v,ct (13.2 m.u.) and ct,cv (6.4 m.u) distances(=19.6 m.u.)?

12. double recombination

13. Aha! – we now realize the smallest classes of recombinants as doubles!
phenotype # of progeny % of progeny recombinant
v ..cv+ . ct ~40% NR
v+ . cv . ct 592 ~41% NR
v .. cv . ct+ 45 ~3% v,cv ; cv,ct
v+ . cv+ . ct 40 ~3% v,cv ; cv,ct
v .. cv . ct 89 ~6% v,cv ; v,ct
v+ . cv+ . ct+ 94 ~6% v,cv ; v,ct
v ..cv+ . ct 3 ~0.2% v,ct ; cv,ct ; v,cv !!
v+ . cv . ct+ 5 ~0.3% v,ct ; cv,ct ; v,cv !!
Aha! – we now realize the smallest classes of recombinants as doubles!
v,cv recombinants: (3+5) = 284 = 19.6%
v,ct recombinants: = 191 = 13.2%
ct,cv recombinants: = 93 = 6.4%
v ct cv
13.2 m.u.
6.4 m.u.
Hmmm…What is the expected # of double recombinants?
A: * = .0084
.0084 * 1448 = 12 expected double recombinants
But… we got only 8 (3+5) Why?
A: Interference! I = 1 - coefficient of coincidence (coc = o/e) = 0.33

17. Tetrad analysis: What it is it good for?
Measuring distance from a gene to the centromere
Determining RF’s

18. First division segregation pattern

20. Second division segregation patterns
RF between gene and its centromere = % of second division
Segregation patterns/2!

21. Tetrad (octad analysis)

22. More than 2 chromatids can cross-over!

23. RF (m.u.) = NPD + ½(T)/total x 100

24. Two Questions: Why is maximum RF between two genes 50%?
Does crossing over (recombination) occur at the two or the four strand stage?

26. If in an arg3 ura2 / ARG3 URA2 meiosis, got 127 PD tetrads, 3 NPD tetrads, and 70
T tetrads, then RF=?

27. Mean # of cross-overs/meiosis
The mapping function… 50 30
Measured
RF (%)
Mean # of cross-overs/meiosis
corrected map units

28. 2 = Σ (O-E)2 E

29. Are two genes linked? Is recombination frequency ~50%? P long, gray
Null hypothesis: the genes are linked and therefore we should get 1:1:1:1 in an F1 testcross
x short, ebony
F1 long, gray x short, ebony from pure line
long, ebony 54
long, gray 47
short, gray 52
short, ebony 47
Total = 200
Expected = 50
2 = (54-50)2 + (47-50)2 + (52-50)2 + (47-50)2 =
= .76 50 50 50 50

31. Cytological mapping of deletions in
polytene chromosomes (Diptera)

34. deletion mapping point mutations within a gene