The document discusses recombination processes in meiosis, namely independent assortment and crossing over, and explains Mendel's law of independent assortment with examples. It covers genetic linkage, types of linkage (cis and trans), recombination frequency, genetic mapping, and includes methods for calculating distances between genes on chromosomes. It also outlines steps for conducting three-point experiments and calculating interference in genetic mapping.

2. The source of recombination
There are two meiotic processes that cause
recombination.
(1) Independent assortment of genes of different
chromosome pairs.
(2) Crossing over between genes on the same
chromosome pairs.

3. Independent assortment

4. Law of Independent Assortment the "Second
Law”
• The Law of Independent Assortment states that alleles
for separate traits are passed independently of one
another i.e from parents to offspring.
• That is, the biological selection of an allele for one trait
has nothing to do with the selection of an allele for any
other trait.
• Mendel found support for this law in his dihybrid cross
experiments .
• In dihybrid crosses, however, he found a 9:3:3:1 ratios.
This shows that each of the two alleles is inherited
independently from the other.

5. No Linkage: Independent Assortment

6. Linkage without Recombination

7. Linkage with Recombination

8. What is Crossing Over?
❖The exchange of chromosomal segments between two non-
sister chromatids.
❖Recall that crossing over occurs in Prophase I of Meiosis I.
❖crossing over provides genetic variation during meiosis.
❖Genetic swapping occurs between paired homologous
chromosomes

9. Single Crossovers: Non-crossover (Parental) and
Crossover (Recombinant) Gametes

10. The total recombination frequency (RF)
❖It is number of new phenotypes compared to the total
phenotypes.
❖It's 25%+25%=50% . This 50% RF is always observed for
genes on different chromosome pairs .However, the loci
that are very far a part on the same chromosome an also
show an RF value of 50%.
❖Ex: AaBb (AB, Ab, aB, ab).
❖map unit: The unit of distance in a genetic map.
❖ 1 map unit is equal to 1 percent recombination.
1 RF = 1 map unit = 1 centimorgan (cM)

11. The total recombination frequency (RF)
⚫In case of crossing over : The recombination frequency is less than
50% due to the linkage between genes.
⚫1 RF = 1 map unit = 1 centimorgen (CM)

13. ⚫Genetic linkage is a term which describes the
tendency of certain loci or alleles to be inherited
together. Genetic loci on the same chromosome are
physically close to one another and tend to stay
together during meiosis.
⚫
Linkage
Bateson and Punnett (1905) in peas
✔No independent assortment = No mendelian ratio

14. ⚫Morgan (1910) carried out an experiment by which he confirmed the
work of Bateson and Punnett
⚫He also used another case of linkage between the eye color ,wing
shape , and body color on the chromosome

15. Types of Linkage
1
-
Cis (Coupling ) linkage
The 2 dominant alleles tend to be together on the
same chromosome ,and 2 recessive alleles tend to
be together on the other chromosome

16. % recombination <50% this is due to cis
linkage.
EX: p Purple ,elongate x red ,round
PE/PE x pe/pe
G PE pe
F1 PE/pe
PE/pe x pe/pe
G PE , Pe,pE, pe pe
F2 P-E-,P-ee, ppE- ,ppee
44
6
6
44

17. 2-Trans (repulsion) linkage
The dominant allele and recessive allele tend
to be together on the same chromosome

18. %
recombination <50% this is due to trans
linkage
.
P Purple, round X red elongate
Pe/Pe X pE/pE
G Pe pE
F1 PE/pe
PE/pe X pe/pe (test cross)
G P-E-, P-ee, ppE-, ppee
6 44 44 6

19. ⚫Morgan determined that the 2 genes of eye color and wing
shape are linked due to the deviation of the phenotypic ratio
( Cis,Trans linkage)
⦿ Red eye W+W+
⦿ White eye ww
⦿ Normal wing m+m+
⦿ Miniature wing mm

20. Genetic Mapping
⚫The linkage of the genes in a chromosome can
be represented in the form of a genetic map,
which shows the linear order of the genes
along the chromosome with the distances
between adjacent genes proportional to the
frequency of recombination between them.

21. ⚫A genetic map is also called a linkage map or a
chromosome map.
⚫The concept of genetic mapping was first developed by
Morgan's student, Alfred H. Sturtevant, in 1913.

23. Mapping using a trihybrid testcross
(Three-point experiments)
⚫Three-point experiments is used to determine the
loci of three genes in an organism's genome.
⚫The two most common phenotypes represent
parental gametes.
⚫ The two least common phenotypes represent
double crossover in gamete formation.
⚫By comparing the parental and double-crossover
phenotypes, the geneticist can determine which
gene is located between the others on the
chromosome.

24. Chromosome Interference
❖ The crossing over in one region prevents the crossing
over in the adjacent regions.
❖ In higher organisms it has been found that one
chiasma formation reduces the probability of another
chiasma formation in an immediately adjacent region
of the chromosome.
❖ The interaction prevents the double crossing over
events

25. Coefficient of coincidence (C.O.C)
❖ In a three point cross the number of observed double
crossovers divided by the number expected based on
the observed occurrence of single crossovers.
❖ The ratio of the observed number of double
recombinants to the expected number
❖ (C.O.C) = observed double recombinants /expected
double recombinants
❖ Interference (I) = 1-(C.O.C)

26. ⚫Draw a map of these 3 genes
(a,b, and d) showing the
distances between all pairs of
genes, and then calculate the
value of interference.
Offspring
Genotypes
Count
a + d 510
+ b + 498
a + + 3
+ b d 1
a b + 61
+ + d 59
a b d 35
+ + + 37
total 1204
❑ 1-What is the order of the genes ?
❑ 2-Calculate the recombination
frequencies and the interference
between the genes ?
❑ 3-Draw a genetic map of the genes ?

27. ⚫Solving :
1-Step 1: identify parental classes
⚫ Parentals are the largest groups:
⚫ Double crossover are the
smallest groups.
The parental class of offspring is
the largest: a + d, and + b +. No
crossovers have occurred here: the
original parents had these
combinations of alleles.
⚫ The double crossovers class (2CO)
is the smallest class: + b d and a
+ +.
⚫ The other two pairs are the two
single crossover classes.
Offspring
Genotypes
Count
a + d 510 NCO
+ b + 498 NCO
a + + 3 DCO
+ d b 1 DCO
a + b 61 SCO
+ d + 59 SCO
a d b 35 SCO
+ + + 37 SCO
total 1204

28. ⚫Since we don’t know the orientation of these genes
relative to the chromosome as a whole, there are only 3
possible orders. These are based on which gene is in
the middle.
⚫The genes could be a--b--d,, or a--d--b.
⚫To determine gene order, set up the parental
chromosomes in the F1, then see what the resulting
double crossover offspring would look like. If the
observed 2CO’s match the expected, then you have the
correct gene order. If not, try a different order.

29. ⚫recombinant offspring(SCO1)= total of
recombinant offspring(SCO1 or 2)+ observed
double recombinants / Total offspring
⚫ a—d, so recombinants1 are a b + and + + d. 61 + 59
+ 3 + 1 = 124 / Total offspring = 1204. So map distance
= 124/1204 * 100 = 10.3 map units.
⚫d--b ,so recombinants 2 are + + + and a d b. There
are 35 + 37 + 3 + 1 = 76 of them. 76 / 1204 * 100 = 6.3
map units.
⚫a—b , so recombinants are a to b . There are 61 + 59
+ 35 + 37 +2(3+1) = 200 of them. 200 / 1204 * 100 =
16.6 map units.

30. Interference and Map
⚫There were 3 + 1 = 4 observed
2CO’s.
⚫Expected 2CO:
= ((10.3/100) x (6.3 /100)
x1204))= 7.81
⚫Coef. of Coincidence = obs
2CO / exp 2CO
= 4 / 7.81
= 0.512
⚫Interference = 1 - C. of C.
= 1 - 0.512
= 0.487
a 10.3 d 6.3 b
16.6

31. Steps to Solving 3-Point experiment
Problems
⚫ 1. Determine which pair is the parental class: it is the LARGEST
class.
⚫ 2. Determine which pair is the double crossover (2CO) class: the
SMALLEST class.
⚫ 3. Determine which gene is in the middle. If you compare the
parentals with the 2COs, the gene which switched partners is in the
middle. If you think two genes have switched sides, it is the other
gene that is in the middle.
⚫ 5. Determine map distances for all three pairs of genes. Count the
number of offspring that have had a crossover between the genes of
interest, then divide by the total offspring and multiply by 100.
⚫ 6. Figure the expected double crossovers: (map distance for interval
I * map distance for interval II / 100 * total / 100 = exp 2CO.
⚫ 6. Figure the interference as: I = 1 - (obs 2CO / exp 2CO). The
observed 2CO class comes from the data.

32. 580
ct+
cv+
v
592
ct
cv
v+
45
ct+
cv
v
40
ct
cv+
v+
89
ct
cv
v
94
ct+
cv+
v+
3
ct
cv+
v
5
ct+
cv
v+
1448
❑ In Drosophila, assuming that you have a cross between
Females heterozygous for all three genes were mated
with wild type males, and the following progeny were
obtained:

33. ❑ 1-What is the order of the genes ?
❑ 2-Calculate the recombination frequencies and
the interference between the genes ?
❑ 3-Draw a genetic map of the genes ?
❑ Vermilion eyes) v ( ,(+V) red eyes
❑ (cv) cross veinless , (+CV) a absence of cross on the
wing or vien
❑ ct (cut or snipped wing edges) ct (+ smooth wing)

34. ❑ In order to draw a genetic map of the gene we need
to calculate the recombinants between each two
adjacent genes (percentage of crossing events
between each two adjacent genes)
❑ The recombinants frequency (RF) between v and ct
loci are:
❑ 89+94+3+5 = 191/ 1448 = 13.2
❑ The recombinants frequency (RF) between ct and cv
loci are:
❑ 45+40+3+5 = 93/ 1448 = 6.4
❑ The recombinants frequency (RF) between v and cv
equal to:

35. ❑ (89+94)+ (45+40)+ 2(3+5) = 284/ 1448 = 19.6
❑ Interference = 1- c.o.c
❑ =1- observed double recombinants /expected double
recombinants
❑ expected double recombinants =
❑ % of region I crossing over x % of region II crossing
over ×total offspring / 100
❑ = (13.2 × 6.4/100 × 1448) /100 = 0.84 x 1448 / 100 =12
❑ Interference = 1- (8/12) = 33
v 13.2 ct 6.4 cv
19.6

36. ⚫Draw a map of these 3
genes (v,w, and z)
showing the distances
between all pairs of
genes, and then
calculate the value of
interference.
❑ 1-What is the order of the genes ?
❑ 2-Calculate the recombination frequencies and
the interference between the genes ?
❑ 3-Draw a genetic map of the genes ?

37. ⚫ In corn, c gives a green plant body, while its
wildtype allele c+
gives a purple plant body.
⚫ bz (bronze) gives brown seeds, while the
wildtype allele bz+
gives purple seeds.
⚫ wx (waxy) gives waxy endosperm in the
seeds; wx+
gives starchy endosperm.
⚫ The genes are arranged on the chromosome
in the order c-bz-wx.
⚫ The cross: c bz wx / + + + x c bz wx.
⚫ Note the +’s are the dominant wildtype
alleles of the corresponding gene.
Genotype Count
c bz wx 318
+ + + 324
c bz + 105
+ + wx 108
c + + 18
+ bz wx 20
c + wx 4
+ bz + 3
total 900
❑ 1-What is the order of the genes ?
❑ 2-Calculate the recombination frequencies and the
interference between the genes ?
❑ 3-Draw a genetic map of the genes ?