Deviations from Mendelian Ratios

1

• Deviations from Mendelian Ratios
• Pleiotropy
• Penetrance
• Expressivity
• Thresholds
• Epistasis
• Linkage

2

• Deviations from Mendelian Ratios
• Pleiotropy one gene affects gt one trait
• example dwarfness in barley (uz uz)
• expect dwarfness to be effected by reduced
  internode lg.
• homozygous recessive results in
• short plants (expected)
• dense spike (might expect this since spike
  internodes also would be shortened)
• short awns (not expected)
• small seeds (not expected)
• short, erect flag leaf (not expected)

3

• Deviations from Mendelian Ratios
• Pleiotropy
• Penetrance the degree of expression of a gene
  (plant/population)
• Ventura lima bean homozygous AA for unifoliate
  leaf marginal albinoism
• but rarely seen in more that 10 of the
  progeny
• THUS trait penetrates or penetrance 10 no
  matter how many generations one selfs the
  line.

4

• Deviations from Mendelian Ratios
• Pleiotropy
• Penetrance
• Expressivity the manner of phenotypic
  expression
• Consider again Ventura Lima Beans and leaf
  marginal albinoism (AA)
• Some unifoliate leaves will be completely
  devoid of chlorophyll
• Some unifoliate leaves will show only marginal
  deficiency
• Some unifoliate leaves will show only tip
  deficiency
• SO, in Ventura, the gene for leaf marginal
  albinoism has incomplete penetrance (10) and
  variability expressivity
• Thankfully these gene-trait relationships are RARE

5

• Deviations from Mendelian Ratios
• Pleiotropy
• Penetrance
• Expressivity
• Thresholds expression of trait (gene(s)) is
  environmental dependent
• Example
• Collins barley albino at temperatures lt 45 F
  but normal chlorophyll development at
  temperatures gt 60 F.

6

• Deviations from Mendelian Ratios
• Pleiotropy
• Penetrance
• Expressivity
• Thresholds
• Epistasis the interaction of alleles at
  different loci interacting to affect the same
  character.
• Another look at gene action and interaction
  (next slide)

7
Gene Action and Interaction
NOTE numbers are values and not ratios
8
Generalized Epistatic Relationships (from Fehr)
F2 Genotypes
NOTE ALL cases of epistasis here assumes
complete dominance when the two loci are
considered separately!!!
9

• Deviations from Mendelian Ratios
• Pleiotropy
• Penetrance
• Expressivity
• Thresholds
• Epistasis
• Linkage phenomenon where a gene (trait) is
  found associate with another gene (trait) more
  often that expected if assortment were
  independent

10
Deviations from Mendelian Ratios Linkage Let A
and B control flower color and pubesence,
respectively, and assume that they occur on the
same chromosome RRSS x rrss ? red-smooth x
white-hairy F1 RrSs ? gametesRS, Rs, rS, and
rs F2 1/16 RRSS 2/16 RrSS 1/16 rrSS 2/16
RRSs 4/16 RrSs 2/16 rrSs 1/16 RRss 2/16
Rrss 1/16 rrss OR phenotypically Red-Smooth 9
Red-hairy 3 white-Smooth 3 white-hairy 1
11
Deviations from Mendelian Ratios Linkage
Following the cross of RRSS x rrss, a Red-Smooth
x white-hirsute, suppose you find in your F2
nursery Expected Observed Red-Smooth 9
12 Red-hairy 3 0 white-Smooth 3
1 white-hairy 1 3 The first expectation
or guess would be linkage between these two
genes. That is, they lie on the same chromosome
in such proximity that cross overs do not occur
as often as expected if they were separated by
greater distance or on different chromosomes.
12
Deviations from Mendelian Ratios Linkage
Following the cross of RRSS x rrss, a Red-Smooth
x white-hirsute, suppose you find in your F2
nursery Expected Observed
P.type G.type Red-Smooth 9 12 Parental
R-S- Red-hairy 3 0 Cross over
R-ss white-Smooth 3 1 Cross over
rrS- white-hairy 1 3 Parental rrss The
degree of linkage is measured in non-physical
distances or recombination frequencies called map
units such that each map unit (mu) 1 cross
over type gametes or cross over phenotypes.
Today, we refer to genetic distances in
Centimorgan units or cM such that 1 cM is
equivalent to a recombination freq. of 1 (THUS
1 mu)
13
Deviations from Mendelian Ratios Linkage
Following the cross of RRSS x rrss, a Red-Smooth
x white-hirsute, suppose you find in your F2
nursery Expected Observed
P.type G.type Red-Smooth 9 12 Parental
R-S- Red-hairy 3 0 Cross over
R-ss white-Smooth 3 1 Cross over
rrS- white-hairy 1 3 Parental rrss NOTE
that in order to get these genotypes that AB, Ab,
aB, and ab F1 gametes would have to occur in
equal freq., i.e. at metaphase I ----R----S-----
----R---S--- 25 ----R----S-----
----R---s--- 25 ----r-----s----- ----r-
--S--- 25 ----r-----s----- ----r---s---
25
50 50mu
14
Linkage ----R----S----- ----R---S--
- 25 ----R----S----- ----R---s---
25 ----r-----s----- ----r---S---
25 ----r-----s----- ----r---s---
25 Note that for this to have occurred that
c.o. had to occur in 100 of the sex cells. What
if c.o. occurred on only 80 of the sex
cells? 20 of sex cells 80 of sex cells
Combined ----R----S----- 5 ----R---S--- 20
RS30 ----R----S----- 5 ----R---s--- 20
Rs20 ----r-----s----- 5 ----r----S--- 20
rS20 ----r-----s----- 5 ----r----s---
20 rs30 SUCH THAT we have A and B linked at
40 m.u. or cM
50 50mu
40 m.u.
15
Deviations from Mendelian Ratios Linkage The
easiest way to calculate linkage units is to use
a test cross where the tester is homozygous
recessive for both traits Pohlman and Sleeper
page 52-53 example with barley Generic example
using RRSS and rrss F1 RrSs x rrss should
produce Test cross population 25
RrSs red-smooth 25 Rrss red-hirsute 2
5 rrSs white-smooth 25 rrss white-hirsute
i.e. cross over types 50 of population
16
Deviations from Mendelian Ratios Linkage F1 RrSs
x rrss should produce-----What are the
gametes???? Test cross population 25
RrSs red-smooth 25 Rrss red-hirsute 2
5 rrSs white-smooth 25 rrss white-hirsute
Suppose we made this cross and found 326
Red-Smooth parental type 282 Red-hirsute
cross over type 290 white-Smooth cross over
type 342 white-hirsute parental type How will
we determine if this is linkage or natural
variation??? Chi Square Analysis also
called the goodness of fit test
17
Deviations from Mendelian Ratios Linkage Chi Sq.
analysis of test cross data X2 Sum
(O-E)2/E Phenotype obs. expected d
d2 d2/expected Red-Smooth 326 310 16
256 0.83 Red-hirsute 282 310 -28
784 2.53 white-Smooth 290 310 -20
400 1.29 white-hirsute 342 310
32 1024 3.30 1240 1240 0 -
7.95 THUS the X2 7.95 Is this significant? Null
hypothesis (Ho) no difference between the obs.
and expected
18
Deviations from Mendelian Ratios Linkage X2
7.95 Null hypothesis (Ho) no difference between
the obs. and expected So go to a Chi Square table
(any stat book) Probabilities df 0.95
0.90 0.70 0.50 0.30 0.10 0.05
0.01 0.001 2 .10 .21 .71 1.39 2.41 4.61 5.
99 9.21 13.82 3 .35 .58 1.42 2.37 3.67 6.25 7.82
11.35 16.27 4 .71 1.06 2.20 3.36 4.88 7.78 9.49
13.28 18.47 Since X2cal at 7.95 gt X2tab at 3 df
and Pr. of 0.05 we reject the null Ho and accept
the HA that the observed ration is different that
the expected ratio in our test cross
population. OR accept that R and S do not
segregate independently (i.e. linked)
19
Deviations from Mendelian Ratios Linkage X2cal, 3
df 7.95 gt X2tab, 3df 7.82 THEREFORE reject
Ho no difference between the obs. and
expected Accept HA populations differ or R and S
are linked So, now the question becomes the
degree of linkage (for the geneticist
anyway) Number of recombinants (i.e. c.o.
types) Total number of progeny in test cross
Red-hirsute white-Smooth 282 290
total number of progeny 1240
m.u.
x 100 recomb.
x 100 46 m.u.
20
Deviations from Mendelian Ratios Linkage c.o.
value or m.u. 46 .46/2 .23 freq. of each
c.o. type 46 2 of sex cells in which c.o.
had to occur In gametic freq. terms RS .27 Rs
.23 rS .23 rs .27 Since the F1 received RS
from 1 original parent and rs from the other, R
and S are linked in coupling, i.e. dom.-dom. and
rec.-rec. Sometimes called cis linkage If the
original parents were Rs and rS then R and S
would have been linked in repulsion or trans
linkage
21
Deviations from Mendelian Ratios Linkage Assuming
you know that R and S are linked in coupling at
46 m.u. What will the F2 look like?
Observed Expected Red-Smooth .5729
.5625 (9) Red-hirsute .1771 .1875
(3) white-Smooth .1771 .1875
(3) white-hirsute .0729 .0625 (1)
22
Deviations from Mendelian Ratios Linkage Assuming
you know that R and S are linked in coupling at
46 m.u. Comparison of F2 and Test Cross
distributions
F2 Distribution Test Cross Distrib
Observed Expected Observed Expected Red-
Smooth .5729 .5625 (9) .2629 .25
(1) Red-hirsute .1771 .1875 (3)
.2274 .25 (1) white-Smooth .1771 .1875
(3) .2338 .25 (1) white-hirsute .0729
.0625 (1) .2758 .25 (1)
We can determine linkage distance or m.u. from F2
data but it is more difficult. It requires
knowledge of parental and cross over phenotypes
and statistical tables developed by Immer.
1930. Formulae and tables for calculating
linkages intensities. Genetics 1581-98
23
Deviations from Mendelian Ratios Linkage Immers
Product Method from F2 data ( c.o.
types)( c.o. types) ( parent 1 types)(
parent 2 types)
product ratio
24
Deviations from Mendelian Ratios Linkage Immers
Product Method from F2 data ( c.o.
types)( c.o. types) ( parent 1 types)(
parent 2 types) In our example, the F2 from
the cross of RRSS x rrss Observed
Red-Smooth .5729 R-S- Parental
Red-hirsute .1771 R-ss co white-Smooth
.1771 rrS- co white-hirsute
.0729 rrss Parental
product ratio
Note that we have established that R and S are
linked in coupling
25
Deviations from Mendelian Ratios Linkage Immers
Product Method from our F2 data (freq. converted
to whole numbers in a population of 10,000.
(1771)(1771 (5729)(729 Enter Immers
table 2 under Ratio of Products for coupling
phase at a value of 0.7510 and table indicates
c.o. value of .460 Red-Smooth .5729 R-S-
P1 Red-hirsute .1771 R-ss co
white-Smooth .1771 rrS- co
white-hirsute .0729 rrss P2
0.7509841