genetics1

1
two independent traits (dihybrid cross)
P0 purple tall x white short
P/P T/T p/p t/t
F1 purple tall
P/p T/t

P/- T/- P/- t/t
p/p T/- p/p t/t F2 purple tall
purple short white tall white short 315
108 101 32
look at traits individually in the F2 purple
white 423133 31 tall short 416140 31
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a mathematical digression probability theory
coin heads vs. tails 106 times
½ heads ½ tails flip coin twice 1st
outcome (H or T) is independent of 2nd (H or
T) 4 possible outcomes HH HT
TH TT pr (HH) pr(H) x pr(H) ½ x ½ ¼
multiply probabilities because outcomes are
independent what is likelihood of at least one
H ? pr (HH or HT or TH) pr(HH) pr(HT)
pr(TH) ¼ ¼ ¼ ¾ add
probabilities because gt1 possible way to
happen pr(a and b) pr(a) x pr(b) pr(a or b)
pr(a) pr(b)
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thinking beyond 9331 flip two coins twice
each coin 1 HH HT TH TT coin 2 HH
HT TH TT probability of a pair of outcomes
pr (1st coin, 2nd coin) ex pr (HH, TT)
pr (HH) x pr (TT) ¼ x ¼ 1/16
pr (HH, H/T) pr (HH) x pr (H/T) pr
(HH) x pr (HT) pr (TH) ¼ x
¼ ¼ ¼ x ½ 1/8 pr (purple, tall)
pr (purple) x pr (tall) pr (P/-) x pr
(T/-) ¾ x ¾ 9/16
4
two independent traits (dihybrid cross)
P0 purple tall x white short
P/P T/T p/p t/t
F1 purple tall
P/p T/t
315 108 101
32 F2 purple tall purple short
white tall white short
P/- T/- P/- t/t p/p T/- p/p t/t
look at traits individually in the F2 purple
white 423133 31 tall short 416140 31
5
Mendels Theory of Particulate Inheritance
--- genes determine the phenotype --- genes
exist as pairs of alleles at a locus --- each
gamete carries a single allele at a locus (i.e.,
A or a) --- gametes unite independent of which
alleles they are carrying different
alleles, same locus --- alleles at different
loci assort independently into the gametes
different loci
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Inheritance in armadillos hairy snout hh
is hairy HH, Hh are smooth droopy ears dd
is droopy DD, Dd are straight red eyes rr
are red RR, Rr are brown banded body bb is
banded BB, Bb are unbanded warty paws WW,
Ww are warty ww are smooth You cross a
homozygous recessive strain to a homozygous
dominant strain to make a heterozygous F1.
P0 h/h d/d r/r b/b w/w x
H/H D/D R/R B/B W/W You then make a
cross between F1 males and females. F1
H/h D/d R/r B/b W/w x H/h
D/d R/r B/b Ww in the F2, what is the
probability of offspring with hairy snouts
straight ears brown eyes banded bodies warty
paws ??
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F1 H/h D/d R/r B/b W/w x
H/h D/d R/r B/b W/w in F2, what is
the probability of getting offspring with hairy
snouts straight ears brown eyes banded bodies
warty paws ?? pr (h/h D/- R/- b/b W/-)
??
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offspring phenotypes Cross parental
pheno black, short black, long
yellow, short yellow, long 1 blk,
sht x blk, sht 89 31
29 11 2
blk, sht x yel, lng 18
19 0 0 3
blk, sht x yel, sht 20
0 21
0 4 yel, sht x yel, sht
0 0 28
9 5 blk, lng x blk, lng
0 32 0
10 6 blk, sht x blk, sht
46 16
0 0 7 blk, sht x
blk, lng 30 31
9 11
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Chi-square tests deciding goodness-of-fit to
mendelian ratios if cross two heterozygotes,
expect 31 phenotypic ratio in offspring
how closely do the numbers need to match???
importance of statistical analysis
See also pp. 40-42 in text
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From the dog problem homework observed
expected short long short long
118 42 120 40 28 9
27.75 9.25 46 16 45.75 15.25 al
l are approximately 31, but none is
exact for the sample size, how unlikely is
result?
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Compare observed and expected numbers with a
goodness of fit chi-square test with n-2 degrees
of freedom c dof
S
2
(observed number - expected number)2
(expected number)
dof degrees of freedom number
of independent classes n - 1
12
observed expected
short long short long 118 42
120 40 28 9 27.75 9.25
46 16 45.75 15.25 c 0.033
0.1 0.133 p gt 0.5
2
(118 120)2 (42 40)2 120
40
1
reject only if p lt 0.05
13
scientific method -- reject null hypothesis
H0 data fit a 31 ratio consistent with
heterozygous parents statistics test how well
data fit hypothesis two possible outcomes -
correct inference - incorrect inference
--reject H0 that is true --accept H0 that
is false
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(No Transcript)
15

• independent assortment and meiosis
• initial cross of strains homozygous at two
  loci
• A/A B/B x a/a b/b
• A/a B/b
• what happens when this individual undergoes
  meiosis?
• four, equally likely, types of gametes
• A B A b a B a b
• Cross two heterozygotesfour possible phenotypes
• A/- B/- A/- b/b a/a B/- a/a b/b

16
Prophase 1
Anaphase 1 reduction division
Anaphase 2
independent assortment of genes on
different chromosomes
17
Figure 3-12
In diploids, recombinants are best detected in a
testcross
18
Figure 3-13
Independent assortment produces 50 percent
recombinants
19
Take-Home Points the pattern of inheritance of
a trait can be determined by -reciprocal
crosses of homozygous lines and their
offspring -analysis of many crosses with parents
of unknown genotype -pedigrees probability
theory can predict the frequency of a particular
phenotype in the progeny of a cross, especially
with many traits it can also predict the risk
of inheriting a hereditary disease from a family
with a history of the disease meiosis in
heterozygotes produces two kinds of gametes that
carry parental combinations of alleles, and two
(or more kinds that carry recombinant
combinations independent assortment in
heterozygotes leads to each type (genetically)
of gamete being equally frequent