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8/13/2019 Genetics Chapter 2
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Comparison of Mitosis and Meiosis
Review :
http://highered.mcgraw-hill.com/olcweb/cgi/pluginpop.cgi?it=swf::535::535::/sites/dl/free/0072437316/120074/bio17.swf::Comparison%20of%20Meiosis%20and%20Mitosishttp://highered.mcgraw-hill.com/olcweb/cgi/pluginpop.cgi?it=swf::535::535::/sites/dl/free/0072437316/120074/bio17.swf::Comparison%20of%20Meiosis%20and%20Mitosis8/13/2019 Genetics Chapter 2
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Chapter 2
Mendelian Genetics
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Mendels Experimental Design
Why are pea plants
so useful?
Self-fertilization(selfing)
Cross-fertilization
(cross-breeding)
discrete,
nonoverlapping traits
- flower and pod color,
height
Pisum sativum
(common garden pea)
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Plant Anatomy
anatomy:
Pistil
Style
Ovary
anatomy:
Stamen
Anther (pollen-producing)
Filament (forsupport)
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7 Phenotypes (traits) of Pea Plants
Pod location
Seed and pod
color and shape
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Flower color
How did Mendel use these characteristics in designing
experiments?
unambiguously described each plant
Stem length(used only 2 extremes:
tall or short)
7 Phenotypes (traits) of Pea Plants
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Crossing plants with only 1 variable
trait
Chose homogeneous plants(pure-breeding: first grew
them for 2 years)
Cross tall and dwarf plants
(P1or parental generation)
Offspring - F1or first filial
generationhybrids
(monohybrids : different in
only 1 trait)
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Examine the traitsof the F1generation
all were tall : dominant
recessive trait was masked
Examine the phenotypes
of the F2generation
3:1 ratio
tall (787)
dwarf (277)
Crossing plants with only 1 variable
trait
(monohybrid)
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Mendels proposed mechanism
a certain inherited unit (gene)
produced 1 trait
Ratios of offspring produced :
each trait is composed of 2factors , one dom inant and 1
recessive alleles
-Homozygotes (pure-bred) ;
heterozygotes (hybrids)
Difference in genotypesand
phenotypesin the P1, F1, and
F2generations
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Mendels First Principle: The Law
of Segregation
Each allele is a different form of a gene that will
separate and randomly distribute into a gamete
- each gamete has an equal probability of receiving
either allele
Fertilization = fusion of two gametes
The Law of Segregation explains
F1genotype: HETEROZYGOTE (hybrid)
F1phenotype: DOMINANT (recessive
trait masked)
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The Law of Segregation: The
Schematic Approach
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The Law of Segregation: Use of
the Punnett Square
critical assumption :
all gametes listed occur (segregate) at equalprobability
every offspring (box) is equally likely
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The Law of Segregation: using
Probability to Predict Genotypes and
Phenotypes
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With only phenotypes, Mendel confirmed
the Law of Segregation by selfing F2plants
F2generationPhenotypic ratio 3:1 dominant to
recessive (recessive trait
reappears)
Genotypic ratio 1:2:1
Self-fertilize F2to produce
F3generation
Predict what you would
expect in
F2: dwarf X dwarf
F2: tall X tall
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Testing the Law of Segregation
How can you determine the genotype of an F2
offspring that expresses the dominant phenotype?
Testcross: Mate the organism with the dominant
phenotype with an organism that expresses therecessive phenotype.
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What Kind of Data Is Obtained from
the Test Cross?
1. Tall (homozygote)
X dwarf
(homozygous
recessive)
2. Tall (heterozygote)
X dwarf
(homozygous
recessive)
Two possible outcomes that are phenotypically distinct :
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P1phenotypes: yellow, round x
green, wrinkled
F1phenotype: yellow, round
(dihybrid: heterozygous for 2 genes)
F2phenotypes:315round, yellow
108round, green
101wrinkled, yellow
32wrinkled, green
divide each category by number in
smallest group:
9.84 : 3.38 : 3.16 : 1.00 (~9:3:3:1)
use of a Punnett square
Crossing plants with 2 variable traits
(dihybrid)
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Punnett square prediction of
genotypes and phenotypes
critical assumption :
all gametes listed occur (segregate) at equalprobability
alleles of 2 different genes are inherited (assorted) independently
(round,
yellow)
(round,
green)
(wrinkled,
yellow)
(wrinkled,
green)
Th L f I d d t
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The Law of Independent
Assortment
Alleles for 1 gene segregate independently of thealleles for other genes
Isolate individual phenotypes :
1. seed shape : 315+108 round
101+ 32 wrinkled423 : 133
round : wrinkled
(3.18:1.00) expect 3:1 for 1 gene
2. seed color : 315+101 yellow108 + 32 green
416 : 140
yellow : green
(2.97:1.00) expect 3:1 for 1 gene
F2phenotypes:
315round, yellow108round, green
101wrinkled, yellow
32wrinkled, green
equalsegregationof alleles for each of the 2 genes
T t f I d d t A t t
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Test of Independent Assortment :
Testcross the Dihybrid Plant
Cross: F1WwGg X wwggPredict the phenotypes in
progeny
1:1:1:1 Round Yellow
Round Green
Wrinkled Yellow
Wrinkled Green
Mendels results verified this and confirmed the Law of
Independent Assortment : alleles for 1 gene segregate
independently of the alleles for other genes
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Multihybrid Crosses
with n genes (each containing 2 segregating alleles),there will be :
differentgametes
differentphenotypes
differentgenotypes
2n
2n
3n
frequency of homozygous recessives = n
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Genetic nomenclature
botany and mammalian genetics :
uppercase letters : dominant alleles,
lowercase letters : recessive alleles
W : round w : wrinkledG : yellow g : green
phenotypes :
shape WW, Ww : roundww : wrinkled
color GG, Gg : yellow
gg : green
Wild t d M t t Ch t i ti i
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Wild-type and Mutant Characteristics in
Drosophila
Wild-type: characteristic most common in nature
Eye color: red
Wings: oblong and flat
Mutant phenotypes: alternatives to the wild-type
Eye color: for example, white
Wings: for example, dumpy (reduced wings)
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Mutations of Drosophila Melanogaster
Genetic Symbol Convention in Drosophila
Wild type allelesymbol of the mutant followed by a superscript +
genes are named after the mutant phenotype
Capital lettermutation is dominant
Lower casemutation is recessive
e.g. : wild type (red) eyes : w+ mutant (white) eyes : w
wing mutations :
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Genetic Symbolism in Drosophila
a +sign can be usedalone for the wild type
(only when referring to
1 gene at a time)
multiple mutant
alleles can exist for a
gene : wa(white
apricot) or wc(whitecrimson)
There are cases where the gene can be designated by
more than one letter: dp = dumpy wings, dp+ = wild type
P b bilit A li ti i
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Probability: Applications in
Mendelian Genetics
Use of ratios : Probability(P) = a/na= number of times an event is observed
n= total number of possible cases
Can be determined by: observation(empirical) - 1 child/10,000 born with
phenylketonuria (PKU)
- probability that next child born will have PKU = a/n= 1/10,000
nature of the event (theoretical): drawinga heart
from a deck of cards 13/52 = 1/4
P b bilit A li ti i
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Probability: Applications in
Mendelian Genetics
P=1 : event is an absolute certainty
P=0 : event is impossible
Probability of all events in an experiment : add
up to 1
B i P i i l f P b bilit
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Basic Principle of ProbabilityIf one event has cpossible outcomes, and a
second event has dpossible outcomes, thenthere are cdpossible outcomes of the two
events.
Mutually exclusive outcomes: events inwhich the occurrences of one possibility
excludes all other possibilities
Independent outcomes:events that are
not influenced by each other
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The Sum Rule
Sum rule (either-or rule)used when eventsare mutually exclusive
Example : what is the probability of pickingeither a heart or a spade from a deck of
cards?
P = 13/52 + 13/52 = 26/52 =
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The Product Rule
Productrule(and rule)when occurrenceof one event is independent of another event
-Example 1: what is the probability of
picking a heart and then a spade from a
deck of cards?
P = 13/52 x 13/52 = 169/2704 = 0.0625
-Example 2 : what is the probability of
throwing 2 quarters with both landing on
tails?
Probability Analysis for Monohybrid and
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Probability Analysis for Monohybrid and
Dihybrid Crosses
Monohybrid cross (tall phenotype): DdX Dd Genotypic Ratio : 1 DD:2 Dd:1 dd
Calculate the probability of DdX Ddproducingthe tall phenotype (DD orDd)
P = 1/4 + 2/4 =3/4
Dihybrid cross between yellow, round plants :
WwGgX WwGg
probability of producing an offspring with wrinkledandyellow seeds :
P of wrinkled phenotype = 1/4
P of yellow seeds = 3/4
P (wrinkled andyellow) = 1/4 x 3/4 = 3/16
(independent
assortment)
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The Branched-Line Method to Calculate
Probabilities: An Alternative to Punnett Squares
Branched-line analysis of dihybrid cross:AaBbXAaBb
Independently calculate the probability of each trait Use the Product Rule
U f th B h d Li M th d f
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Use of the Branched-Line Method for a
Trihybrid CrossAaBbCcXAabbCc
Calculate the probability of each phenotypic class.use the product rule with 3 independently assorting genes
Hypothesis Testing: Mendels Crosses
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Hypothesis Testing: Mendel s Crosses
Examination of tall F1plants that are self-crossed
F2plants yielded 787 tall and 277 dwarf
2.84:1 (Mendel interpreted this as 3:1)
Based on probability, can you expect some deviation?
How much deviation is acceptable?
what if F2plants yielded 709 tall and 355 dwarf (2:1 ratio) ?
Where do you draw the line?
Statisticshelps to develop Confidence Limits
- not absolute certainty, but within random chance
- allows us to determine whether results confirm or refute
a hypothesis
Chi Di t ib ti
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Chi-square Distribution
= Greek letter chi
O = observed number for a category
E= expected number for that category
= sum of calculations for all categories
F Progeny from a Monohybrid Cross
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F2Progeny from a Monohybrid Cross
P1: tall (DD) X dwarf (dd)
F1progeny: all tall (Dd)
F1self cross (Dd X Dd)
F2progeny
1064 total
787 tall
277 dwarf
What ratio did you expect?
Test 3:1 and 1:1
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Chi-square Analysis of the F2Progeny
These 2values are meaningless. They need to be
interpreted using the Probability value (p)
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Determination of Probability Values (p)
I t t ti f 2: E i th
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Interpretation of 2: Examine the
Probability Table
The rows represent the degrees of freedomDegrees of freedom equals the number of phenotypic
categories1
This column tells us : the probability is 0.05 of obtaining a 2 value of3.841 or greater bychance alone cannot reject hypothesis
- if p
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