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EVOLUTION:Molecular Genetics
Biology 11 Enriched
Mendelian GeneticsTuesday, December 4th,2 012
Introduction
Human have observed changes in population for many centuries. Humans have domesticated animals and
plants and have artificially selected for various traits in our livestock and crops.
Character – an observable physical feature
Trait – a particular form of a character
Introduction
In the 19th Century, 2 theories on how traits are passed on from one generation to the next:1. THEORY OF BLENDING
INHERITANCE2. THEORY OF PARTICULATE
INHERITANCE
19th Century theories on inheritance
BLENDING INHERITANCE That “hereditary
determinants” (genes) were found in gametes
Physical traits from both parents are blended and the offspring will exhibit an intermediate trait between the two.
PARTICULATE INHERITANCE That both
hereditary determinants remain in fertilized zygote.
Both parental traits will be exhibited in the offspring.
Gregor Mendel
He studied biology, physics, and mathematics
Developed the fundamental laws of heredity Took some ideas from blended
model and particulate model
Mendel chose garden peas (Pisum sativum) as his subjects as they are easily grown and their pollination is easily controlled. He controlled pollination by
manually moving pollen between plants
Funfact: Mendel originally wanted to breed mice, but wasn't allowed to because it was considered scandalous
Gregor Mendel
Pea flowers have both male and female sex organs so Mendel was able to developed “true-breeding” plants by self-pollination.
Mendel examined varieties of peas for inheritable characters and traits for his study. (stem length, pod shape, seed shape, seed color, etc.)
Gregor Mendel
In 1865, Mendel published his findings in a paper called Experiments on Plant Hybridization, which was mostly ignored at the time due to a number of reasons. First, Mendel was not well known in
scientific community. Second, his theory ran against the popular
model of blended inheritance.
Mendel’s Three Laws of Inheritance
Mendel’s Three Laws1. Law of Dominance2. Law of Segregation3. Law of Independent
Assortment
Monohybrid Cross
A monohybrid cross involves one (mono) character and different (hybrid) traits.
Mendel crossed a parental generation with 2 different traits of the same character (in this example, flower color) The F1 seeds were all purple The white flower trait failed to
appear at all. There was also no “blending” of
colors
Monohybrid Cross
Mendel then took purple flowers from the F1 generation and allowed them to self-fertilize.
The flowers in the F2 generation showed a 3:1 ratio of purple:white flowers
Monohybrid Cross
The purple trait completely masks the white trait when true-breeding plants are crossed the purple flower trait
is called dominant the white flower trait is
called recessive. The Law of Dominance
Monohybrid Cross
When Mendel repeated the procedure but for other characters and traits, he would observe a F1 generation with only 1 trait and the 3:1 ratio in the F2 generation.
Mendel’s Conclusion / Law of Segregation
The Law of Segregation Mendel concluded that each gamete
must hold 1 copy of a gene and the zygote will hold 2 copies (1 from each parent) Character = genes traits = alleles of the gene.
But why that 3:1 ratio?
Why 3:1 Ratio?
An organism has 2 alleles of a gene (1 from each parent). If both alleles are the same
homozygous If alleles are different heterozygous
Let’s assign the each allele a letter/symbol.P = purple p = white
Why 3:1 Ratio?
BY CONVENTION:
The dominant trait is given a capital letter, the lowercase of that same letter is the recessive trait. DO NOT MIX LETTERS. Pick one and stick to it.
Also, some letters are better than others. Capital S looks a lot like a lowercase (s). Pick a different letter...
Okay Better (use H for hair)
Short hair = SS HHShort hair = Ss HhLong hair = ss hh
Why 3:1 Ratio?
Both parents are homozygous.
Generation
ParentalGenotype PP x ppPhenotype
Purple x white
Why 3:1 Ratio?
We can use a Punnett Square to show us the allele combinations
Genotype
Male gametes
P P
Female
gametes
p Pp Pp
p Pp Pp
All offspring in the F1 generation are heterozygous dominant
Why 3:1 Ratio?
In his experiment, he then crossed the F1 generation to produce F2 generation
Genotype
Male gametes
P p
Female
gametes
P PP Pp
p Pp pp
3 out of 4 are purple• 1 is
homozygous• 2 are
heterozygous
1 out 4 is white• Homozygous
recessive
Why 3:1 Ratio?
Practice picking letters.... the following traits are found in the common Shirtus canadianus.1. Polka dots are dominant to stripes.
2. Long sleeves are dominant to short sleeves.
3. Collared shirts are recessive.
4. Buttons are dominant over snaps.
5. Pockets are recessive.
Practice with Punnett Squares
1. A round seeded plant (RR) is crossed with a wrinkle seeded plant (rr). What are the phenotypes of the offspring?
2. Two heterozygous purple flowered pea plants are crossed. What are the phenotypes of their offspring and in what proportion?
3. A plant with green seeds (yy) is crossed with a heterozygous plant. What percentage of their offspring have yellow seeds?
In dragons...wings are a dominant trait, but some dragons are born wingless.
1. What are the chances that two heterozygous dragons have a whelp that is wingless?
2. If a wingless dragon is crossed with one that is heterozygous, how many of its offspring will also be wingless?
What is a Test cross?
Help, help! I don't know what my genotype is!!
Am I Dd or DD?
I can help you! Let's have offspring!
D = wingedd = wingless
What is a Test Cross?
Because we know wingless dragons are homozygous recessive, we can breed wingless dragons with a winged dragon. By looking at the ratios, we can tell if the winged
dragon is homozygous or heterozygous
• If female dragon is PP then 100% heterozygous winged dragons
Genotype
Male gametes
p p
Female gametes
P Pp Pp
P Pp Pp
What is a Test Cross?
If female dragon is Pp then a 1:1 ratio is observed.
Genotype
Male gametes
p p
Female gametes
P Pp Pp
p pp pp
Dihybrid Cross
Mendel's Law of Independent Assortment is illustrated by the dihybrid cross The second law describes the outcome of dihybrid
(two character) crosses, or hybrid crosses involving additional characters.▪ A dihybrid is an individual that is a double
heterozygote (e.g., with the genotype RrYy - round seed, yellow seed).R = round/r = bumpy, Y = yellow/y = green
▪ What are the gametes that can be produced by an individual that is RrYy?
▪RY, Ry, rY, ry
Dihybrid Cross (RrYy x RrYy)
RY Ry rY ry
RYRRYY
round, yellow
RRYyround, yellow
RrYYround, yellow
RyYyround, yellow
RyRRYy
round, yellow
RRyyround, green
RrYyround, yellow
Rryyround, green
rYRrYY
round, yellow
RrYyround, yellow
rrYYbumpy, yellow
rrYybumpy, yellow
ryRrYy
round, yellow
Rryyround, green
rrYybumpy, yellow
rryybumpy, green
Dihybrid Cross
The ratio that is seen is a 9:3:3:1 ratio a total of 4 phenotypes are
observed:▪ 9 round, yellow▪ 3 round, green▪ 3 bumpy, yellow▪ 1 bumpy, green (double recessive)
Dihybrid Cross – Practice
You have 30 minutes to complete as much of this package as you can.
At 2:20, we are moving to notes.
Probability and Inheritance
A Punnet Square for a dihybrid cross is pretty epic! a Punnett Square is helpful for 1 or 2
genes but a little troublesome for more characters▪ A trihybrid cross needs 64 boxes▪ A tetrahybrid cross needs 256 boxes▪ TOO MUCH EPIC!
Probability and Inheritance
Mendel’s laws of segregation and independent assortment reflect the rules of probability
Probability and Inheritance
Remember that the alleles of different (and unlinked) traits ending up in a gamete is independent of the chances of other alleles.
To find the probability of an series of INDEPENDENT events happening together, the individual probabilities of the events are multiplied together:
P(A and B) = P(A) x P(B)
Probability and Inheritance
If an event is absolutely certain to happen, its probability is 1.
If it cannot possibly happen, its probability is 0.
All other events have a probability between 0 and 1.
Probability and Inheritance
An example of independent events is the flipping of a coin.
Each flip of a coin is independent from the previous or next flips. They don’t influence each other.
Probability and Inheritance
P(A and B) = P(A) x P(B)
What is the probability of getting 5 “tails” in a row?
If P(tails) = 0.5 (or ½)
P(5 tails) = ½ x ½ x ½ x ½ x ½ = 1/32 or 0.03125 (never use
%)
Rr RrSegregation of
alleles into eggs
Sperm
Segregation ofalleles into sperm
Eggs
R
R
R RR
R rrr
r
r
r1/2
1/2
1/2
1/2
1/41/4
1/41/4
Heads ¼ + ¼ + ¼ = ¾
Tails1/4
Probability and Inheritance We used the rule of multiplication to find
the probability of a certain genotype. We used the rule of addition to find the
probability of a certain phenotype.
In calculating the chances for various genotypes, each character is considered separately, and then the individual probabilities are multiplied together
Calculate the probability that an F2 seed will be spherical and yellow.
Remember, calculate each trait separately. (create 2 monohybrid squares rather than 1 dihybrid square)
P(yellow,round) = ¾ yellow x ¾ round = 9/16
P(yellow, wrinkled) = ¾ yellow x ¼ wrinkled = 3/16P(green, round) = ¼ green x ¾ round = 3/16P(green,wrinkled) = ¼ green x ¼ wrinkled = 1/16
A 9:3:3:1 ratio!!
For any gene with a dominant allele A and recessiveallele a, what proportions of the offspring from anAA x Aa cross are expected to be homozygous dominant, homozygous recessive, and heterozygous?
Two organisms, with genotypes BbDD and BBDd are mated. Assuming independent assortment of the B/b and D/d genes, write the genotypes of all possible offspring from this cross and use the rules of probability to calculate the chance of each genotype occurring.
Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIi x ppYyii).
What fraction of offspring are predicted to be homozygous recessive for at least two of the three characters?
Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIi x ppYyii).
What fraction of offspring are predicted to be homozygous recessive for at least two of the three characters?P p
pPp (1/4)
pp (1/4)
pPp (1/4)
pp (1/4)
Pp = ¼ + ¼ = ½pp = ¼ + ¼ = ½
Y y
YYY(1/4)
Yy (1/4)
yYy (1/4)
Yy (1/4)
YY = ¼Yy = ¼ + ¼ = ½yy = ¼
I i
iIi (1/4)
ii (1/4)
iIi (1/4)
ii (1/4)Ii = ¼ + ¼ = ½
ii = ¼ + ¼ = ½
Inheritance patterns are often more complex than predicted by simple Mendelian genetics
Extension on Mendel
The relationship between genotype and phenotype is rarely as simple as in the pea plant characters Mendel studied Many heritable characters are not
determined by only 1 gene with 2 alleles.
However, the basic principles of segregation and independent assortment apply even to more complex patterns of inheritance
Extending Mendelian Genetics for a Single Gene
Inheritance of characters by a single gene may deviate from simple Mendelian patterns in the following situations: When alleles are not completely
dominant or recessive When a gene has more than two
alleles When a gene produces multiple
phenotypes
Degrees of Dominance
1. Complete dominance occurs when phenotypes of the heterozygote and dominant homozygote are identical
2. In incomplete dominance, the phenotype of F1 hybrids is somewhere between the phenotypes of the two parental varieties
3. In codominance, two dominant alleles affect the phenotype in separate, distinguishable ways
Fig. 14-10-1
Red
P Generation
Gametes
WhiteCRCR CWCW
CR CW
Fig. 14-10-2
Red
P Generation
Gametes
WhiteCRCR CWCW
CR CW
F1 GenerationPinkCRCW
CR CWGametes 1/21/2
Fig. 14-10-3
Red
P Generation
Gametes
WhiteCRCR CWCW
CR CW
F1 GenerationPinkCRCW
CR CWGametes 1/21/2
F2 Generation
Sperm
Eggs
CR
CR
CW
CW
CRCR CRCW
CRCW CWCW
1/21/2
1/2
1/2
Frequency of Dominant Alleles
Dominant alleles are not necessarily more common in populations than recessive alleles For example, one baby out of 400 in the
United States is born with extra fingers or toes▪ The allele for this unusual trait is dominant to
the allele for the more common trait of five digits per appendage▪ In this example, the recessive allele is far more
prevalent than the population’s dominant allele
Multiple Alleles
Most genes exist in populations in more than 2 allelic forms
For example, blood typesThe 4 blood types are: Type A Type B Type AB Type O
Multiple Alleles: Blood Types
the 4 phenotypes of the ABO
blood group in humans are
determined by 3 alleles for
the enzyme (I) that attaches
A or B carbohydrates to red
blood cells: IA, IB, and i. A alleles = Carbohydrate A
B alleles = Carbohydrate B
i alleles = no carbohydrates
IA
IB
i
A
B
none(a) The three alleles for the ABO blood groups and their associated carbohydrates
Allele Carbohydrate
GenotypeRed blood cell
appearancePhenotype
(blood group)
IAIA or IA i A
BIBIB or IB i
IAIB AB
ii O
(b) Blood group genotypes and phenotypes
Blood types is also an example of codominance.
Pleiotropy
Most genes have multiple phenotypic effects a property called pleiotropy For example, pleiotropic alleles are
responsible for the multiple symptoms of certain hereditary diseases, such as cystic fibrosis and sickle-cell disease
Extending Mendelian Genetics for Two or More Genes
Some traits may be determined by two or more genes
Epistasis
In epistasis, a gene at one locus alters the phenotypic expression of a gene at a 2nd locus For example, in mice and many other
mammals, coat color depends on two genes▪ One gene determines the pigment color
(BB/Bb = black & bb = brown)▪ The other gene determines whether the pigment
will be deposited in the hair(CC/Cc = color & cc = no color)
Fig. 14-12
BbCc BbCc
Sperm
EggsBC bC Bc bc
BC
bC
Bc
bc
BBCC
1/41/4
1/41/4
1/4
1/4
1/4
1/4
BbCC BBCc BbCc
BbCC bbCC BbCc bbCc
BBCc BbCc
BbCc bbCc
BBcc Bbcc
Bbcc bbcc
9 : 3 : 4
Polygenic Inheritance
Quantitative characters are those that vary in the population along a continuum/spectrum Quantitative variation usually indicates
polygenic inheritance, an additive effect of two or more genes on a single phenotype
Fig. 14-13
Eggs
Sperm
Phenotypes:
Number ofdark-skin alleles: 0 1 2 3 4 5 6
1/646/64
15/6420/64
15/646/64
1/64
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/81/8
1/81/8
1/81/8
1/81/8
AaBbCc AaBbCc
Skin color in humans is an example of polygenic inheritance
Nature and Nurture: The Environmental Impact on Phenotype
Another departure from Mendelian genetics arises when the phenotype for a character depends on environment AND genotype The norm of reaction is the phenotypic
range of a genotype influenced by the environment
Fig. 14-14
For example, hydrangea flowers of the same genotype range from blue-violet to pink, depending on soil acidity
Nature and Nurture
An organism’s phenotype includes its: physical appearance internal anatomy physiology behavior
An organism’s phenotype reflects its overall genotype and unique environmental history
Gene Linkage
Some genes do not sort independently They are often inherited together
because they are on the same chromosome.
Gene Linkage
New allele combinations can be produced if chromosomes cross-over during meiosis Recombinant
chromosomes and gametes are produced
Sex-Linkage
Females: XX Males: XY
Some genes are found only on the X chromosome. Because males are heterogametic (X
and Y) and hemizygous (only 1 X), whatever genes are on the X chromosome will be expressed
Sex Linkage – Color blindness The gene for seeing color has 2
alleles: B and b. The loci for this gene is on the X chromosome
Genotype Phenotype
XBXB Color
XBXbColor / carrier for color blindness
XbXb Color-blind
Genotype Phenotype
XBY Color
XbY Color-blind
Mendelian Genetics and Humans
Humans are not good subjects for genetic research Generation time is too long Parents produce relatively few
offspring Breeding experiments are
unacceptable
However, basic Mendelian genetics endures as the foundation of human genetics
Pedigree Analysis
A pedigree is a family tree that describes the interrelationships of parents and children across generations
Pedigree analysis allows us to figure out whether an allele controlling a particular phenotype is dominant or recessive.
KeyMale
Female
AffectedmaleAffectedfemale
Mating
Offspring, inbirth order(first-born on left)
1st generation(grandparents)
2nd generation(parents, aunts,and uncles)
3rd generation(two sisters)
Ww ww ww Ww
Ww ww wwWw Ww ww
ww
Ww
WWor
Widow’s peak No widow’s peak(a) Is a widow’s peak a dominant or recessive trait?
1st generation(grandparents)
2nd generation(parents, aunts,and uncles)
3rd generation(two sisters)
Ff Ff Ff
Ff FfFfFF or ff ff
ff
ff
ff FForFf
Attached earlobe Free earlobe
(b) Is an attached earlobe a dominant or recessive trait?
Dominant or Recessive?
Dominant trait:Every affected individual (G2) has
an affected parent (G1)About ½ of offspring (G3) are
affected
Fig. 14-15a
KeyMale
Female
AffectedmaleAffectedfemale
Mating
Offspring, inbirth order(first-born on left)
Fig. 14-15b
1st generation(grandparents)
2nd generation(parents, aunts,and uncles)
3rd generation(two sisters)
Widow’s peak No widow’s peak
(a) Is a widow’s peak a dominant or recessive trait?
Ww ww
Ww Wwww ww
ww
wwWw
Ww
wwWW
Wwor
Dominant or Recessive?
Recessive trait:Affected people have parents that
are not affected (skip generations)
Fig. 14-15c
Attached earlobe
1st generation(grandparents)
2nd generation(parents, aunts,and uncles)
3rd generation(two sisters)
Free earlobe
(b) Is an attached earlobe a dominant or recessive trait?
Ff Ff
Ff Ff Ff
ff Ff
ff ff ff
ff
FF or
orFF
Ff
Pedigree Analysis
Pedigrees can also be used to make predictions about future offspring
We can use the multiplication and addition rules to predict the probability of specific phenotypes
The Hardy-Weinberg TheoremBiology 11 EnrichedTuesday, December 18th, 2012
The Hardy-Weinberg Theorem The Hardy-Weinberg Equilibrium is a
mathematical theory that describes, in detail, the conditions that must be met for evolution to not occur (for allele frequencies to remain the same) Thus, it is a null hypothesis with which
natural populations (that are NEVER at H-W equilibrium) can be compared to.
Useful model to measure if forces are acting on a population▪ Measuring evolutionary change
H-W Equilibrium
For a population to NOT evolve, the following conditions MUST be met:1. Mating is random2. Large population3. No movement in to or out of
population4. No mutation5. No natural selection
H-W Equilibrium
If any of the 5 conditions for maintaining a Hardy-Weinberg equilibrium are not met, then evolution must be occurring.
Of course, none of these conditions is ever permanently met in any known natural population of organisms: Mutations occur at a slow but steady rate in all known
populations. Many organisms, especially animals, enter (immigration)
and leave (emigration) populations. Most populations are not large enough to be unaffected
by random changes in allele frequencies. Survival is virtually never random. Reproduction in organisms that can choose their mates
is also virtually never random.
H-W Equilibrium
Therefore, according to the Hardy-Weinberg Equilibrium Law, evolution (defined as changes in allele frequencies over time) must be occurring in virtually every population of living organisms.
In other words, “Evolution is as ubiquitous and inescapable as gravity.”
H-W Equilibrium
Hardy-Weinberg Theorem: Assumes 2 alleles (B,b)▪ Frequency of the dominant allele (B) = p▪ Frequency of the recessive allele (b) = q
Frequencies in a gene pool must add up to 1 so:
p + q = 1
p + q = 1
B
B BB
Bb
Bb
b
b bb
p = B alleleq = b allele
p + q = 1
H-W Equilibrium
If we break down frequencies of individual genotypes, then: Frequency of homozygous dominant = p x p = p2
Frequency of homozygous recessive = q x q = q2
Frequency of heterozygous = (p + q) + (q + p) = 2pq
Frequencies of individuals must add up to 1:
p2 + 2pq + q2 = 1
p2 + 2pq + q2 = 1
Homozygous
dominant
Homozygous
recessiveHeterozygous
H-W formulas
Alleles in a gene pool: p + q = 1
Individuals: p2 + 2pq + q2
= 1 bbBbBB
BB
B b
Bb bb
What are the genotype frequencies?What are the genotype frequencies?
Using Hardy-Weinberg Equation
q2 (bb): q (b): p (B):
q2 (bb): q (b): p (B):
population: 100 cats84 black, 16 whiteHow many of each allele?
population: 100 cats84 black, 16 whiteHow many of each allele?
bbBbBB
p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16
Must assume population is in H-W equilibrium!
Must assume population is in H-W equilibrium!
Using Hardy-Weinberg Equation
The following 4 examples will allow you to practice solving HW Equilibrium questions.
q2 q p p2 2pq
p + q = 1p2 + 2pq + q2 = 1
In a population of pigs color is determined by one gene. If the black allele (b) is recessive and the white allele (B) is dominant, what is the frequency of the black allele in this population?
q2 q p p2 2pq
p + q = 1q2 + 2pq + q2 = 1
In a population of 1000 fruit flies, 640 have red eyes and the remainder have sepia eyes. The sepia eye trait is recessive to red eyes. How many individuals would you expect to be homozygous for red eye color?
q2 q p p2 2pq
p + q = 1p2 + 2pq + q2 = 1
In a population of squirrels, the allele that causes bushy tail (B) is dominant, while the allele that causes bald tail (b) is recessive. If 91% of the squirrels have a bushy tail, what is the frequency of the dominant allele?
q2 q p p2 2pq
p + q = 1p2 + 2pq + q2 = 1
In the U.S. 1 out of 10,000 babies are born with Phenylketonuria, a recessive disorder that results in mental retardation if untreated. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele?
Using Hardy-Weinberg equation
bbBbBB
p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16
Assuming H-W equilibrium
Assuming H-W equilibrium
Sampled data Sampled data bbBbBB
p2=.74p2=.74 2pq=.102pq=.10 q2=.16q2=.16
How do you explain the data?
How do you explain the data?
p2=.20p2=.20 2pq=.642pq=.64 q2=.16q2=.16
How do you explain the data?
How do you explain the data?
Null hypothesis Null hypothesis
Application of HW Theorem:
Sickle cell anemia inherit a mutation in gene coding for
hemoglobin▪ oxygen-carrying blood protein▪ recessive allele = HsHs / normal allele = Hb
low oxygen levels causes RBC to sickle breakdown of RBC clogging small blood vessels damage to organs
often lethal
Sickle cell frequency
High frequency of heterozygotes 1 in 5 in Central Africans = HbHs
unusual for allele with severe detrimental effects in homozygotes▪ 1 in 100 = HsHs
▪ usually die before reproductive age
Why is the Hs allele maintained at such high levels in African populations?Why is the Hs allele maintained at such high levels in African populations?
Suggests some selective advantage of being heterozygous…Suggests some selective advantage of being heterozygous…
Malaria Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells
Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells
1
2
3
Heterozygote Advantage In tropical Africa, where malaria is common:
homozygous dominant (normal) : HbHb
▪ die or reduced reproduction from malaria
homozygous recessive: HsHs
▪ die or reduced reproduction from sickle cell anemia
heterozygote carriers are relatively free of both: HbHs
▪ survive & reproduce more, more common in population
Hypothesis:In malaria-infected cells, the O2 level is lowered enough to cause sickling which kills the cell & destroys the parasite.
Hypothesis:In malaria-infected cells, the O2 level is lowered enough to cause sickling which kills the cell & destroys the parasite.
Frequency of sickle cell allele & distribution of malaria