Upload
allison-mckenna
View
215
Download
0
Tags:
Embed Size (px)
Citation preview
Mendelian Genetics
Mendel’s Experiments
Gregor Mendel Joined the Augustinian Monastery at the age of
21 Taught in a secondary school, was fascinated with
science and nature (physics, evolution, botany and natural sciences)
Attended the University of Vienna to study physics and biology
Returned to the monastery and began his experiments with the common garden plant (Pisum sativum)
Mendel’s Experimental, Quantitative Approach
• Advantages of pea plants for genetic study: Many varieties with distinct heritable features, or
characters (such as flower color); character variants (such as purple or white flowers) are called traits
Mating of plants can be controlled
Each pea plant has sperm-producing organs (stamens) and egg-producing organs (carpels)
Cross-pollination (fertilization between different plants) can be achieved by dusting one plant with pollen from another
Fig. 14-2
TECHNIQUE
RESULTS
Parentalgeneration(P) Stamens
Carpel
1
2
3
4
Firstfilialgener-ationoffspring(F1)
5
Mendel’s Experiments
Mendel’s Experimental Design Crossed only peas with desired traits, two methods
Self-fertilization – pollen from anther falls onto the stigma of the same flower before it opens
Cross-fertilization – pollen from one plant fertilizes another Mendel opened the keel and removed the anther
before self-fertilization could occur Selected 7 discrete, nonoverlapping characteristics
Flower color – purple or white
Table 14-1
Mendel’s Experiments
Mendel’s Experimental Design Grew the plants for two years to identify
homogenous, pure-breeding characteristics Started by crossing two pure-breeding plants (the
parent generation, P), one purple, one white The offspring (first filial generation, F1), were
referred to as hybrids (mixture of both parents) Monohybrids – hybrid of only one characteristic
All the F1 plants were purple – purple flower color is a dominant trait Thus the white flower color trait is recessive
Mendel’s Experiments
Mendel’s Experimental Design He allowed the F1 to self-fertilize
The resulting second filial generation (F2) showed both dwarf and tall characteristics 705 purple, 224 white, a ratio of 3:1
Mendel did not recognize that the traits were controlled by genes (term coined in 1909)
He did propose that each trait relied on two related, but different, determinants. Alleles represent different forms of a gene
Mendel’s Experiments
Mendel’s Experimental Design Phenotype – observable characteristics
Allele for purple flower plants (P) is dominant over white flower plants (p)
Genotype - combination of alleles an organism possesses Homozygous purple– PP (Both alleles are the same) Heterozygous purple– Pp (Two alleles are different)
Fig. 14-3-3EXPERIMENT
P Generation
(true-breeding parents) Purple
flowers Whiteflowers
F1 Generation
(hybrids) All plants hadpurple flowers
F2 Generation
705 purple-floweredplants
224 white-floweredplants
Fig. 14-5-3
P Generation
Appearance:Genetic makeup:
Gametes:
Purple flowers White flowersPP
P
pp
p
F1 Generation
Gametes:
Genetic makeup:Appearance: Purple flowers
Pp
P p1/21/2
F2 Generation
Sperm
Eggs
P
PPP Pp
p
pPp pp
3 1
Segregation
Law of Segregation During gamete formation the alleles will separate
randomly Fertilization is the fusion of two gametes,
reestablishing the two copies of a gene
Allele for purple flowers
Homologouspair ofchromosomesLocus for flower-color gene
Allele for white flowers
Segregation
Law of segregation explains Mendel’s experiments with four related concepts: 1. Alternate versions of a gene accounts for
variations in inherited characters 2. For each character, an organism inherits
two alleles, one from each parent
Segregation
Law of segregation explains Mendel’s experiments with four related concepts:
3. If two alleles at a locus differ, then one, the dominant allele, determines the organisms appearance, the other, the recessive allele, has no noticeable effect on the organisms appearance.
4. The two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes.
How are each of the 4 parts of Mendel’s law of segregation portrayed in his experiment with crossing peas plants with different flower colors?
• Mendel’s segregation model accounts for the 3:1 ratio he observed in the F2 generation of his numerous crosses
• The possible combinations of sperm and egg can be shown using a Punnett square, a diagram for predicting the results of a genetic cross between individuals of known genetic makeup
• A capital letter represents a dominant allele, and a lowercase letter represents a recessive allele
Segregation
Segregation
Testing the Law of Segregation Along with the 3:1 phenotype ratio, there should also
be a 1:2:1 genotypic ratio 1 PP, 2 Pp, 1 pp
This can be tested by self-fertilizing the F2 to create an F3
The resulting whites should be homozygous The purple F2 plants should be 1/3 homozygous and 2/3
heterozygous The homozygous should produce only purple flower
plants The heterozygous should produce 3 purple flower
plants for each 1 white flower plant (3:1)
Fig. 14-6Phenotype
Purple
Purple3
Purple
Genotype
1 White
Ratio 3:1
(homozygous)
(homozygous)
(heterozygous)
(heterozygous)
PP
Pp
Pp
pp
Ratio 1:2:1
1
1
2
Segregation
Testing the Law of Segregation Another way to test is by using a testcross – cross
any organism with a homozygous recessive If the organism in question is homozygous dominant,
then all progeny will have the dominant phenotype If the organism is heterozygous, then the progeny will be
50% phenotypically dominant, and 50% phenotypically recessive
Practicing a Test Cross
Draw punnett square crossing homozygous white flower pea plant with heterozygous purple flower pea plant. What are the possible offspring? Can you determine the genotypes of the offspring?
Draw punnett square crossing homozygous white flower pea plant with homozygous purple flower pea plant. What are the possible offspring? Can you determine the genotypes of the offspring?
Fig. 14-7
TECHNIQUE
RESULTS
Dominant phenotype, unknown genotype:
PP or Pp?
Predictions
Recessive phenotype, known genotype: pp
If PP If Ppor
Sperm Spermp p p p
P
P
P
p
Eggs Eggs
Pp
Pp Pp
Pp
Pp Pp
pp pp
or
All offspring purple 1/2 offspring purple and1/2 offspring white
Independent Assortment
Mendel analyzed the inheritance of two different traits Homozygous round yellow seeds (YY) crossed with
homozygous wrinkled(rr) green seeds The F1 (dihybrid) were all round yellow seeds
Dihybrid-individuals that are heterozygous for two characters
When the F1 was self-fertilized, the resulting F2 had all four combinations of characteristics
The ratio is very close to 9:3:3:1
Fig. 14-8
EXPERIMENT
RESULTS
P Generation
F1 Generation
Predictions
Gametes
Hypothesis ofdependentassortment
YYRR yyrr
YR yr
YyRr
Hypothesis ofindependentassortment
orPredictedoffspring ofF2 generation
Sperm
Sperm
YR
YR
yr
yr
Yr
YR
yR
Yr
yR
yr
YRYYRR
YYRR YyRr
YyRr
YyRr
YyRr
YyRr
YyRr
YYRr
YYRr
YyRR
YyRR
YYrr Yyrr
Yyrr
yyRR yyRr
yyRr yyrr
yyrr
Phenotypic ratio 3:1
EggsEggs
Phenotypic ratio 9:3:3:1
1/21/2
1/2
1/2
1/4
yr
1/41/4
1/41/4
1/4
1/4
1/4
1/43/4
9/163/16
3/161/16
Phenotypic ratio approximately 9:3:3:1315 108 101 32
Independent Assortment
Punnett squares are used to visualize possible gamete fusions Assumption: Four types of gametes from each dihybrid
parent will be produced in equal numbers
WG Wg wG wg
WG WWGG
WWGg
WwGG
WwGg
Wg WWGg WWgg
WwGg
Wwgg
wG WwGG WwGg
wwGG
wwGg
wg WwGg Wwgg wwGg wwgg
Independent Assortment
Law of Independent Assortment Alleles for one gene can segregate independently of
alleles for other genes Each phenotypic class is made of several different
genotypes Except the homozygous recessive
The genotypic ratio is 1:2:1:2:4:2:1:2:1WG Wg wG wg
WG WWGG
WWGg
WwGG
WwGg
Wg WWGg WWgg
WwGg
Wwgg
wG WwGG WwGg
wwGG
wwGg
wg WwGg Wwgg wwGg wwgg
Independent Assortment
Testing the Law of Independent Assortment This law can be tested by doing a dihybrid testcross
Review: monohybrid heterozygous testcross resulted in a 1:1 phenotypic ratio
Testcross WwGg with wwgg The result is a 1:1:1:1 phenotypic ratio
• Strictly speaking, this law applies only to genes on different, nonhomologous chromosomes
• Genes located near each other on the same chromosome tend to be inherited together
Independent Assortment
Laws of Probability
Mendel’s laws of segregation and independent assortment reflect the rules of probability
Probability
Types of Probability Probability (P) = Number of times an event is
observed (a) / the total number of possible cases (n) P=a/n
Examples: Probability of rolling a 4 on a six-sided dice
P=1/6 Probability of drawing a 7 of clubs from a card deck
P=1/52
Probability
Types of Probability An event that is certain has a probability of 1
P=1/1 An event that is impossible has a probability of 0 An event has a probability of P, the likelihood of the
alternative event is Q=1-P Probability of rolling a 4 on a six-sided dice
P=1/6 Probability of rolling anything else
Q=1-1/6= 5/6 The probability of all possible events must equal 1
P+Q=1
Probability
Types of Probability Mutually exclusive outcomes
Event in which the occurrence of one possibility excludes all other possibilities Rolling a dice, only one side can face up
Independent outcomes Events that do not influence one another
Rolling two dice, the face value of one does not influence the other
Two rules of probability that affect genetics Sum rule and product rule
Probability
Sum Rule When events are mutually exclusive The probability that one of several mutually exclusive
events will occur is the sum of the probabilities Probability of a dice showing either a 4 or 6
P=1/6 + 1/6 = 2/6 = 1/3 The probability increases as the number of possible
outcomes increase Probability of a dice showing any number
P=1/1 Not used for traits expressed on a continuum (human
heights)
Probability
Product Rule When one event is independent of other events The probability that two events will both occur is the
product of their separate probabilities Probability of throwing a die two times and getting a 4 and
then a 6 P=1/6 x 1/6 = 1/32
The probability will decrease as you increase the number of independent events Much like the lottery, the more numbers you need the less
likely you are to win
Probability
Using Probabilities In a monohybrid cross, Dd X Dd (1:2:1), what is the
probability of getting either a homozygous dominant or heterozygous? P= ¼ + ½ = ¾
In a dihybrid cross WwGg X WwGg, what is the probability of getting a round and green seed? P=3/4 x 1/4 = 3/16
Solving Complex Genetics Problems
We can apply the rules of probability To predict the outcome of crosses involving multiple characters
A dihybrid or other multicharacter cross Is equivalent to two or more independent monohybrid crosses occurring simultaneously
In calculating the chances for various genotypes from such crosses Each character first is considered separately and then the individual probabilities are multiplied together
Probability
Branch-Line Approach to Calculate Probabilities Punnett squares require 16 squares for a dihybrid
cross, 64 squares for a trihybrid Punnett squares are useful, but hard to use with more
complex crosses The branch-line approach is based on the law of
independent assortment In branch-line each trait is examined independently
Probability
Branch-Line Approach to Calculate Probabilities AaBb x AaBb To determine the probability of this dihybdrid cross
calculate the probability of each trait and apply the product ruleA phenotype
3/4 (either AA or Aa)
B phenotype ¾ (either BB or Bb)
AB phenotype 9/16 (A-B- genotype)
b phenotype 1/4 (bb genotype)
Ab phenotype 3/16 (A-bb genotype)
a phenotype 1/4 (aa genotype)
B phenotype ¾ (either BB or Bb)
aB phenotype 3/16 (aaB- genotype)
b phenotype 1/4 (bb genotype)
ab phenotype 1/16 (aabb genotype)
Probability
Branch-Line Approach to Calculate Probabilities Use branch-line to determine the probabilities of this
trihybrid cross: AaBbCc x AabbCc
A phenotype 3/4
a phenotype 1/4
B phenotype 1/2
b phenotype 1/2
B phenotype 1/2
b phenotype 1/2
C phenotype 3/4
c phenotype 1/4
C phenotype 3/4
c phenotype 1/4
C phenotype 3/4
c phenotype 1/4
C phenotype 3/4
c phenotype 1/4
ABC phenotype 9/32
ABc phenotype 3/32
AbC phenotype 9/32
Abc phenotype 3/32
aBC phenotype 3/32
aBc phenotype 1/32
abC phenotype 3/32
abc phenotype 1/32
Probability
Branch-Line Approach to Calculate Probabilities The branch-line approach can also be used to determine
genotypes
Statistics
When dealing with probabilistic events, there is a chance the data will cause us to support a bad hypothesis
Mendel’s F2 heterozygous pea plants yielded 788 tall and 277 dwarf. 2.84:1 not exactly 3:1
Is 2.84:1 close enough to represent 3:1?
Statistics
Hypothesis Testing Statistics are used by scientists to summarize data
and test their hypothesis by comparing data predicted results
To determine if the data is consistent with the hypothesis we generate a null hypothesis The null hypothesis assumes the difference between the
observed and expected results are due to chance
Statistics
Chi-Square This test is used when data is distributed among
discrete categories (tall and dwarf plants) Formula for the chi-square
χ is the Greek letter chi, O is the observed number for a category, E is the expected number for the category, and ∑ means to sum the calculations for all categories
χ2=∑(O – E)2
E
Statistics
Chi-Square Example: Mendel observed 787 tall plants and 277
dwarf. Total of 1064. The expected results for a 3:1 ratio are 798 tall and 266 dwarf. Start with figuring the chi-square for tall plants
Then figure it for the dwarf plants
Now sum the two results
(787 – 798)2
798= 0.15
(277 – 266)2
266= 0.45
0.45 + 0.15 = 0.60
Statistics
Chi-Square Example: Use the same data, but assume we were
testing for a 1:1 ratio. How does the chi-square change? Mendel observed 787 tall plants and 277 dwarf. Total of 1064. Start with figuring the chi-square for tall plants
Then figure it for the dwarf plants
Now sum the two results
(787 – 532)2
532= 122.23
(277 – 532)2
532= 122.23
122.23 + 122.23 = 244.45
Statistics
Chi-Square To properly use chi-square values we need to convert
them to a probability value (p) As the number of categories increases so will the chi-
square value (because we sum each category) To help solve this we use degrees of freedom (# of
categories minus 1) Example: two categories would have 1 degree 0f
freedom Consult the chi-square table
Statistics
Chi-Square Table
Probabilities
Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01
1 0.000 0.004 0.064 0.455 1.642 3.841 6.635
2 0.020 0.103 0.446 1.386 3.219 5.991 9.210
3 0.115 0.352 1.005 2.366 4.642 7.815 11.345
4 0.297 0.711 1.649 3.357 5.989 9.488 13.277
5 0.554 1.145 2.343 4.351 7.289 11.070 15.086
6 0.872 1.635 3.070 5.348 8.558 12.592 16.812
Statistics
Chi-Square Table
We will look a p value of 0.05We have only 1 degree of freedom
Probabilities
Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01
1 0.000 0.004 0.064 0.455 1.642 3.841 6.635
2 0.020 0.103 0.446 1.386 3.219 5.991 9.210
3 0.115 0.352 1.005 2.366 4.642 7.815 11.345
4 0.297 0.711 1.649 3.357 5.989 9.488 13.277
5 0.554 1.145 2.343 4.351 7.289 11.070 15.086
6 0.872 1.635 3.070 5.348 8.558 12.592 16.812
Statistics
Chi-Square Table
There is a 0.05 probability of getting a χ2 value of 3.841 or larger by chance alone, given the hypothesis is correct
Probabilities
Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01
1 0.000 0.004 0.064 0.455 1.642 3.841 6.635
2 0.020 0.103 0.446 1.386 3.219 5.991 9.210
3 0.115 0.352 1.005 2.366 4.642 7.815 11.345
4 0.297 0.711 1.649 3.357 5.989 9.488 13.277
5 0.554 1.145 2.343 4.351 7.289 11.070 15.086
6 0.872 1.635 3.070 5.348 8.558 12.592 16.812
Inheritance Patterns
Inheritance patterns are often more complex than predicted by simple Mendelian genetics
The relationship between genotype and phenotype is rarely simple
Extending Mendelian Genetics for a Single Gene
The inheritance of characters by a single gene may deviate from simple Mendelian patterns
The Spectrum of Dominance Complete dominance
Occurs when the phenotypes of the heterozygote and dominant homozygote are identical
Codominance Two dominant alleles affect the phenotype in separate,
distinguishable ways Ex. The human blood group MN is an example of codominance
The Spectrum of Dominance
Incomplete dominance The phenotype of F1
hybrids is somewhere between the phenotypes of the two parental varieties
P Generation
F1 Generation
F2 Generation
RedCRCR
Gametes CR CW
WhiteCWCW
PinkCRCW
Sperm
CR
CR
CR
Cw
CR
CRGametes
1⁄2 1⁄2
1⁄2
1⁄2
1⁄2
Eggs1⁄2
CR CR CR CW
CW CWCR CW
The Relation Between Dominance and Phenotype
Dominant and recessive alleles do not really interact; it is in the pathway from genotype to phenotype that dominance and recessiveness come into play
Lead to synthesis of different proteins that produce a phenotype
Examples: Mendel’s Pea Shape, Tay-Sachs disease
Dominant alleles are not always the most common in a population
• Example: polydactyly (+5 digits)
Multiple Alleles
Most genes exist in populations In more than two allelic
forms
The ABO blood group in humans Is determined by
multiple alleles
Table 14.2
Blood Types
Glycoproteins on surface determine blood type; Important in transfusions/transplants
IA and IB are codominant ii (type O) is recessive to A or BType O = universal donorType AB= universal recipient
Differences in Rh factor (Mom Rh- and baby Rh+) can result in erythroblastosis fetalis
Pleiotropy
Pleiotropy-a gene has multiple phenotypic effects Ex. Hereditary diseases such
as cystic fibrosis & sickle cell disease
Extending Mendelian Genetics for Two or More Genes
Some traits may be determined by two or more genes
epistasis- a gene at one locus alters the phenotypic expression of a gene at a second locus
polygenic inheritance- an additive effect of two or more genes on a single phenotype
EPISTASIS
EX: Coat color in miceB = Black b = brown
C = color deposited in coatc = color NOT deposited
cc-mouse looks white eventhough it has color genes
Image from Biology; Campbell and Reece; Pearson Prentice Hall publishing as Benjamin Cummings © 2005
An example of epistasis
BC bC Bc bc1⁄41⁄41⁄41⁄4
BC
bC
Bc
bc
1⁄4
1⁄4
1⁄4
1⁄4
BBCc BbCc BBcc Bbcc
Bbcc bbccbbCcBbCc
BbCC bbCC BbCc bbCc
BBCC BbCC BBCc BbCc
9⁄163⁄16
4⁄16
BbCc BbCc
Sperm
Eggs EX: Coat color in miceB = Black b = brown
C = color deposited in coatc = color NOT deposited
cc-mouse looks white eventhough it has color genes
Polygenic Inheritance
Many human characters Vary in the population along a continuum and are called
quantitative characters
Quantitative variation usually indicates polygenic inheritance An additive effect of two or more genes on a single
phenotype
POLYGENIC traits are recognizable by their expression as a gradation of small differences (a continuous variation).
The results form a bell shaped curve.
Nature and Nurture: The Environmental Impact on Phenotype
Another departure from simple Mendelian genetics arises When the phenotype for a character depends on environment as well as
on genotype
The norm of reaction Is the phenotypic range of a particular genotype that is influenced by
the environment
Environment influences Phenotype“Nature vs Nurture”
Siamese cats and Himalayan rabbits have dark colored fur on their extremities
Allele that controls pigment production is only able to function at the lower temperatures of those extremities.
Integrating a Mendelian View of Heredity and Variation
Multifactorial characters Are those that are influenced by both genetic and
environmental factors
An organism’s phenotype Includes its physical appearance, internal anatomy,
physiology, and behavior Reflects its overall genotype and unique
environmental historyEven in more complex inheritance patterns
Mendel’s fundamental laws of segregation and independent assortment still apply
Sex Linked Genes
Genes carried on the X chromosome are called X-linked traits.
Red-green colorblindness, hemophilia, an Duchenne muscular dystropy are examples of X-linked traits.
Y-LINKED GENES: Genes carried on the Y chromosome
Y-linked genes only show up in MALES
Hairy pinnaeSRY geneinitiates male sexdetermination
X and y chromosomes
NON-HOMOLOGOUS partners
Pedigrees are diagrams that show how genes are passed on in families over several generations
Pedigrees can be used to predict future offspring in families with genetic disorders
Drawing a pedigree chart
A mutation in an allele that causes a protein to be NON-FUNCTIONAL would appear
recessive to the normal working allele.
Examples of autosomal recessive GENETIC DISORDERS:Phenylketonuria (PKU)Tay-Sachs DiseaseCystic Fibrosis
Phenylketonuria (PKU)
CAUSE: Mutation in gene for an enzyme the breaks down an amino acid called phenylalanine
Build up causes mental retardation
Phenylketonuria (PKU)
ALL babies are tested for PKU before they leave the hospital.
Treatment: Need a diet low in phenylalanine to extend life and prevent mental retardation
CYSTIC FIBROSISCAUSE: Loss of 3 DNA bases in a gene for the ion channel
protein that transports Cl- ions Salt balance is upsetCauses a build up of thick mucous in lungs and
digestive organs, Leads to: Respiratory and digestive complications, increased susceptibility to infections
thick mucous
Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006
http://www.biochem.arizona.edu/classes/bioc460/spring/rlm/RLM36.1.html
Carrier Heterozygous individualThat carries one recessive allele for a genetic disorder
Doesn’t show the disorder themselves,but can pass it on tooffspring
TAY-SACHS DISEASEAutosomal Recessive
CAUSE: Mutation in gene for an enzyme the breaks down a kind of lipid in the developing brain
As these lipids build up in brain infant suffers seizures, blindness, loss of motor & mental function > > > leads to early death.
Found more frequently in people with Jewish, Mediterranean, or Middle Eastern ancestry
Disorders caused by autosomal codominant alleles: SICKLE CELL DISEASE
CAUSE: A changed to T in gene for
Hemoglobin (protein in red blood cells that carries oxygen in blood)
SICKLE CELL DISEASE
SYMPTOMS:Red blood cells become sickle shaped under low oxygen condition in persons with two sickle cell alleles (ss)
Ss=Sickle cell trait Normally healthy, but can suffer some sickle cell episodes
SICKLE CELL DISEASE
Circulatory problemsCells stick in capillariesLoss of blood cells (anemia)Organ damage (brain, heart, spleen)Can lead to DEATH
HUNTINGTON’S DISEASE is AUTOSOMAL DOMINANT
CAUSE:
Extra 40-100 CAG repeats at end of gene on chromosome 4
The more repeats . . . the more severe the symptoms.
HUNTINGTON’S DISEASE
Begins in middle ageCauses progressive loss
of muscle control and mental function
1 in 10,000 people in U.S. have Huntington’s disease
http://www.scielo.br/img/revistas/bjmbr/v39n8/html/6233i01.htm
Huntington’s brain
Normal brain
A person with Huntington’s disease has a _____ chance of passing the disorder on totheir offspring.
Problem:Symptoms of disorder usually don’t show until ____________ . . .
so you don’t know you have it until ________ you have had children.
50%
MIDDLE AGE
AFTER
ACHONDROPLASIA(One kind of Dwarfism)
CAUSE: Autosomal Dominant gene
200,000 “little people” worldwide
One of oldest known disorders – seen in Egyptian art
1 in 25,000 births
DD = lethalDd = dwarf phenotypedd= = normal height
ACHONDROPLASIA(One kind of Dwarfism)
Normal size head and torso; short arms and legs
Problem with way cartilage changes to bone as bones grow
Image from Biology; Campbell and Reece; Pearson Prentice Hall publishing as Benjamin Cummings © 2006