chapter 2
TRANSCRIPT
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Chapter 02
MENDELIANMENDELIANINHERITANCE
Many theories of inheritance have been proposed to explain t i i f
INTRODUCTION
transmission of hereditary traits
The most prominent are
Theory of pangenesis Theory of preformationism Blending theory
Factors that control hereditary traits are maleable
They can blend together generation after generation
AverageOffspring
Parent 1 Parent 2
Mendel’s Laws of Inheritance
Gregor Mendel’s pioneering experiments with garden peas refuted all of the above
Mendel was successful, when his predecessors were not, because he
1. Chose a very suitable organism:
Pisum sativum
2. Took time to ensure pure breeding
3. Concentrated on one or a few discontinuous characters at a time
4. Adopted quantitative form of analysis
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Mendel’s Experimental System
Used pure-breeding lines of garden peas
Selected 7 traits to study, each of which had two distinguishable appearancesFigure 2.4
He carefully designed his experiments and gathered mathematical data
Figure 2.4
1. Cross-fertilization
Pollen and egg are derived from different plants
Mendel’s Experimental System
Refer to Figure 2.3
2. Self-fertilization
Pollen and egg are derived from the same plan
Figure 2.3
MONOHYBRID CROSSES
Mendel crossed individuals from two different lines that differ in only one traitParental generation = Pure bred lines
Fi t fili l ti Off i fFirst filial generation = Offspring of cross-fertilization of parentals
Second filial generation = Offspring of self-fertilization of F1 plants
Third filial generation = Offspring of self-fertilization of F2 plants
Figure 2.5
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Mendel also performed the reciprocal cross, where dwarf plants were pollinated using pollen from tall plants The results were the same
For all seven traits studied
1. The F1 generation showed only one of the two parental traits
2. The F2 generation showed an ~ 3:1 ratio of the two parental traits
3. Reciprocal crosses yielded similar results
P Cross F1 generation F2 generation Ratio
Tall X dwarf stem
All tall 787 tall, 277 dwarf
2.84:1
Round X wrinkled seeds
All round 5,474 round, 1,850 wrinkled
2.96:1
Yellow X Green seeds
All yellow 6,022 yellow, 2,001 green
3.01:1
DATA FROM MONOHYBRID CROSSES
Purple X white flowers
All purple 705 purple, 224 white
3.15:1
Axial X terminal flowers
All axial 651 axial, 207 terminal
3.14:1
Smooth X constricted pods
All smooth 882 smooth, 229 constricted
2.95:1
Green X yellow pods
All green 428 green, 152 yellow
2.82:1
These results refuted a blending mechanism of heredity
MONOHYBRID CROSSES
Indeed, the data suggested a particulate theory of inheritance
Mendel postulated the following
1. A pea plant contains two discrete hereditary factors, one from each parent
These factors are now referred to as alleles
2 Th t f t b id ti l 2. The two factors may be identical or different
If the alleles are identical the organism is referred to as homozygous
If the alleles are different the organism is referred to as heterozygous
3. When the two factors of a single trait are different, one is dominant and its effect can be seen, while the other is recessive and is not expressed
4. During gamete formation, the paired factors segregate randomly so that half of the gametes received one factor and half of the gametesreceived one factor and half of the gametes received the other This is Mendel’s First Law, or the Law
(Principle) of Segregation
5. The union of male and female gametes is a random process that reestablishes the paired units
Alleles are designated by letters derived from the specific characteristic they control
Usually the letter chosen is the first letter of the dominant (or recessive) trait
For example,
T = Tall plants
t = dwarf plants
Alternatively,
D = Tall plants
d = dwarf plants
Let’s introduce a few more terms
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Genotype = The specific allelic composition of an individual
With respect to seed shape, a pea plant may have different genotypes3
TT ; Tt ; tt
Phenotype = The physical appearance of anPhenotype The physical appearance of an individual
With respect to seed shape, a pea plant may have different phenotypes2
Tall ; Dwarf
Note that the phenotype is a product of the genotype and the environment
Genotype Notation
Genotype Description
Phenotype
TTHomozygous
dominantTall
Tt Heterozygous Tall
ttHomozygous
recessiveDwarf
Information about parents can be used to predict genotypes and phenotypes of offspring
Punnett square
A grid that enables one to calculate the results of simple genetics crosses
Gametes
Genotypes
Refer to p. 24
We will illustrate the Punnett square approach using the cross of heterozygous tall plants as an example
1. Write down the genotypes of both parentsM l t Tt Male parent = Tt
Female parent = Tt
2. Write down the possible gametes each parent can make. Male gametes: T or t Female gametes: T or t
3. Create an empty Punnett square
4. Fill in the Punnett square with the possible genotypes of the offspring
5. Determine the relative proportions of genotypes and phenotypes of the offspring
Genotypic ratio
TT : Tt : tt
1 : 2 : 1
Phenotypic ratio
Tall : dwarf
3 : 1
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Mendel verified his principle of segregation by performing a test-cross In a test-cross an organism with known
phenotype but unknown genotype is crossed with an organism that has the recessive phenotype (and hence is homozygous recessive)
F l if l t i t ll it ld b For example, if a pea plant is tall it could be homozygous dominant (true-breeding) or heterozygous This unknown genotype is designated
T –
A test-cross will determine its correct genotype
Tall X Dwarf
If all offspring are tall
=> the genotype is TT
If the offspring are in a ratio of 1 tall:1 dwarf
=> the genotype is Tt
To solve problems dealing with standard genetic crosses, follow these general rules:
1. Lay out the problem as a simple diagram
2 Assign letters for the alleles of the2. Assign letters for the alleles of the genes in question
3. Determine the genotypes of the parentals
4. Do a Punnett square
Application problem
In humans, free earlobes are dominant to attached ear lobes. This characteristic is under the control of a single gene.
1. Elias and his wife Nadine have attachedearlobes. What percentage of their childrenis expected to have attached earlobes?
2. Peter and his wife Aida have freeearlobes. Peter’s mom and Aida’s dad haveattached earlobes. What percentage of Peterand Aida’s children are expected to haveattached ear lobes?
DIHYBRID CROSSES
Mendel crossed individuals from two different lines that differ in two traits
Refer to Experiment 2B Refer to Experiment 2B
P Round, yellow X Wrinkled, green
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Trait 1 = Seed texture
Trait 2 = Seed color
Figure 2.8
P Cross F1 generation F2 generation
R d All d 315 d ll d
DATA FROM DIHYBRID CROSSES
Round,
Yellow seeds X wrinkled, green seeds
All round, yellow
315 round, yellow seeds
101 wrinkled, yellow seeds
108 round, green seeds
32 green, wrinkled seeds
2.97 : 1
Mendel looked at each trait in the F2 separately
3.18 : 1
These values are very close to segregation expectations
If the genes do assort independently, what will the predicted phenotypic ratio be?
9 3 3 1
Thus, Mendel proposed his second law, the law (principle) of independent assortment During gamete formation, the segregation of any
pair of hereditary determinants is independent of the segregation of other pairs
P Cross F1 generation F2 generation Ratio
Round,
Yellow seeds x Wrinkled, green seeds
All round, yellow 315 round, yellow seeds
101 wrinkled, yellow seeds
108 round, green seeds
32 green, wrinkled seeds
9.8
3.2
3.4
1.0
9:3:3:1
Figure 2.8
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Figure 2.9 Mendel verified this principle by, again,
performing a test cross
Cross F2 yellow, round plant with a
Green, Wrinkled
If the offspring are 1 yellow, round plant :
YYRr
If the offspring are 1 yellow, round plant : 1 yellow, wrinkled plant, what is the genotype of the dominant parent?
Note: Offspring of a test-cross are always in equal proportions!
Crosses that involve three pairs of traits
TRIHYBRID CROSSES P
Gametes
F1
Gametes
A Punnett square in this case will have squares! 64
Therefore it will be too cumbersome!
An alternative is the forked-line method Treat each gene separately Consider phenotypic (or genotypic) differences
In the F2 generation, 3/4 of all organisms will have the A phenotype,
and 1/4 ill ha e the a phenot peand 1/4 will have the a phenotype 3/4 of all organisms will have the B phenotype,
and 1/4 will have the b phenotype 3/4 of all organisms will have the C
phenotype, and 1/4 will have the c phenotype
Refer to Solved Problem 4 (pp. 37-38)
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linked One caveat: The genes must not be
General Rules for Multihybrid Crosses
If n = # of heterozygous gene pairs involved
Number of possible parental gametes
2n
Number of different genotypes in zygotes
3n
Number of different phenotypes in zygotes
2n
Proportion of homozygous recessive in F2
1/4n
PEDIGREE ANALYSIS
Refer to pp. 29-30 and 602-609
Determining the inheritance pattern of human traits is somewhat difficulthuman traits is somewhat difficult 1. Controlled matings are not possible
2. Sibship sizes are relatively small 3. Complete penetrance is sometimes
lacking 4. Degrees of expressivity sometimes vary
Pedigrees
Diagrams that show the relationship among members of a family, as well
PEDIGREE ANALYSIS
among members of a family, as well as their status with respect to a particular hereditary condition
Figure 2.11
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Traits that are determined by a single gene are of five main types
Autosomal dominant
Autosomal recessive Autosomal recessive
X-linked dominant
X-linked recessive
Y-linked
AUTOSOMAL DOMINANT TRAITS
1. Trait usually present in every generation 2. Members of both sexes affected equally 3. Unaffected parents produce unaffected
offspringoffspring 4. Two affected parents can produce
unaffected offspring 5. On average half the children of an
affected parent will be affected
Polydactyly
Autosomal Dominant Disorders Brachydactyly
Syndactyly Achondroplasia
Autosomal Dominant Traits
1. Trait tends to skip generations 2. Members of both sexes affected equally 3. Unaffected parents can have an affected
child
AUTOSOMAL RECESSIVE TRAITS
child 4. Two affected parents cannot have an
unaffected child 5. Traits are more likely to occur in
consanguineous marriages
Albinism
Absence or partial deficiency of melanin in the skin, eyes and hair
It occurs in a wide variety of animals
Autosomal Recessive Disorders
It occurs in a wide variety of animals
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Cystic fibrosis The most common
lethal genetic disease among caucasians
Autosomal Recessive Disorders
Mainly characterized by serious digestive and respiratory problems
Average life expectancy ~ 30 yrs
1. Trait may skip generations 2. Most affected individuals are male 3. Affected males result from affected
mothers or mothers who are carriers
X-LINKED RECESSIVE TRAITS
mothers or mothers who are carriers 4. Affected females come from affected
fathers and affected or carrier mothers 5. Affected females will have affected
sons 6. Trait can never be transmitted directly
from father to son
Red-green color blindness
X-linked Recessive Disorders
Hemophilia
1. Trait does not skip generations 2. Affected males must come from affected
mothers 3. Affected females come from affected
mothers or fathers
X-LINKED DOMINANT TRAITS
4. Affected males have no normal daughters and no affected sons
5. Affected heterozygous females transmit the trait to approximately half of their children of either sex
6. Affected homozygous females transmit the trait to all their children
Vitamin D-resistant ricketsRi k t i di d
X-linked Dominant Disorders
Faulty tooth enamel
Rickets is a disorder involving softening and weakening of the bones (of children) primarily caused by lack of vitamin D
This form of rickets is caused by lack of phosphate
1. Trait only affects males
2. Affected males get it from their fathers and give it to their sons
Y-LINKED TRAITS
Hypertrichosisof the ear
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Penetrance
The proportion of individuals with a specific genotype who manifest the corresponding phenotype
% Penetrance = # of indiv. with correct phenotype
# of individuals with genotype coding
for phenotype
Complete penetrance
Incomplete penetrance
The ABO blood group genes show almost complete penetrance
You will find out later!
Penetrance
Neurofibromatosis shows 50-80% penetrance
Polydactyly also shows incomplete penetrance
Note: Incomplete penetrance may cause a phenotype to skip a generation
Consider this pedigree that traces polydactyly in a family
Expressivity The degree to which a penetrant gene is
phenotypically expressed
Identical known genotype
A range of phenotypes
is produced
Variable Expressivity
If Neurofibromatosis penetrates, it may be exhibited in various forms
Mildest form
Café-au-lait spots
Intermediate form Intermediate form
Café-au-lait spots
Freckling
Severe form
Neurofibromas of various sizes
Speech impediments
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If polydactyly penetrates, it may be exhibited in various forms
What are the reasons for incomplete
Note: Penetrance and expressivity are used most often with autosomal dominant traits
penetrance and variable expressivity?
Gene interactionsGene interactions
Environmental effectsEnvironmental effects
Siamese cats have darker fur at their extremities
Phenylketonuria
Refer to pp. 8-9
An autosomal recessive disorder An autosomal recessive disorder
Defect is in the gene that encodes phenylalanine hydroxylase
Melanin
The metabolic pathway of phenylalanine breakdownFigure 13.1
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Affected individuals cannot metabolize phenylalanine
Phenylalanine will accumulate and be converted to phenylpyruvic acid
These are toxic and will cause mental and physical abnormalities
Newborns are now routinely screened for PKU
Individuals with the disease are put on a strict dietary regimen
Their diet is very low on phenylalanine
These individuals tend to develop normally
Phenocopy
An organism whose phenotype has been modified to resemble the phenotype of a different mutant organismg
Phocomelia
Thalidomide
Phocomelia Suppression of limb
development
Caused by a dominant allele with variable expressivityp y
Thalidomide Sedative
Between 1959-1962 children with shortened limbs were born to mothers who took the drug early in their pregnancy
PROBABILITY AND STATISTICS
Probability deals with chance or random events
The laws of probability allow us to predict the outcomes of crosses, in addition to many other processes
Probability =
# of favorable casestotal # of cases
It ranges fromg
0 to 1
If an event has a probability of p, the probability of all alternative events is q
Rules For Combining Probabilities
1. Product rule
2. Sum rule2. Sum rule
3. Binomial expansion equation (binomial theorem)
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1. Product rule
The probability of the occurrence of independent events is the product of their respective probabilities
Example: Probability of rolling a 3 ANDExample: Probability of rolling a 3 AND then a 4 with a die?
P(3 and 4) =
P(3) X P(4)
1/6 X 1/6
1/36
2. Sum rule
The probability of the occurrence of one of several mutually exclusive events is the sum of their respective probabilities
Example: Probability of rolling a 3 OR aExample: Probability of rolling a 3 OR a 4 with a die?
P(3 or 4) =
P(3) + P(4)
1/6 + 1/6
1/3
2. Sum rule Note: If the 2 events are not mutually
exclusive, then
P(A or B) = P(A) + P(B) - [P(A) X P(B)]
Example: Probability of drawing an ace OR a heart from a deck of cardsa heart from a deck of cards
P(A or ) =
P(A) + P() – [P(A) X P()]
4/52 + 13/52 – (4/52 X 13/52)
17/52 – (1/52)
16/52
3. Binomial Expansion Equation
The probability of the occurrence of unordered mutually exclusive events is defined by the binomial theorem
n!! ( )!
px qn – x P = x! (n – x)!
p q
where
p = probability that the unordered number of events will occur
n = total number of events
x = number of events in one category
p = individual probability of x
q = individual probability of the other category
Refer to the example on pp. 32-33
Question
Two heterozygous brown-eyed (Bb) yg y ( )individuals have five children
What is the probability that two of the couple’s five children will have blue eyes?
Applying the binomial expansion equation
Step 1: Calculate the individual probabilities
This can be obtained via a Punnett square
1/4 P(blue eyes) = p =
3/4 P(brown eyes) = q =
Step 2: Determine the number of events
n = total number of children = 5
x = number of blue-eyed children = 2
Step 3: Substitute the values for p, q, x, and nin the binomial expansion equation
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n!
x! (n – x)!px qn – x
P =
5!
2! (5 – 2)!(1/4)2 (3/4)5 – 2
P =
5 X 4 X 3 X 2 X 1
(2 X 1) (3 X 2 X 1)(1/16) (27/64)P =
Therefore 26% of the time, a heterozygous couple’s five children will contain two with blue eyes and three with brown eyes
(2 X 1) (3 X 2 X 1)
P = 0.26 or 26%
Use Of Rules
1. What is the probability of rolling two dice together and obtaining a 3 and a 6?
2. Consider two parents heterozygous for albinism. a) What is the probability that they will
have an albino child?
b) What is the probability that they will have 3 normal children in a row?
c) What is the probability that they will have, in order, 2 normal children, an albino child, and then a normal child?
d) What is the probability that they will have 3 normal children and an albino child, in any order?
CHI-SQUARE ANALYSIS
The 3:1 and 9:3:3:1 ratios are hypothetical ratios based on the following assumptions
Dominance/recessiveness of alleles
Complete segregation
Independent assortment
Random fertilization
Independent assortment and random fertilization are chance events
Therefore, they are subject to random
CHI-SQUARE ANALYSIS
y jfluctuations
To minimize the possibility that chance deviation will significantly alter the outcome, we need to look at very large numbers
How can we evaluate the likelihood that deviations are NOT significant, but due to chance?
The chi-square test is a statistical method e c squa e test s a stat st ca et odused to determine “goodness of fit”
This refers to how close the observed data are to those predicted from a hypothesis
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The basic steps are the following:
1. Establish a NULL hypothesis
It assumes that there is no real difference between the measured values and the predicted values
2. Apply 2 formula
(O – E)2
E
3. Interpret 2 value
This will ultimately lead to the acceptance or rejection of the null hypothesis
Consider the following example in Drosophila melanogaster (pp. 33-35)
Gene affecting wing shape
c+ = Normal wing
c = Curved wing
Gene affecting body color
e+ = Normal (gray)
e = ebony
Note:
Th ild t ll l i d i t d ith + i The wild-type allele is designated with a + sign
Recessive mutant alleles are designated with lowercase letters
The Cross:
A cross is made between two true-breeding flies (c+c+e+e+
and ccee). The flies of the F1 generation are then allowed to mate with each other to produce an F2 generation.
The outcome F1 generation
All offspring have straight wings and gray bodies
F2 generation
193 straight wings, gray bodies
69 straight wings, ebony bodies
64 curved wings, gray bodies
26 curved wings, ebony bodies
352 total flies
Applying the chi square test
Step 1: Propose a hypothesis that allows us to calculate the expected values based on Mendel’s laws
The two traits are independently assorting
Step 2: Calculate the expected values of the four phenotypes, based on the hypothesis
According to our hypothesis, there should be a 9:3:3:1 ratio on the F2 generation
Phenotype
straight wingsstraight wings, gray bodies
straight wings, ebony bodies
curved wings, gray bodies
curved wings, ebony bodies
Step 2: Calculate the expected values of the four phenotypes, based on the hypothesis
According to our hypothesis, there should be a 9:3:3:1 ratio on the F2 generation
Phenotype Expected probability
Expected number
straight wings 9/16straight wings, gray bodies
9/16
straight wings, ebony bodies
3/16
curved wings, gray bodies
3/16
curved wings, ebony bodies
1/16
Step 2: Calculate the expected values of the four phenotypes, based on the hypothesis
According to our hypothesis, there should be a 9:3:3:1 ratio on the F2 generation
Phenotype Expected probability
Expected number
straight wings 9/16 9/16 X 352 = 198straight wings, gray bodies
9/16 9/16 X 352 = 198
straight wings, ebony bodies
3/16 3/16 X 352 = 66
curved wings, gray bodies
3/16 3/16 X 352 = 66
curved wings, ebony bodies
1/16 1/16 X 352 = 22
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Step 3: Apply the chi square formula
(O1 – E1)2
E1
(O2 – E2)2
E2
(O3 – E3)2
E3
(O4 – E4)2
E4
+ + +
(193 – 198)2
198
(69 – 66)2
66
(64 – 66)2
66
(26 – 22)2
22+ + +
198 66 66 22
0.13 + 0.14 + 0.06 + 0.73
1.06
Step 4: Interpret the chi square value
The calculated chi square value can be used to obtain probabilities, or P values, from a chi square table
These probabilities allow us to determine the likelihood that the observed deviations are due to random chance alone
Low chi square values indicate a high probability that the observed deviations could be due to random chance alone
High chi square values indicate a low probability that the observed deviations are due to random chance alone
By convention, the “normal” level of significance is 5 %
If p ≥ 0.05, the deviation is considered to be statistically insignificant
=> Null hypothesis is accepted > Null hypothesis is accepted
If p < 0.05, the deviation is considered to be statistically significant
=> Null hypothesis is rejected
Before we can use the chi square table, we have to determine the degrees of freedom (df)
The df is a measure of the number of categories that are independent of each other
df = n – 1
where n = total number of categoriesg
In our experiment, there are four phenotypes/categories
Therefore, df = 4 – 1 = 3
Refer to Table 2.1
1.06
With df = 3, the chi square value of 1.06 is slightly greater than 1.005 (which corresponds to P = 0.80)
A P = 0.80 means that values equal to or greater than 1.005 are expected to occur 80% of the time based on random chance alone Thus, it is quite probable that the deviations
between the observed and expected values in this expt. can be explained by random sampling error
If the NULL hypothesis is rejected, then what???