Wheeler High School
The Center for Advanced Studies in Science, Math & Technology
Course Introduction
Genetics Lecture 3: Mendelian Genetics
Post-AP DNA/Genetics – Ms. Kelavkar
A Little Background on our
Homeboy Gregor Mendel?
•1822-1884
•Studied Physics &
Botany at the U of Vienna
•Influential publication was
Experiments with Plant
Hybrids
–This publication went
unnoticed until well after
his death!
Post-AP DNA/Genetics – Ms. Kelavkar
Was the first to discover
the basis for the
transmission of heredity
traits!
Result of Mendel’s monohybrid crosses of Pisum sativum.
Post-AP DNA/Genetics – Ms. Kelavkar
Mendel’s Suggestions
• Mendel suggested that heredity resulted in
discontinuous variation, as opposed to the existing
continuous variation hypothesis of his time—in
which offspring were thought to be a blend of the
parental phenotypes.
• Discontinuous variation is controlled by alleles of
a single gene or a small number of genes. The
environment has little effect on this type of
variation.
– Basic Mendelian Genetics (one trait, one gene)
• Continuous variation is a complete range of
measurements from one extreme to the other.
– Polygenic Inheritance
Post-AP DNA/Genetics – Ms. Kelavkar
Mendel’s First Three Postulates
(collectively considered his 1st law)
1. Unit Factors In Pairs
1. (Ex: AA or Aa or aa)
2. Dominance/Recessiveness
1. TT = tall, Tt = tall, tt = short
3. Segregation
Post-AP DNA/Genetics – Ms. Kelavkar
The Monohybrid Cross Reveals How One Trait
Is Transmitted from Generation to
Generation…
Thus confirming Mendel’s first 3
postulates!
Post-AP DNA/Genetics – Ms. Kelavkar
An Analytical Approach
Aaahhhh….the Punnett Square!
• Reginald D. Punnett was a biologist who
came up with this rather simple approach of
determining the probability of a cross.
Post-AP DNA/Genetics – Ms. Kelavkar
The Testcross
This is a one character test cross
Testcross = organism in question x
homozygous recessive
•Simple method used
today to determine the
genotype of plants &
animals.
You can cross
the offspring
with a
homozygous
recessive dwarf
plant to
determine the
genotypes of the
parents.
Independent Assortment
• Independent Assortment
– States that allele pairs separate independently
during the formation of gametes. Therefore,
traits are transmitted to offspring independently
of one another.
This can be shown using a dihybrid cross.
The Testcross: Two Characters
Post-AP DNA/Genetics – Ms. Kelavkar
Some Key Terms
• Genotype – genetic makeup of organism
• Phenotype – physical result of genetic
makeup (what you see)
• Testcross - used to determine if an
individual exhibiting a dominant trait is
heterozygous or homozygous for
– Heterozygous = Hh
– Homozygous = HH or hh
Any Questions?
Wheeler High School
The Center for Advanced Studies in Science, Math & Technology
Course Introduction
Genetics Lecture 4: Probability and Genetics
Post-AP DNA/Genetics – Ms. Kelavkar
Laws of Probability Help to Explain
Genetic Events
• The probability of two independent events
occurring at the same time can be calculated
using the product law:
– the probability of both events occurring is the
product of the probability of each individual
event
– Look for the word ‘and’ in the problem!
Post-AP DNA/Genetics – Ms. Kelavkar
Laws of Probability Help to Explain
Genetic Events
• The sum law is used to calculate the
probability of a generalized outcome that
can be accomplished in more than one way.
The sum law states that the probability of
obtaining any single outcome, where that
outcome can be achieved in two or more
events, is equal to the sum of the individual
probabilities of all such events.
– Look for the word ‘or’ in the problem!
Post-AP DNA/Genetics – Ms. Kelavkar
Otherwise Known as the ‘and’ ‘or’
rules!
1) Product rule = ‘and’ =multiply the events
1) Remember, the product rule describes the
probability that event A and event B will occur
1) Sum rule = ‘or’ = add the events
1) The sum rule describes the probability that
event A or event B will occur
Post-AP DNA/Genetics – Ms. Kelavkar
The Product Rule (multiply)
Suppose we roll one die followed by another
and want to find the probability of rolling a 4
on the first die and rolling an even number
on the second die.
P(4) = 1/6
P(even) = 3/6
P 4 and even = (1/6)(3/6) = 3/36 or 1/12
Chi Square & The Null Hypothesis
• Data from genetic crosses are quantitative.
– Geneticists use statistics to understand the significance
of their results.
• Hypothesis developed is called the null hypothesis
• Null Hypothesis states there is no real difference
between the observed data and the predicted data.
– Is the genetic event due to chance? If not, then the null
hypothesis is rejected and a new hypothesis must be
developed to explain the data.
Post-AP DNA/Genetics – Ms. Kelavkar
Null Hypothesis Example (H0)
In a clinical trial of a new drug, the null
hypothesis might be that the new drug is no
better, on average, than the current drug.
We would write:
H0: There is no difference between the two drugs
on average.
Chi-Square Analysis Evaluates the
Influence of Chance on Genetic Data
• A simple statistical tool used to test the null
hypothesis is called the chi-square (2) test.
– “goodness of fit” test
Post-AP DNA/Genetics – Ms. Kelavkar
Table 3-3 shows the steps in 2 calculations
for the F2 generation of a monohybrid cross.
Post-AP DNA/Genetics – Ms. Kelavkar
Degree of Freedom
• Chi-square analysis requires that the degree
of freedom (df) be taken into account, since
more deviation is expected with a higher
degree of freedom.
– df = n-1
Post-AP DNA/Genetics – Ms. Kelavkar
Chi-Square Probabilities
• Once the number of degrees of freedom is
determined, the 2 value can be interpreted
in terms of a corresponding probability value
(p).
Post-AP DNA/Genetics – Ms. Kelavkar
Example
Let’s analyze the theoretical progeny data from
a testcross of smooth, yellow double
heterozygote (SsYy) with a wrinkled, green
homozygote (ssyy). The progeny data:
154 smooth yellow
124 smooth green
144 wrinkled yellow
146 wrinkled green
Total 568
Post-AP DNA/Genetics – Ms. Kelavkar
We hypothesize that a testcross should yield a 1:1:1:1
ratio of the 4 phenotypic classes (that is if independent
assortment is taking place).
(1)
Phenotype
s
(2)
(3)
Observed # Expected #
(o)
(e)
(4)
(5)
(6)
d
(= o-e)
d2
d2/e
Smooth,
yellow
154
142
+12
144
1.01
Smooth,
green
124
142
-18
324
2.28
Wrinkled,
yellow
144
142
+2
4
0.03
Wrinkled,
green
146
142
+4
16
0.11
Total
568
568
0
n/a
3.43
(7) 2 = 3.43
(8) df = 3
Post-AP DNA/Genetics – Ms. Kelavkar
Chi-Square Probability Chart
2 = 3.43
Fail to reject
Null
Reject Null
This means that with the hypothesis being tested, 30 to 50 out
of 100 trials (so…30-50% of the time) we could expect chisquare values due to chance.
Any Questions?
Post-AP DNA/Genetics – Ms. Kelavkar
3.2 The Monohybrid Cross Reveals
How One Trait Is Transmitted from
Generation to Generation
Figure 3-4
Copyright © 2006 Pearson Prentice Hall, Inc.
3.3 Mendel’s Dihybrid Cross Revealed
His Fourth Postulate: Independent
Assortment
Figure 3-6
Copyright © 2006 Pearson Prentice Hall, Inc.
Figure 3-7
Copyright © 2006 Pearson Prentice Hall, Inc.
3.3 Mendel’s Dihybrid Cross Revealed
His Fourth Postulate: Independent
Assortment- Two Characters
Figure 3-9
Copyright © 2006 Pearson Prentice Hall, Inc.
3.4 The Trihybrid Cross Demonstrates
That Mendel’s Principles Apply to
Inheritance of Multiple Traits
3.4.1 The Forked-Line Method, or Branch Diagram
Figure 3-11
Copyright © 2006 Pearson Prentice Hall, Inc.
Figure 3-11a
Copyright © 2006 Pearson Prentice Hall, Inc.
Figure 3-11b
Copyright © 2006 Pearson Prentice Hall, Inc.
Figure 3-11c
Copyright © 2006 Pearson Prentice Hall, Inc.
Table 3-1
Copyright © 2006 Pearson Prentice Hall, Inc.
Table 3-2
Copyright © 2006 Pearson Prentice Hall, Inc.