Transcript Slide 1

Human Genetics
Concepts and Applications
Ninth Edition
RICKI LEWIS
14
Constant
Allele
Frequencies
PowerPoint® Lecture Outlines
Prepared by Johnny El-Rady, University of South Florida
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
Genetic ratios are most properly expressed
as probabilities Ex ¾ tall ¼ short, which
predict the outcome of each fertilization
event, such as a zygote having the genetic
potential to be tall.
Probabilities range from 0, where an event is
certain not to occur, to 1 where an event is
certain to occur.
TWO INDEPENDENT, NONGENETIC EVENTS:
• Product law : Holds that the probability
of 2 independent events occurring
simultaneously is the product of the
event’s independent probability.
• Sum law: Holds that the probability of
one or the other of two mutually
exclusive events occurring is the sum
of the event’s individual probabilities.
Example:
• If 2 coins are tossed simultaneously
each having a ½ chance of heads and
½ tails the probability of 2 heads would
be ¼, the probability of tails would be ¼
and the probability of one head and one
tail would be (1/4 HT, ¼ TH) 1/2
 IMPORTANT: Remember
when dealing with
probability, predictions of
possible outcomes are
usually realized only with
large sample sizes.
 Any deviation from the
predicted ratio in small
sample sizes is attributed to
chance.
•
• So if 2 coins are tossed for,
say 50 tosses simple
assumptions about the coins
must first be made before
probabilities are calculated:
 coins must be unbiased:
constructed physically so
that each coin has an equal
chance of landing either
head or tail
 Coin tosses are
independent of each other:
does each coin toss have an
equal chance of coming to
rest heads or tails.
 Successive tosses (events)
are also independent of
each other.
binomial theorem
 The binomial theorem can be used to predict the
likely outcome of such a number of coin tosses.
 The binomial formula is: ( a + b)n = 1
 Example: In the 100 coin tosses, coins will either be
heads or tails
•
Where: a = probability of a head for any
coin = ½
•
b= probability of a tail for any coin
=½
•
n = 4 because 4 coins were tossed
Solution
• So we get: ( a + b)4 = 1
•
Expanded: a4 + 4a3b + 6 a2b2
+ 4ab3 + b4
•
• Problem setup: (1/2)4 + 4[( ½)3 (1/2)]
+ 6[(1/2)2 (1/2)2] + 4[(1/2) (1/2)3] +
(1/2)4 = 1
• So that: 1/16 + 4/16 + 6/16 + 4/16 + 1/16
=1
GENETIC APPLICATION OF THE
BINOMIAL
• A man with ptsosis, whose father also
displays the trait but whose mother did not,
wishes to marry a woman with normal
eyelids. They consult a physician to
determine the likelihood of the occurrence of
the defect in their children.
• If they have 4 children, what is the probability
of 3 of those being normal and 1 having
ptosis?
Solution
•
•
•
•
•
•
•
•
MAN is heterozygous because he has ptosis but mother was normal…. So he is:
The woman is normal so she is: pp
Parents (the man and woman above) are: Pp x pp
Gametes: P, p
p, p
F1
Pp
Pp, Pp, pp, pp
The probability of a normal child is ½ and the probability for an afflicted child is
½
•
•
•
•
•
•
•
So:
So:
a = probability of a normal child = ½
b = probability of a child with ptosis = ½
now use the expansion of the second term, 4a3b
4(1/2)3 (1/2) = 4/16 = ¼ = .25
This is used because it contains a3 (representing 3 children of the phenotype
denoted by a) and b ( for one child o phenotype represented by b)
Results
• There is a probability in this case of 1
in 4 or .25, that if they have four
children, 3 will be normal and 1 will
have ptosiss. There is a 1 chance in 16
(a4) or .0625, that non of the 4 will
exhibit ptosis or (b4) that all 4 will have
ptosis.
EXAMPLE #2:
• A young man and woman, each of
whom has ptosis and is heterozygous.
If such a couple should marry and have
4 children, what is the probability of 3
being normal and one having ptosis?
• Parents (couple): Pp x Pp
•
•
Gametes:
P, p P, p
•
F1:
PP, Pp, Pp, pp
3:1 ( ¾ having the condition)
•
•
Let :
a = probability of a
normal child = ¼
•
b = probability of a
child with ptosis = ¾
Solution
• Because this case again concerns a family of
4 children, the binomial (a + b)4 would have
been used. With the 2nd binomial expression
used.
•
4a3b = 4[(1/4)3 x (3/4)]
12/256
=
• NOTE: Had the question been 2 normal and 2
afflicted children, the third term ( 6a2b2) would have
been used.
Introduction
• Population = An interbreeding group of the
same species in a given geographical area
• Gene pool = The collection of all alleles in
the members of the population
• Population genetics = The study of the
genetics of a population and how the
alleles vary with time
• Gene Flow = Movement of alleles between
populations when people migrate and mate
Allele Frequencies
• Allele
frequency =
# of particular allele
Total # of alleles in
the population
Count both chromosomes of each individual
Allele frequencies affect the genotype
frequencies
- The frequency of the two homozygotes
and the heterozygote in the population
Phenotype Frequencies
• Frequency of a trait varies in different
populations
Table 14.1
Microevolution ( adaptation)
• The small genetic changes due to
changing allelic frequencies in populations
• Five factors can change genotypic
frequencies:
•
1) Nonrandom mating
•
2) Migration
•
3) Genetic drift
•
4) Mutation
•
5) Natural selection
Macroevolution
• Refers to the formation of new species
Hardy-Weinberg Equation
• Developed independently by an English
mathematician and a German physician
• Used algebra to explain how allele
frequencies predict genotypic and
phenotypic frequencies in a population of
diploid, sexually-reproducing species
• Disproved the assumption that dominant
traits would become more common, while
recessive traits would become rarer
Hardy-Weinberg Equation
p = allele frequency of one allele
q = allele frequency of a second allele
All of the allele frequencies
together equals 1
p+q=1
p2 + 2pq + q2 = 1
p2 and q2
2pq
All of the genotype frequencies
together equals 1
Frequencies for each homozygote
Frequency for heterozygotes
Source of the Hardy-Weinberg
Equation
Figure 14.3
Figure 14.3
Solving a Problem
Figure 14.4
Solving a Problem
Figure 14.4
The allele and genotypic frequencies do not
change from one generation to the next
Thus, this gene is in Hardy-Weinberg equilibrium
Table 14.2
Applying the Hardy-Weinberg
Equation
Used to determine carrier probability
For autosomal recessive diseases, the
homozygous recessive class is used to
determine the frequency of alleles in a
population
- Its phenotype indicates its genotype
Figure 14.3
Calculating the Carrier Frequency
of an Autosomal Recessive
Figure 14.5
Figure 14.3
Calculating the Carrier Frequency
of an Autosomal Recessive
Table 14.3
Calculating the Carrier Frequency
of an Autosomal Recessive
What is the probability that two unrelated
Caucasians will have an affected child?
Probability that both are carriers =
1/23 x 1/23 = 1/529
Probability that their child has CF = 1/4
Therefore, probability = 1/529 x 1/4 =
1/2,116
Figure 14.3
Calculating the Risk with
X-linked Traits
For females, the standard Hardy-Weinberg
equation applies
p2 + 2pq + q2 = 1
However, in males the allele frequency is
the phenotypic frequency
p + q= 1
Figure 14.3
Calculating the Risk with
X-linked Traits
Figure 14.6
HARDY-WEINBER THEREM is:
P2
+
2pq
frequency
AA
+ q2 = 1
frequency frequency
Aa
aa
sum must equal
100%
EXAMPLE PROBLEM:
• 1 in every 100 babies in USA is born with PKU
resulting in mental retardation if untreated.
• Allele for PKY is recessive, so babies with the
disorder are homozygous = q2
• Thus q2 = .0001 ( 1/1000) with .01 = q (square root
of .0001)
• Frequency of p can be determined since p = 1-q
• P= 1- .01 = .99
• Frequency of carriers (heterozygous ) in population
is 2pq
• 2pq = 2 (.99) (.01) = .0198
• Thus 2 % of USA population are carriers.
Hardy-Weinberg Equilibrium
• Hardy-Weinberg equilibrium is rare for
protein-encoding genes that affect the
phenotype
• However, it does apply to portions of the
genome that do not affect phenotype
• These include repeated DNA segments
•
- Not subject to natural selection
CONDITIONAL PROBABILITY:
• Sometimes we may wish to calculate the probability
of an event or outcome that is dependent on a
specific condition related to that outcome.
• Example: in the F2 generation of a monohybrid
cross involving tall and dwarf plants, what would
the probability be that a tall plant is heterozygous?
• The condition we set is to consider only tall F2
offspring since we know that all dwarf plants are
homozygous.
• Of any F2 tall plant, what is the probability of it being
heterozygous?
Conditional Probability
Pedigrees and Punnett squares apply
Mendel’s laws to predict the recurrence
risks of inherited conditions
Example:
- Taneesha’s brother Deshawn has sickle cell
anemia, an autosomal recessive disease.
- What is the probability that Taneesha’s child
inherits the sickle cell anemia allele from her?
• Because the outcome and specific condition
are not independent, we cannot apply the
product law of probability….Instead we use
conditional probability.
• Conditional probability deals with the
probability of one out come occurring, given
the specific condition upon which this
outcome depends.
X
Taneesha and Deshawn’s
parents must be
heterozygous
Taneesha is not affected
and cannot be ss
Probability Taneesha is a carrier = 2/3
Probability child inherits sickle cell allele = 1/2
Probability child carries sickle cell allele from her
= 2/3 x 1/2 = 1/3
DNA Repeats
• Short repeated segments are distributed
all over the genome
• The repeat numbers can be considered
alleles and used to classify individuals
• Two types of repeats are important
•
- Variable number of tandem
repeats (VNTRs)
•
- Short tandem repeats (STRs)
DNA Repeats
DNA Profiling
• A technique that detects differences in
repeat copy number
• Calculates the probability that certain
combinations can occur in two sources
of DNA by chance
• DNA evidence is more often valuable in
excluding a suspect
•
- Should be considered along with
other types of evidence
Comparing DNA Repeats
Figure 14.7
Figure 14.7
DNA Profiling
• Developed in the 1980s by British
geneticist Sir Alec Jeffreys
• Also called DNA fingerprinting
• Identifies individuals
• Used in forensics, agriculture, paternity
testing, and historical investigations
Differing number of copies of the same repeat
migrate at different speeds on a gel
Figure 14.8
Figure 14.8
• Jeffreys used his
technique to
demonstrate that
Dolly was truly a
clone of the 6year old ewe
that donated her
nucleus
Figure 14.9
DNA Profiling Technique
1) A blood sample is collected from
suspect
2) White blood cells release DNA
3) Restriction enzymes cut DNA
4) Electrophoresis aligns fragments by size
5) Pattern of DNA fragments transferred to
a nylon sheet
DNA Profiling Technique
6) Exposed to radioactive probes
7) Probes bind to DNA
8) Sheet placed against X ray film
9) Pattern of bands constitutes DNA profile
10) Identify individuals
Box Figure 14.1
DNA Fingerprinting Animation
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Figure 2.3
DNA Sources
DNA can be obtained from any cell with a
nucleus
STRs are used when DNA is scarce
If DNA is extremely damaged, mitochondrial
DNA (mtDNA) is often used
For forensics, the FBI developed the
Combined DNA Index System (CODIS)
- Uses 13 STRs
CODIS
Figure 14.10
The probability that any two individuals have same
thirteen markers is 1 in 250 trillion
Population Statistics Is Used to
Interpret DNA Profiles
The power of DNA profiling is greatly
expanded by tracking repeats in different
chromosomes
The number of copies of a repeat are
assigned probabilities based on their
observed frequency in a population
The product rule is then used to calculate
probability of a certain repeat combination
To Catch A Thief With A Sneeze
Table 14.6
To Solve A Crime
Table 14.6
Figure 14.11
Table 14.6
Figure 14.11
Using DNA Profiling to Identify
Victims
Recent examples of large-scale disasters
- World Trade Center attack (2001)
- Indian Ocean Tsunami (2004)
- Hurricane Katrina (2005)
Challenges to DNA Profiling
Figure 14.12
Genetic Privacy
Today’s population genetics presents a
powerful way to identify individuals
Our genomes can vary in more ways than
there are people in the world
DNA profiling introduces privacy issues
- Example: DNA dragnets