Genetic ratios are most properly expressed as probabilities Ex tall 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 events independent probability.Sum law: Holds that the probability of one or the other of two mutually exclusive events occurring is the sum of the events individual probabilities.

Example:If 2 coins are tossed simultaneously each having a chance of heads and tails the probability of 2 heads would be , the probability of tails would be and the probability of one head and one tail would be (1/4 HT, 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 tailCoin 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 theoremThe 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 = b= probability of a tail for any coin = n = 4 because 4 coins were tossed

SolutionSo we get: ( a + b)4 = 1 Expanded: a4 + 4a3b + 6 a2b2 + 4ab3 + b4 Problem setup: (1/2)4 + 4[( )3 (1/2)] + 6[(1/2)2 (1/2)2] + 4[(1/2) (1/2)3] + (1/2)4 = 1So 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?

SolutionMAN is heterozygous because he has ptosis but mother was normal. So he is: PpThe 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 The probability of a normal child is and the probability for an afflicted child is So: a = probability of a normal child = b = probability of a child with ptosis = So: now use the expansion of the second term, 4a3b 4(1/2)3 (1/2) = 4/16 = = .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 ( having the condition) Let : a = probability of a normal child = b = probability of a child with ptosis =

SolutionBecause 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)] = 12/256NOTE: Had the question been 2 normal and 2 afflicted children, the third term ( 6a2b2) would have been used.

IntroductionPopulation = An interbreeding group of the same species in a given geographical areaGene pool = The collection of all alleles in the members of the populationPopulation genetics = The study of the genetics of a population and how the alleles vary with timeGene Flow = Movement of alleles between populations when people migrate and mate

Allele FrequenciesAllele frequency =# of particular alleleTotal # of alleles in the populationCount both chromosomes of each individual Allele frequencies affect the genotype frequencies - The frequency of the two homozygotes and the heterozygote in the population

Frequency of a trait varies in different populations

Table 14.1Phenotype Frequencies

Microevolution ( adaptation)The small genetic changes due to changing allelic frequencies in populationsFive factors can change genotypic frequencies:1) Nonrandom mating2) Migration3) Genetic drift4) Mutation5) Natural selection

MacroevolutionRefers to the formation of new species

Hardy-Weinberg EquationDeveloped independently by an English mathematician and a German physicianUsed algebra to explain how allele frequencies predict genotypic and phenotypic frequencies in a population of diploid, sexually-reproducing speciesDisproved the assumption that dominant traits would become more common, while recessive traits would become rarer

p + q = 1p = allele frequency of one alleleq = allele frequency of a second allelep2 + 2pq + q2 = 1p2 and q2 Frequencies for each homozygote 2pq Frequency for heterozygotesAll of the allele frequencies together equals 1All of the genotype frequencies together equals 1 Hardy-Weinberg Equation

Figure 14.3Source of the Hardy-Weinberg EquationFigure 14.3

Solving a ProblemFigure 14.4

Solving a ProblemFigure 14.4

Table 14.2The allele and genotypic frequencies do not change from one generation to the nextThus, this gene is in Hardy-Weinberg equilibrium

Figure 14.3Applying the Hardy-Weinberg EquationUsed 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 14.3Calculating the Carrier Frequency of an Autosomal RecessiveFigure 14.5

Table 14.3Calculating the Carrier Frequency of an Autosomal Recessive

Figure 14.3Calculating the Carrier Frequency of an Autosomal RecessiveWhat is the probability that two unrelated Caucasians will have an affected child?

Probability that both are carriers =1/23 x 1/23 = 1/529Probability that their child has CF = 1/4 Therefore, probability = 1/529 x 1/4 = 1/2,116

Figure 14.3Calculating the Risk with X-linked TraitsFor females, the standard Hardy-Weinberg equation appliesp2 + 2pq + q2 = 1However, in males the allele frequency is the phenotypic frequencyp + q = 1

Calculating the Risk with X-linked TraitsFigure 14.6

HARDY-WEINBER THEREM is:

P2 + 2pq + q2 = 1 frequency frequency frequency sum must equal AA Aa aa 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 = q2Thus q2 = .0001 ( 1/1000) with .01 = q (square root of .0001) Frequency of p can be determined since p = 1-qP= 1- .01 = .99Frequency of carriers (heterozygous ) in population is 2pq2pq = 2 (.99) (.01) = .0198Thus 2 % of USA population are carriers.

Hardy-Weinberg EquilibriumHardy-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 ProbabilityPedigrees and Punnett squares apply Mendels laws to predict the recurrence risks of inherited conditions

Example:- Taneeshas brother Deshawn has sickle cell anemia, an autosomal recessive disease.- What is the probability that Taneeshas 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.

Taneesha and Deshawns parents must be heterozygousTaneesha is not affected and cannot be ssProbability Taneesha is a carrier = 2/3Probability child inherits sickle cell allele = 1/2 Probability child carries sickle cell allele from her = 2/3 x 1/2 = 1/3X

DNA RepeatsShort repeated segments are distributed all over the genomeThe 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 ProfilingA technique that detects differences in repeat copy numberCalculates the probability that certain combinations can occur in two sources of DNA by chanceDNA evidence is more often valuable in excluding a suspect- Should be considered along with other types of evidence

Comparing DNA RepeatsFigure 14.7Figure 14.7

DNA ProfilingDeveloped in the 1980s by British geneticist Sir Alec Jeffreys

Also called DNA fingerprinting

Identifies individuals

Used in forensics, agriculture, paternity testing, and historical investigations

Figure 14.8Differing number of copies of the same repeat migrate at different speeds on a gelFigure 14.8

Jeffreys used his technique to demonstrate that Dolly was truly a clone of the 6-year old ewe that donated her nucleus

Figure 14.9

1) A blood sample is collected from suspect2) White blood cells release DNA3) Restriction enzymes cut DNA4) Electrophoresis aligns fragments by size5) Pattern of DNA fragments transferred to a nylon sheet

DNA Profiling Technique

6) Exposed to radioactive probes7) Probes bind to DNA8) Sheet placed against X ray film9) Pattern of bands constitutes DNA profile10) Identify individuals

DNA Profiling Technique

Box Figure 14.1

Figure 2.3DNA Fingerprinting AnimationPlease note that due to differing operating systems, some animations will not appear until the presentation is viewed in Presentation Mode (Slide Show view). You may see blank slides in the Normal or Slide Sorter views. All animations will appear after viewing in Presentation Mode and playing each animation. Most animations will require the latest version of the Flash Player, which is available at http://get.adobe.com/flashplayer.

DNA can be obtained from any cell with a nucleusSTRs are used when DNA is scarceIf DNA is extremely damaged, mitochondrial DNA (mtDNA) is often usedFor forensics, the FBI developed the Combined DNA Index System (CODIS)- Uses 13 STRsDNA Sources

The probability that any two individuals have same thirteen markers is 1 in 250 trillionCODISFigure 14.10

The power of DNA profiling is greatly expanded by tracking repeats in different chromosomesThe number of copies of a repeat are assigned probabilities based on their observed frequency in a populationThe product rule is then used to calculate probability of a certain repeat combination Population Statistics Is Used to Interpret DNA Profiles

To Catch A Thief With A SneezeTable 14.6

To Solve A CrimeTable 14.6Figure 14.11

Table 14.6Figure 14.11

Recent examples of large-scale disasters- World Trade Center attack (2001)- Indian Ocean Tsunami (2004)- Hurricane Katrina (2005)Using DNA Profiling to Identify Victims

Challenges to DNA ProfilingFigure 14.12

Todays population genetics presents a powerful way to identify individualsOur genomes can vary in more ways than there are people in the worldDNA profiling introduces privacy issues- Example: DNA dragnetsGenetic Privacy

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