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Mendelian Genetics. Genetic Definitions. Genes - genetic material on a chromosome that codes for a specific trait Genotype - the genetic makeup of the organism Phenotype- the expressed trait Allel- an alternative form of a gene. Dominance Mechanism Two alleles are carried for each trait - PowerPoint PPT Presentation
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Mendelian Genetics
• Genes- genetic material on a chromosome that codes for a specific trait
• Genotype- the genetic makeup of the organism
• Phenotype- the expressed trait• Allel- an alternative form of a
gene
Genetic Definitions
Dominance Mechanism
• Two alleles are carried for each trait• In true-breeding individuals, both
alleles are the same (homozygous).• Hybrids, on the other hand, have one
of each kind of allele (heterozygous).• One trait is dominant, the other trait is
recessive
Mendel’s Three Principles
• Dominance
• Segregation
• Independent Assortment
The foundation of “classical” science
(1822-1884)
• Frequency 150 purple
• Proportion 150:200
• Percentage 75% purple
• Ratio 3:1
150 purple kernels50 yellow kernelsTotal 200 kernels
This laboratory activity is designed tofamiliarize you with the statistical nature ofMendel's model.
We will attempt to understand theprobabilistic aspects of monohybrid crosses(Mendel’s Law of Segregation) by "randomly"tossing special "coins" designed to simulate thegenotypes of parents, focusing on a singlegenetic character, seed color.
Part I
We will attempt to understand the probabilistic aspects of dihybrid crosses (Mendel’s Law of Independent Assortment) by “randomly” tossing special “dice” designed to simulate the genotypes of parents, focusing on two genetic characters, seed color and plantheight.
Part II
Dihybrid Cross
In Part III, we will investigate various human genetic characters.
In Part IV, we will also evaluate the results of monohybrid and dihybrid crosses involving the characteristics of corn seeds.
In Part V, we will learn about the use of the Chi-Square test for statistical hypothesistesting.
c2 =∑(o – e)2
e
Parts I-III of this lab activity will involvecollection of class data. So we will proceedthrough these parts together as a class. Buteach round of data collection will involve workingin pairs.
Parts IV & V will not involve collection ofclass data. You will proceed through these partsworking in pairs.
Tongue Roller
R = Tongue Rollerr = Unable to Roll Tongue
Widow’s Peak
W = Widows Peakw = Lack of Widow’s Peak
Free Ear Lobe Attached Ear Lobe
E = Free Ear Lobee = Attached Ear Lobe
Hitchhiker’s Thumb
Hi = Straight Thumbhi = Hitchhiker’s Thumb
Bent Little Finger
Bf = Bent Little Fingerbf = Straight Little Finger
Mid-digital Hair
M = Mid-Digital Hairm = Absence of Mid-Digital Hair
Dimples
D = Dimplesd = Absence of Dimples
Short Hallux
Ha = Short Halluxha = Long Hallux
Short Index Finger
Ss = Short Index FingerS1 = Long Index Finger
*Sex-Influenced Trait
http://www.youtube.com/watch?v=gCPuHzbb5hA
The Chi Square Test
• A statistical method used to determine goodness of fit
• Goodness of fit refers to how close the observed data are to those predicted from a hypothesis
• Note:• The chi square test does not prove that a
hypothesis is correct• It evaluates to what extent the data and the
hypothesis have a good fit
The Chi Square Test
• The general formula is
c2 =∑(o – e)2
e
• where – o = observed data in each category– e = observed data in each category based on the
experimenter’s hypothesis– ∑ = Sum of the calculations for each category
We start with a theory for how the offspring will be distributed: the “null hypothesis”.
The null hypothesis is that the offspring will appear in a ratio of 3/4 dominant to 1/4 recessive.
Example of its use:Observed Expected*Frequency Frequency
(o) (e)red flowers 73 75white flowers 27 25
*The expected frequency depends upon the hypothesis.
o - e (o - e )2 (o - e ) 2
ered flowers -2 4 0.053white flowers 2 4 0.16
c2 =∑(O – E)2
E = 0.053 + 0.16 = 0.213
Degrees of Freedom
• A critical factor in using the chi-square test is the “degrees of freedom”, which is essentially the number of independent random variables involved.
• Degrees of freedom is simply the number of classes of offspring minus 1 (d.f. = n-1).
• For our example, there are 2 classes of offspring: red and white. Thus, degrees of freedom (d.f.) = 2 -1 = 1.
Critical Chi-Square
• Critical values for chi-square are found on tables, sorted by degrees of freedom and probability levels. Be sure to use p = 0.05.
• If your calculated chi-square value is greater than the critical value from the table, you “reject the null hypothesis”.
• If your chi-square value is less than the critical value, you “fail to reject” the null hypothesis (that is, you accept that your genetic theory about the expected ratio is correct).
d.f. 0.95 0.90 0.80 0.70 0.50 0.30 0.10 0.05 0.01
1 .004 .016 .064 .148 .455 1.07 2.71 3.84 6.642 .103 .211 .446 .713 1.38 2.41 4.6 5.99 9.203 .352 .584 1.00 1.42 2.37 3.66 6.25 7.85 11.3
probability
Chi-Square Table 2 = 0.213
Your value
Do you accept or reject the null hypothesis?