Test Content 30-40 pts. mult. choice and short answer from chapter two.
Chi-square analysis
- Mendel's Ratios
- Monohybrid 3:1
- Dihybrid 9:3:3:1
- Ratios are based on assumptions:
- each allele is either dominant or recessive
- normal segregation
- independent assortment
- fertilization is random
- Chi square analysis allows you to test your assumptions
- Does your data match your assumptions over a number of cases?
- Begin with a null hypothesis (Ho). What should your data look like based on your assumptions. This is determined by working out your problem.
- EX. (8:00)
- Yellow, round x green, wrinkled
- 9/16 yellow, round -->587
- 3/16 yellow, wrinlked--> 197
- 3/16 green, round --> 168
- 1/16 green, wrinkled --> 56
- Total of offspring ------>1,008
- Multiply the ratios by the total number of offspring to determine the number of expected offspring
# of offspring expected (E) (D) Deviation D^2 D^2/E 587 567 20 400 .71 197 189 8 64 .34 168 189 -21 441 2.33 56 63 -7 49 .78 Total 1008 x^2=sum of d^2/E which equals = 4.16 - --
- What does the x^2 value tell us?
- Determine the degrees of freedom by taking n (where n is = to the number of unique individuals) and subtracting 1.
- In the above problem df=4-1 . . . . 3.
- You can calculate the P-value. A P-value of .05 --> 1 Fail to reject your original hypothesis. A p-value below .05 you reject your original hypothesis. What you are saying is that there is a greater than 95% chance that your variation is random.
- We failed to reject the Ho because 4.16 is less than 7.82.
- hw 8,11,19,24,28
- Concepts
- Wildtype --> the typical phenotype for a particular trait in a population. Typically a wildtype trait is dominant.
- Mutant - trait caused by a mutation. typically recessive
- New letters to be used
- e+/e
- e+ = wildtype trait (most often dominant)
- e = mutant trait
- multiple alleles: I^a I^v I^o
- Incomplete dominance - two alleles for a trait and neither allele overtakes the other one completely. "A form of intermediate inheritance in which heterozygous alleles are both expressed, resulting in a combined phenotype." (http://biology.about.com/bldefincomdom.htm)
- Snapdragon flowers R=red r=white
- homozygous red, homozygous white, heterozygous pink
- Typical F1 cross: Rr x Rr
- 1/4 RR (red 1/4), 1/2 Rr (pink 1/2), 1/4rr (white 1/4)
- Tay - sach's disease
- Homozygous recessive. fail to produce enzyme hexosaminidase. Individuals die by age 3.
- Heterozygous 50% of the WT level of the hexosaminidase your phenotype is normal. (Het. are the carriers)
- Homozygous dominant you will make 100% of the WT level of enzyme and are phenotypically normal.
- Codominance - Two alleles of a trait and the heterozygote expresses both alleles. the heterozygote displays the phenotypic characteristics of both alleles. MN blood type is governed by two alleles, M and N. Individuals who are homozygous for the M allele have a surface molecule (called the M antigen) on their red blood cells. The gene that governs ABO blood types has three alleles: IA, IB, and IO. IA and IB are codominant, but IO is recessive.
- Example is the MN blood groups in humans. a glycoprotein found on the surface of red blood cells
- M and N are different versions of this glycoprotein.
- Lm Ln. Cross a heterozygote by M blood group
- Lm Ln x Lm Lm.
- 1/2 Lm Lm --> M blood group
- 1/2 Lm Lm --> M + N blood groups
- ABO bloodtypes you can have more than 2 alleles in a population. The gene that governs ABO blood types has three alleles: IA, IB, and IO. IA and IB are codominant, but IO is recessive.
- Ia --> presence of the antigen
- Ib --> presence of the B antigen
- Io --> absence of antigen
- Jen is "O" Ben is "B" blood type. Could they produce an offspring with "O" blood type?
- Jen is Io Io Ben is IbIo or IbIb
- If ben is IbIb then all offspring are IbIo
- IbIo x Io Io then 1/2 (O blood type) and 1/2 IbIo (B blood type)
- AB x AB. results:
- 1/4 aa is ab blood
- 1/2 ab which is ab
- 1/4 bb is b blood
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