Knowt
AP Bio Unit 4 Genetics
Chapter 9 Mitosis
Chapter 10 Meiosis
Chapter 11 Mendelian Genetics
Character: A heritable feature that varies among individuals (i.e. color)
Trait: Each variant for a character (i.e. white flowers vs purple flowers)
Generations: P generation—parents; F1 generation—first filial generations cross from parents; F2 generation—second filial generation crossing F1 generations
Mendel’s Model
1) Alleles/alternative versions of genes account for variations in inherited characters
Results from slight variations in nucleotide sequence along chromosomes
2) For each character, an organism inherits TWO versions of a gene, one from each
3) If two alleles in an organism differ (heterozygous), then one is a dominant allele that determines the organism’s appearance and the other recessive allele has no noticeable effect
4) Law of segregation—the two alleles for a heritable character segregate (separate from each other) during gamete formation and end up in different gametes
Egg/sperm gets only one of the two alleles present in somatic cells of the organism (i.e. with heterozygous 50% of gametes receive dominant and 50% receive recessive allele)
Heterozygote—has two different alleles for a gene (heterozygous); homozygote—same alleles for a gene (homozygous)
Phenotype- the appearance/observable traits (purple or white)
Genotype- the genetic makeup (PP, Pp, pp)
Testcross
Used to determine the genotype of an unknown dominant trait (whether it is homozygous or heterozygous) by crossing with a recessive
If homozygous all offspring will show the dominant trait because dominant allele is always present; if heterozygous, half of the offspring will show the recessive trait
Law of Independent Assortment
Law of Segregation based on a SINGLE character with all F1 generations being monohybrids—heterozygous for one particular being followed in the cross (Monohybrid cross)
Dihybrids- individual heterozygous for TWO characters being followed in the cross (Dihybrid cross)
Law of Independent Assortment—two of more genes assort independently—each pair of alleles segregates independently during gamete formation
Statistics of Inheritance—laws of probability govern mendelian inheritance
Multiplication Rule-
Probability that 2+ INDEPENDENT events will occur together in a specific combination → multiply probabilities of each event
Ex. Crossing AABbCc x AaBbCc probability of AaBbcc
Probability of each TRAIT (denoted by the different letters A, B, C ) → ½ (A) x ½ (B) x ¼ (C) = 1/16
AA x Aa = 50% chance of Aa or AA; Bb x Bb = 50% chance of Bb and 25% chance of BB or bb; Cc x Cc = 50% chance of Cc and 25% chance of cc or CC
Ex. Crossing Rr x Rr to get RR or rr
½ probability for carrying dominant allele (R) or recessive (r ) probability that you will get TWO recessive alleles present is ½ x ½ = ¼
Addition Rule-
Probability that 2+ MUTUALLY EXCLUSIVE events will occur → add together individual probabilities
Ex. Throwing a die landing on a 4 OR 5 → 1/6 + 1/6 = 1/3
Ex. Crossing Rr x Rr to get Rr
½ probability of rR and ½ probability of Rr → ¼ + ¼ = ½
More Complex Genetics
Incomplete Dominance- When hybrids (heterozygous) have an appearance BETWEEN that of 2 parents (i.e. red x white = pink)
Complete Dominance- heterozygote & homozygote for dominant allele are indistinguishable
Codominance- phenotype of BOTH alleles is expressed (i.e. red hair x white hair = roan horses)
Indicated not by capital/lowercase but by exponent (i.e. L^M vs L^N)
Multiple Alleles- Gene has 2+ alleles for the trait
Ex. Human ABO blood groups
I^A, I^B, i: I^A & I^B are Codominant creating type AB blood; i is recessive creating type O blood
Chi-Squared (X²) test
Used to determine if there is a significant difference between the expected and observed data
Null Hypothesis- NO statistically significant different between expected and observed data
Formula:
X² = the sum of (O - E)²/E; Observed frequencies, Expected frequencies
Steps:
Determine the null hypothesis
Use formula to calculate the X² value
n = number of categories, e = expected frequency/value, o= observed frequency/value
Calculate expected frequency/count—multiple total by expected percentage to get numbers
Plug into formula (observed value - expected value)²/expected value
Add all together for X² value
Find df with number of categories - 1
Find critical value using table (Use p = 0.05 (default) or p = 0.01)
P-value probability- how often our results could happen due to change
Degrees of freedom (df) = n - 1
If X² < critical value…FAIL to reject the null hypothesis
Differences in data due to change
If X² > critical value…reject the null hypothesis
Differences in data NOT due to chance
Example
Total M&M’s—100; 6 types of M&M’s each color has 20; Claimed percentage - 14% yellow
Expected frequencies- Total * (Percent/100)
Calculation:
(O-E)²/E = (20 - 14)²/14 = 2.57…. add all M&M’s together to get SUM(X²) = 21.01
df = n - 1 = 6 - 1 = 5
p = 0.05
critical value = 11.07 (look on table using p value and degrees of freedom)
RESULT: 21.01 > 11.07 we REJECT the null hypothesis because X² > critical value
General Rules
Crossing 2 heterozygous Xx & Xx: 50% chance of heterozygous (Xx) 25% chance of homozygous dominant/recessive (XX or xx)
Crossing homozygous and heterozygous: 50% chance of heterozygous (Xx) 50% of homozygous same as homozygous parent alleles either dominant or recessive
Heterozygous Dihybrid cross PpRr x PpRr = 9:3:3:1; 9 Pp/PP/Rr/RR, 3Pp/PP/rr 3pp/Rr/rr 1pprr