GENETICS
Genetics The study of heredity. Gregor Mendel (1860 s) discovered the fundamental principles of genetics by breeding garden peas.
Genetics Alleles 1. Alternative forms of genes. 2. Units that determine heritable traits. 3. Dominant alleles (TT - tall pea plants) a. homozygous dominant 4. Recessive alleles (tt - dwarf pea plants) a. homozygous recessive 5. Heterozygous (Tt - tall pea plants)
Phenotype Outward appearance Physical characteristics Examples: 1. tall pea plant 2. dwarf pea plant
Genotype Arrangement of genes that produces the phenotype Example: 1. tall pea plant TT = tall (homozygous dominant) 2. dwarf pea plant tt = dwarf (homozygous recessive) 3. tall pea plant Tt = tall (heterozygous)
Punnett square A Punnett square is used to show the possible combinations of gametes.
Breed the P generation tall (TT) vs. dwarf (tt) pea plants T T t t
tall (TT) vs. dwarf (tt) pea plants T T t Tt Tt produces the F 1 generation t Tt Tt All Tt = tall (heterozygous tall)
Breed the F 1 generation tall (Tt) vs. tall (Tt) pea plants T t T t
tall (Tt) vs. tall (Tt) pea plants T t T TT Tt produces the F 2 generation t Tt tt 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype
Monohybrid Cross A breeding experiment that tracks the inheritance of a single trait. Mendel s principle of segregation a. pairs of genes separate during gamete formation (meiosis). b. the fusion of gametes at fertilization pairs genes once again.
Homologous Chromosomes eye color locus B = brown eyes eye color locus b = blue eyes This person would have brown eyes (Bb) Paternal Maternal
Meiosis - eye color B B B sperm Bb diploid (2n) b b haploid (n) meiosis I meiosis II b
Monohybrid Cross Example: BB = brown eyes Bb = brown eyes bb = blue eyes Bb x Bb Cross between two heterozygotes for brown eyes (Bb) B B b male gametes b female gametes
Monohybrid Cross B b Bb x Bb B BB Bb 1/4 = BB - brown eyed 1/2 = Bb - brown eyed 1/4 = bb - blue eyed b Bb bb 1:2:1 genotype 3:1 phenotype
Dihybrid Cross A breeding experiment that tracks the inheritance of two traits. Mendel s principle of independent assortment a. each pair of alleles segregates independently during gamete formation (metaphase I) b. formula: 2 n (n = # of heterozygotes)
Independent Assortment Question: How many gametes will be produced for the following allele arrangements? Remember: 2 n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq
Answer: 1. RrYy: 2 n = 2 2 = 4 gametes RY Ry ry ry 2. AaBbCCDd: 2 n = 2 3 = 8 gametes ABCD ABCd AbCD AbCd abcd abcd abcd abcd 3. MmNnOoPPQQRrssTtQq: 2 n = 2 6 = 64 gametes
Dihybrid Cross Example: cross between round and yellow heterozygous pea seeds. R r Y y = round = wrinkled = yellow = green RrYy x RrYy RY Ry ry ry x RY Ry ry ry possible gametes produced
Dihybrid Cross RY Ry ry ry RY Ry ry ry
Dihybrid Cross RY Ry ry ry RY RRYY RRYy RrYY RrYy Round/Yellow: 9 Ry RRYy RRyy RrYy Rryy Round/green: 3 wrinkled/yellow: 3 ry RrYY RrYy rryy rryy wrinkled/green: 1 ry RrYy Rryy rryy rryy 9:3:3:1 phenotypic ratio
Test Cross A mating between an individual of unknown genotype and a homozygous recessive individual. Example: bbc x bbcc BB = brown eyes Bb = brown eyes bb = blue eyes bc b CC = curly hair Cc = curly hair cc = straight hair bc
Test Cross Possible results: bc b C bc b c bc bbcc bbcc or bc bbcc bbcc
Incomplete Dominance F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. Example: snapdragons (flower) red (RR) x white (rr) R R RR = red flower rr = white flower r r
Incomplete Dominance R R r Rr Rr produces the F 1 generation r Rr Rr All Rr = pink (heterozygous pink)
Codominance Two alleles are expressed (multiple alleles) in heterozygous individuals. Example: blood 1. type A = I A I A or I A i 2. type B = I B I B or I B i 3. type AB = I A I B 4. type O = ii
Codominance Example: homozygous male B (I B I B ) x heterozygous female A (I A i) I B I B I A i I A I B I B i I A I B I B i 1/2 = I A I B 1/2 = I B i
Codominance Example: male O (ii) x female AB (I A I B ) I A I B i I A i I B i 1/2 = I A i 1/2 = I B i i I A i I B i
Codominance Question: If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents. boy - type O (ii) X girl - type AB (I A I B )
Codominance Answer: I A i I B i I A I B ii Parents: genotypes = I A i and I B i phenotypes = A and B
Sex-linked Traits Traits (genes) located on the sex chromosomes Example: fruit flies (red-eyed male) X (white-eyed female)
Sex-linked Traits Sex Chromosomes fruit fly eye color XX chromosome - female Xy chromosome - male
Sex-linked Traits Example: fruit flies (red-eyed male) X (white-eyed female) Remember: the Y chromosome in males does not carry traits. RR = red eyed Rr = red eyed rr = white eyed X r X R y Xy = male XX = female X r
Sex-linked Traits X R y X r X r X R X r X R X r X r y X r y 1/2 red eyed and female 1/2 white eyed and male
Population Genetics The study of genetic changes in populations. The science of microevolutionary changes in populations. Hardy-Weinberg equilibrium: the principle that shuffling of genes that occurs during sexual reproduction, by itself, cannot change the overall genetic makeup of a population. Hardy-Wienberg equation: 1 = p 2 + 2pq + q 2
Question: How do we get this equation? Answer: Square 1 = p + q 1 2 = (p + q) 2 1 = p 2 + 2pq + q 2
Hardy-Wienberg equation Five conditions are required for Hardy-Wienberg equilibrium. 1. large population 2. isolated population 3. no net mutations 4. random mating 5. no natural selection
Important Need to remember the following: p 2 = homozygous dominant 2pq = heterozygous q 2 = homozygous recessive
Question: Iguanas with webbed feet (recessive trait) make up 4% of the population. What in the population is heterozygous and homozygous dominant.
Answer: 1. q 2 = 4% or.04 q 2 =.04 q =.2 2. then use 1 = p + q 1 = p +.2 1 -.2 = p.8 = p 3. for heterozygous use 2pq 2(.8)(.2) =.32 or 32% 4. For homozygous dominant use p 2.8 2 =.64 or 64%
Hardy-Wienberg equation 1 = p 2 + 2pq + q 2 64% = p 2 = homozygous dominant 32% = 2pq = heterozygous 04% = q 2 = homozygous recessive 100%