Genetics. The study of heredity. discovered the. Gregor Mendel (1860 s) garden peas.

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1 GENETICS

2 Genetics The study of heredity. Gregor Mendel (1860 s) discovered the fundamental principles of genetics by breeding garden peas.

3 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)

4 Phenotype Outward appearance Physical characteristics Examples: 1. tall pea plant 2. dwarf pea plant

5 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)

6 Punnett square A Punnett square is used to show the possible combinations of gametes.

7 Breed the P generation tall (TT) vs. dwarf (tt) pea plants T T t t

8 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)

9 Breed the F 1 generation tall (Tt) vs. tall (Tt) pea plants T t T t

10 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

11 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.

12 Homologous Chromosomes eye color locus B = brown eyes eye color locus b = blue eyes This person would have brown eyes (Bb) Paternal Maternal

13 Meiosis - eye color B B B sperm Bb diploid (2n) b b haploid (n) meiosis I meiosis II b

14 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

15 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

16 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)

17 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

18 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

19 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

20 Dihybrid Cross RY Ry ry ry RY Ry ry ry

21 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

22 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

23 Test Cross Possible results: bc b C bc b c bc bbcc bbcc or bc bbcc bbcc

24 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

25 Incomplete Dominance R R r Rr Rr produces the F 1 generation r Rr Rr All Rr = pink (heterozygous pink)

26 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

27 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

28 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

29 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 )

30 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

31 Sex-linked Traits Traits (genes) located on the sex chromosomes Example: fruit flies (red-eyed male) X (white-eyed female)

32 Sex-linked Traits Sex Chromosomes fruit fly eye color XX chromosome - female Xy chromosome - male

33 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

34 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

35 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

36 Question: How do we get this equation? Answer: Square 1 = p + q 1 2 = (p + q) 2 1 = p 2 + 2pq + q 2

37 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

38 Important Need to remember the following: p 2 = homozygous dominant 2pq = heterozygous q 2 = homozygous recessive

39 Question: Iguanas with webbed feet (recessive trait) make up 4% of the population. What in the population is heterozygous and homozygous dominant.

40 Answer: 1. q 2 = 4% or.04 q 2 =.04 q =.2 2. then use 1 = p + q 1 = p = p.8 = p 3. for heterozygous use 2pq 2(.8)(.2) =.32 or 32% 4. For homozygous dominant use p =.64 or 64%

41 Hardy-Wienberg equation 1 = p 2 + 2pq + q 2 64% = p 2 = homozygous dominant 32% = 2pq = heterozygous 04% = q 2 = homozygous recessive 100%

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