1. 1 Genetics copyright cmassengale

2. Gregor Mendel Mendel (1822): crossed pea plants to study the inheritance of traits. –Father of Genetics Why did Mendel use pea plants? –They reproduce sexually –They have two distinct, male and female, sex cells called gametes –Their traits are easily to isolate

3. Mendel Cont. Mendel crossed (fertilized) the pea plants and studied the results Mendel concluded 2 things: 1.Inheritance is determined by factors that are passed from one generation to the next 2.Some alleles are dominant and others are recessive

4. Phenotype and Genotype Phenotype: They way an organism looks –EX: red hair, brown hair, short, tall Genotype: the gene combination of an organism –EX: AA or Aa or aa Alleles – copies of genes from parents

5. Principle of Dominance Dominant Trait: the trait that is expressed –traits always appear; uppercase letter (A) Recessive Trait: the trait that is unexpressed –traits that dissappear; lowercase letter (a) –for a recessive trait to appear, both parents must give the recessive allele Homozygous: The two alleles for a trait is the same; EX: (AA or aa) Heterozygous: The two alleles for a trait is different; EX: (Aa)

6. 6 Punnett Square Used to help solve genetics problems copyright cmassengale

7. Solving Crosses Dominant = A (AA or Aa) Recessive = a (aa) Solve crosses using Punnett Square Punnett Square: diagram used to determine the probability of an offspring having a particular genotype  Does not show what the offspring will be. Answer questions based on results from the Punnett Square _______

8. 8copyright cmassengale

9. Monohybrid Crosses Monohybrid Cross: Cross that involves one pair of contrasting traits. Ex: hair color  Solve using Punnett Square Sample problems: Red hair (R)is dominant to white hair (r) 1.RR x rr → red hair x white hair 2.Rr x rr → red hair x white hair 3.Rr x Rr → red hair x red hair 4.Rr x RR → red hair x red hair

10. Sample Problems _________ _______ _________ _______ RR r r 1.2. 3. 4. Rr Rr r r rr Rrrr Rr R r Rr R R

11. Sample Question Short hair (L) is dominant to long hair (l) in mice. What is the genotype and phenotype ratio of a heterozygous short- haired mouse crossed with a long-haired mouse?

12. 12 Generation “Gap” Parental P 1 Generation = the parental generation in a breeding experiment.Parental P 1 Generation = the parental generation in a breeding experiment. F 1 generation = the first-generation offspring in a breeding experiment. (1st filial generation)F 1 generation = the first-generation offspring in a breeding experiment. (1st filial generation) –From breeding individuals from the P 1 generation F 2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation)F 2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation) – From breeding individuals from the F 1 generation copyright cmassengale

13. 13 Following the Generations Cross 2 Pure Plants TT x tt Results in all Hybrids Tt Cross 2 Hybrids get 3 Tall & 1 Short TT, Tt, tt copyright cmassengale

14. 14 Law of Dominance In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation. All the offspring will be heterozygous and express only the dominant trait. RR x rr yields all Rr (round seeds) copyright cmassengale

15. 15 Law of Dominance copyright cmassengale

16. 16 Law of Segregation During the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other.During the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other. Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspringAlleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring. copyright cmassengale

17. 17 Applying the Law of Segregation copyright cmassengale

18. 18 Law of Independent Assortment Alleles for different traits are distributed to sex cells (& offspring) independently of one another.Alleles for different traits are distributed to sex cells (& offspring) independently of one another. This law can be illustrated using dihybrid crosses.This law can be illustrated using dihybrid crosses. copyright cmassengale

19. Dihybrid Crosses Cross that involves two pairs of contrasting traits Examples: –Pea shape and pea color –Coat length and coat color in rodents –Plant height and flower color Solve using Punnett square –Punnett square has 16 boxes

20. Sample Problem In guinea pigs, the allele for short hair (S) is dominant to long hair (s), and the allele for black hair (B) is dominant over the allele for brown hair (b). What is the probable offspring phenotype ratio for a cross involving two parents that are heterozygotes for both traits?

21. Steps to solving Dihybrid Crosses 1.Set up 16 box punnett square 2.Figure out parent genotypes and phenotypes 3.FOIL both parents (first, out, in, last) 4.Fill out punnett square

22. Sample Problem Short hair = dominant = SS or Ss Long Hair = recessive = ss Black coat = dominant = BB or Bb Brown coat = recessive = bb Parents are heterozygous (SsBb) SsBb x SsBb (gametes done by the FOIL method) –SB, Sb, sB, sb and SB, Sb, sB, sb

23. Sample Problem Cross SBSbsBsb SBSSBBSSBBSSBbSsBBSsBbSsBb SbSSBbSSBbSSbbSsBbSsbbSsbb sBSsBBSsBBSsBbssBBssBbssBb sbSsBbSsBbSsbbssBbssbbssbb

24. Sample Problem Cross SBSbsBsb SBSSBB short, black SSBb short, black SsBB short, black SsBb short, black SbSSBb short, black SSbb short, brown SsBb short, black Ssbb short, brown sBSsBB short, black SsBb short, black ssBB long, black ssBb long, black sbSsBb short, black Ssbb short, brown ssBb long, black ssbb long, brown

25. Sample Answer What is the probable offspring phenotype ratio for a cross involving two parents that are heterozygotes for both traits? –9/16 Short coats, black –3/16 Long coats, black –3/16 Short coats, brown –1/16 Long coats, brown

26. Incomplete Dominance Incomplete Dominance: one allele is not completely dominant over another allele –Heterozygous is intermediate (blending)

27. Codominance Codominance: both alleles of a gene contribute to the phenotype –Heterozygous shows both traits

28. Multiple Alleles: Multiple genes control trait Human Blood Type Codominance: Two genes are shown as dominant A, B alleles govern presence of different surface antigens on blood cells. O allele is absence of A or B antigen; O is recessive. A, B alleles are codominant; if both alleles are present, AB phenotype occurs

29. Sex Linked Traits Genes located on the X or Y chromosomes are sex linked If it is X-linked it is carried on the X chromosome, the same process applies to the Y chromosome Colorblindness, hemophilia, and male pattern baldness are x-linked traits

30. Hemophelia Hemophelia is a blood clotting disorder that is a recessive trait found on the X chromosome. Let’s say that H = normal blood clotting and h = hemophelic …

31. Mother Without hemophilia = X X With hemophilia = X X Father Without hemophilia = X y With hemophilia = X y H H h h CARRIER H h

32. X H y XH XH X h Make a cross with an X-linked gene Carrier Mom X Normal dad X H X H X H y X H X h X h y GIRLS _______ = normal _______ = look normal but are CARRIERS BOYS _______ = normal _______ = hemophilia 1/2

33. Sex chromosomes can carry other genes Y-LINKED GENES: Genes carried on Y chromosome EX: Hairy pinna _________genes only show up in _______ Y linkedmales.

34. Make a cross with a y-linked gene Hairy ears is a ________________ ________________ trait Y linked dominant Use ______ for hairy ears. Use ______ for recessive normal ears. On y chromosome so write it as ________ H h yHyH

35. Make a cross with a y-linked gene X X X yHyH X X y H ALL GIRLS = ____________ Normal ears ALL BOYS = ____________ Hairy ears

36. karyotype Karyotype: diagram of chromosomes by size and number –Humans have 22 pairs of autosomes and 1 pair of sex chromosomes (XX or XY) Normal Human Male Karyotype

37. Genetic Disorders Nondisjunction: failure of chromosomes to separate during meiosis –EX: Down syndrome and Turner syndrome Down syndrome: (trisomy 21) Genetic disorder in which there is an extra copy of the 21 st chromosome –Mild to severe physical and mental defects –Occurs in about 1 in every 700 people Turner Syndrome: (monosomy X) Genetic disorder in which females only have 1 X chromosome –reproductively sterile (can’t reproduce) –underdeveloped ovaries –Occurs in about 1 in every 2,500 females

38. Down Syndrome

39. Turner Syndrome

40. Mutations Mutation: change in a DNA sequence that affects genetic information Most mutations have little to no effect on the functions of proteins Some mutations are bad & some are good

41. Mutations Cont. Harmful mutations are mutations that produce dramatic changes in protein structure –Causes many genetic disorders such as cystic fibrosis and sickle cell anemia Beneficial mutations can be useful to organisms in changing environments  EX: 32 base pair deletion in human protein CCR5 (located on chromosome 3)  mostly in Northern European descendents of the bubonic plague –CCR5 homozygous: immune to AIDS & smallpox – CCR5 heterozygous: delays onset of AIDS

42. 2 Types of Mutations 1.Gene mutations: produce change in a single gene –EX: TAC GCA to TAT GCA 2.Chromosomal mutations: produce changes in whole chromosomes –EX: deletions, duplications, inversions, & translocations

43. Chromosomal Mutations Deletion: loss of all or parts of a chromosome Duplication: produce extra copies of parts of a chromosome Inversion: reverse the direction of parts of chromosomes Translocation: part of one chromosome breaks off and attaches to another

44. Chromosomal Mutations Deletion Duplication Inversion Translocation