GENETICS Mendelian & Human Review

1. GENETICSMendelian & HumanReview Modified with permission from Robert Goodman

2. Genetics The study of heredity Gregor Mendel (1860’s) discovered the fundamental principles of genetics by breedinggardenpeas http://www.jic.bbsrc.ac.uk/germplas/pisum/zgs4f.htm

3. Alleles • Alternative forms of genes • Units that determine heritable traits • Dominant alleles (T - tall pea plants) • Recessive alleles (t - dwarf pea plants) • Homozygous dominant • TT - tall pea plants • Homozygous recessive • tt - dwarf pea plants • Heterozygous • Tt - tall pea plants

4. Phenotype Outward appearance Physical characteristics Examples: Tall pea plant Dwarf pea plant

5. Genotype Arrangement of genes that produces phenotype Example: • Tall pea plant • TT = tall (homozygous dominant) • Dwarf pea plant • tt = dwarf (homozygous recessive) • Tall pea plant • Tt = tall (heterozygous)

6. Tongue rollers vs Non-tongue rollers

7. Punnett Square A Punnett square is used to show the combinations of gametes and possible offspring

8. T T produces the F1 generation Tt Tt t Tt Tt All Tt = tall (heterozygous tall) t Breed the P Generation Tall (TT) vs. dwarf (tt) pea plants

9. produces the F2 generation T t 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt T 1:2:1 genotype 3:1 phenotype t Breed the F1 generation Tall (Tt) vs. Tall (Tt) pea plants Tt TT Tt tt

10. Monohybrid Cross • A breeding experiment that tracks the inheritance of a single trait • Mendel’s “Principle of Segregation” • Pairs of genes separate during gamete formation (meiosis) • Fusion of gametes at fertilization re-pairs genes

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

13. B sperm BB B BBbb haploid (n) b bb diploid (2n) b meiosis II meiosis I Meiosis - Eye Color

14. male gametes B b 1/4 = BB - brown eyed 1/2 = Bb - brown eyed 1/4 = bb - blue eyed B Bb x Bb 1:2:1 genotype 3:1 phenotype b female gametes Monohybrid Cross Example: Cross between two heterozygotesfor brown eyes (Bb) BB = brown eyes Bb = brown eyes bb = blue eyes BB Bb Bb bb

15. Dihybrid Cross • A breeding experiment that tracks the inheritance of two traits • Mendel’s “Principle of Independent Assortment” • Each pair of alleles segregates independently during gamete formation (metaphase I) • Formula: 2n (n = # of heterozygotes)

16. BT sperm Bt BbTt haploid (n) bT diploid (2n) bt meiosis II Independent Assortment

17. Independent Assortment Question: How many gametes will be produced for the following allele arrangements? Remember: 2n (n = # of heterozygotes) • RrYy • AaBbCCDd • MmNnOoPPQQRrssTtQq

18. Answer: • RrYy: 2n = 22 = 4 gametes • RY Ry rY ry • AaBbCCDd: 2n = 23 = 8 gametes • ABCD ABCd AbCD AbCd • aBCDaBCDabCDabCd • MmNnOoPPQQRrssTtQq: • 2n = 26 = 64 gametes

19. BT sperm Bt BbTt haploid (n) bT diploid (2n) bt meiosis II Independent Assortment

20. RY Ry rY ryx RY Ry rY ry possible gametes produced Dihybrid Cross Example: cross between round and yellow heterozygous pea seeds R = round r = wrinkled Y = yellow y = green RrYy x RrYy

21. RY Ry rY ry RY Round/Yellow: 9 Round/green: 3 Ry wrinkled/Yellow: 3 wrinkled/green: 1 rY ry Dihybrid Cross RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy 9:3:3:1 phenotypic ratio RrYy Rryy rrYy rryy

22. Multiple Alleles – Polygenics • Skin color is a polygenic trait, additive effects (essentially, incomplete dominance) of multiple genes on a single trait • Multiple genes produce a continuous distribution in a “Bell Shape” curve of degrees of light to dark • Early models suggested 2 or 4 major genes • Recent work suggests many more genes working together in very complex, additive and non-additive combinations • AaBbCcDdEeFf http://www.as.ua.edu/ant/bindon/ant570/topics/Skincolor.PDF

23. Polygenic – Skin Color • Each gene has two forms • An allele for high melanin production, or dark skin (A,B,C) • An allele for low melanin production, or light skin (a,b,c) • Each dark skin allele (A,B,C) in the genotype adds a small but equal amount of pigment to the skin

26. C ___ c Test Cross A mating between to determine genotype of an individual ofunknown genotypeand a homozygous recessive individual • Example:C__ x cc CC = curly hair Cc = curly hair cc = straight hair

27. c C ___ C C ___ c Cc Cc or c Cc cc Test Cross Possible results:

28. Codominance In Codominance,Multiple Allelesareexpressed in heterozygous individuals Example: Blood ABO system 1. type A = IAIA or Iai (AA or AO) 2. type B = IBIB or Ibi (BB or BO) 3. type AB = IAI B (AB) 4. type O = ii (OO)

29. IB IB IAIB IAIB IA 1/2 = IAIB 1/2 = IBi i IBi IBi Codominance Example: Homozygous male B (IBIB) x Heterozygous female A (IAi)

30. IA IB IAi IBi i 1/2 = IAi 1/2 = IBi i IAi IBi Codominance Example:male O (ii) x female AB (IAIB)

31. 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 (IAIB) OO AB

32. IA i IAIB IB i ii Codominance Answer: Parents: genotypes = IAi and IBi phenotypes = A and B

33. Question #7 • Charlie Chaplin, a film star, was involved in a paternity case. The woman bringing suit had two children, on whose blood type was A and the other whose blood type was B. • Her blood type was O, the same as Charlie ’s! • The judge in the case awarded damages to the woman, saying that Charlie had to be the father of at least one of the children.

34. Answer #7A • Obviously, the judge should be sentenced to Biology. For Charlie to have been the father of both children, his blood type would have had to be what? IA IB Answer AB i IAi IBi i IAi IBi

35. produces the F1 generation Rr Rr R R r Rr Rr All Rr = pink flower (heterozygous) r Incomplete Dominance F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties Example:snapdragons (flower) red flower (RR) x white flower (rr)

36. Sex Determination Sex Chromosomes XX chromosome - female Xy chromosome - male

37. X y X X Sex Determination XX Xy XX Xy

38. OtherSex Determination • The Y chromosome sometimes does not dictate its maleness • Absence of a second X • XY fruit fly is male • XXY fruit fly is female

39. Environmental Sex Determination • Sex may be determined after fertilization • Determined by temperature during early embryonic development • Turtles produce more females at a higher temperature • Alligators and many lizards produce more males at a higher temperature

41. Sex-linked Traits Traits (genes) located on the sex chromosomes • Hemophiliacs (X-linked) • Male Pattern Baldness (X-linked) • Color-blindness (X-linked) • Male Ear Hair (y-linked)

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

43. XN y XN Xn XN XN XN y XN Xn Xny N = normal n = Hemophilia

44. XN y XN Xn Male Pattern Baldness XN XN XN y XN Xn Xny N = normal n = Bald

45. XN y XN Xn Colorblindness XN XN XN y XN Xn Xny N = normal n = Colorblind http://www.toledo-bend.com/colorblind/Ishihara.html

46. X y X X Y-linked Ear-Hair XX Xy XX Xy y = Ear Hair

47. Y linked • 3 muslim brothers in South India

48. Barr Bodies • Barr discovered the Barr Body • An inactive X chromosome • a darkly staining body in the nuclei of females • Mary Lyon, identified the Barr body as an inactive X chromosome • Inactivation is random, with a 50 - 50 chance of inactivating the maternal or paternal X • The mammalian female is a genetic mosaic • some cells have the XP active • some have the XM active

49. Barr Bodies • Fur coloration of calico cats is governed by two alleles (black and orange - multiple alleles ) • Both attached to the same loci on a homologous pair of X chromosomes • In black fur cells – orange allele is inactive • In orange fur cells –black allele is inactive Explained in more detail at Barr Bodies and Gender Verification