Oogenesis

1. Oogenesis • As embryo until menopause... • Ovaries • Primordial germ cells (2N) • Oogonium (2N) • Primary oocyte (2N) • Between birth & puberty; prophase I of meiosis • Puberty; FSH; completes meiosis I • Secondary oocyte (1N); polar body • Meiosis II; stimulated by fertilization • Ovum (1N); 2nd polar body

2. Spermatogenesis • Puberty until death! • Seminiferous tubules~ location • Primordial germ cell (2n)~ differentiate into…. • Spermatogonium (2n)~ sperm precursor • Repeated mitosis into…. • Primary spermatocyte (2n) • 1st meiotic division • Secondary spermatocyte (n) • 2nd meiotic division • Spermatids (n)~Sertoli cells…. • Sperm cells (n)

3. Comparing Mitosis & Meiosis

4. Ch. 14 - Mendelian Genetics and the Inheritance of Genetic Traits • Modern genetics began with Gregor Mendel’s quantitative experiments with pea plants Stamen Carpel Figure 9.2A, B

5. Gregor Mendel (Father of Genetics) • Discovered the fundamentals of Genetics in the 1860’s • Lived in Austria and studied in Vienna • Worked with Garden Peas (Pisum sativum) • Gathered a huge amount of numerical data • Discovered the frequency of how traits are inherited • Established basic principles of Genetics

6. MENDEL’S PRINCIPLES • The science of heredity dates back to ancient attempts at selective breeding • Until the 20th century, however, many biologists erroneously believed that • characteristics acquired during lifetime could be passed on • characteristics of both parents blended irreversibly in their offspring

7. Reason Mendel worked with Garden Peas • Easy to grow • Many variations were available • Easy to control pollination (self vs cross) • Flower is protected from other pollen sources (reproductive structures are completely enclosed by petals) • Plastic bags can be used for extra protection

8. White 1 Removed stamensfrom purple flower • Mendel crossed pea plants that differed in certain characteristics and traced the traits from generation to generation Stamens Carpel 2 Transferred pollen from stamens of white flower to carpel of purple flower PARENTS(P) Purple 3 Pollinated carpel matured into pod • This illustration shows his technique for cross-fertilization 4 Planted seeds from pod OFF-SPRING(F1) Figure 9.2C

9. FLOWER COLOR Purple White • Mendel studied seven pea characteristics FLOWER POSITION Axial Terminal • He hypothesized that there are alternative forms of genes (although he did not use that term), the units that determine heredity SEED COLOR Yellow Green SEED SHAPE Round Wrinkled POD SHAPE Inflated Constricted POD COLOR Green Yellow STEM LENGTH Figure 9.2D Tall Dwarf

10. Introductory Questions #4 Traits: Flower Position: Axial & terminal Seed Color: Yellow & green Height: Tall & short 1) Monohybrid cross: Two hybrid plants are tall. How many of the offspring would you predict will be short if there were 400 produced? 2) Dihybrid cross: Two hybrid plants with yellow seeds and axial flowers are crossed. How many of the offspring would you predict will have axial flowers with green seeds if 3750 are produced? 3) Trihybrid cross: Both parents are heterozygous for all three traits. How many will be tall with terminal flowers and yellow seeds if 250 are produced?

11. Genetic Vocabulary • Punnett square: predicts the results of a genetic cross between individuals of known genotype • Homozygous: pair of identical alleles for a character • Heterozygous: two different alleles for a gene • Phenotype: an organism’s traits • Genotype: an organism’s genetic makeup • Testcross: breeding of a recessive homozygote X dominate phenotype (but unknown genotype)

12. Mendelian Genetics Character (heritable feature, i.e., fur color) Trait (variant for a character, i.e., brown) True-bred (all offspring of same variety) Hybridization (crossing of 2 different true-breeds) P generation (parental-beginning gen.) F1 generation (first filial generation) F2 generation (second filial generation)

13. Homologous chromosomes bear the two alleles for each characteristic • Alternative forms of a gene (alleles) reside at the same locus on homologous chromosomes GENE LOCI DOMINANT allele P a B P a b RECESSIVE allele GENOTYPE: PP aa Bb HOMOZYGOUSfor thedominant allele HOMOZYGOUSfor therecessive allele HETEROZYGOUS Figure 9.4

14. Mendel’s principle of segregation describes the inheritance of a single characteristic P GENERATION(true-breedingparents) • From his experimental data, Mendel deduced that an organism has two genes (alleles) for each inherited characteristic • One characteristic comes from each parent Purple flowers White flowers All plants have purple flowers F1generation Fertilization among F1 plants(F1 x F1) F2generation 3/4 of plantshave purple flowers 1/4 of plantshave white flowers Figure 9.3A

15. GENETIC MAKEUP (ALLELES) P PLANTS PP pp Gametes All P All p • A sperm or egg carries only one allele of each pair • The pairs of alleles separate when gametes form • This process describes Mendel’s law of segregation • Alleles can be dominant or recessive F1 PLANTS(hybrids) All Pp Gametes 1/2P 1/2p P P Eggs Sperm PP F2 PLANTS p p Pp Pp Phenotypic ratio3 purple : 1 white pp Genotypic ratio1 PP : 2 Pp : 1 pp Figure 9.3B

16. The principle of independent assortment is revealed by tracking two characteristics at once • By looking at two characteristics at once, Mendel found that the alleles of a pair segregate independently of other allele pairs during gamete formation • This is known as the principle of independent assortment

17. Remember…….. • Every Trait within a diploid organism will have two alleles • These alleles are separated during Meiosis Traits: Seed color, seed shape, height Alleles: Y or y R or r T or t Parents are diploid Gametes are produced: only one allele is present for each trait in each gamete

18. Parents GenotypePossibleAlleles combinations for one gamete Example: monohybrid Yy Y & (y) Dihybrid YyRr YR (3 OTHERS) Trihybrid YyRrTt yRt (7 OTHERS)

19. Mendel’s principles reflect the rules of probability • Inheritance follows the rules of probability • The rule of multiplication and the rule of addition can be used to determine the probability of certain events occurring F1 GENOTYPES Bb female Bb male Formation of eggs Formation of sperm 1/2 B B 1/2 B B 1/2 b b 1/2 1/4 B b b B 1/4 1/4 b b F2 GENOTYPES 1/4 Figure 9.7

20. HYPOTHESIS: DEPENDENT ASSORTMENT HYPOTHESIS: INDEPENDENT ASSORTMENT RRYY rryy Pg. 257 PGENERATION RRYY rryy ry ry Gametes RY Gametes RY F1GENERATION RrYy RrYy Eggs 1/2 RY 1/2 RY Sperm Eggs 1/4 RY 1/4 RY 1/2 ry 1/2 ry 1/4 rY 1/4 rY RRYY 1/4 Ry 1/4 Ry RrYY RrYY F2GENERATION 1/4 ry 1/4 ry RRYy rrYY RrYy Yellow round RrYy RrYy RrYy RrYy 9/16 Actual resultscontradict hypothesis Green round rrYy RRyy rrYy 3/16 ACTUAL RESULTSSUPPORT HYPOTHESIS Yellow wrinkled Rryy Rryy 3/16 Yellow wrinkled rryy 1/16 Figure 9.5A

21. The chromosomal basis of Mendel’s Principles Figure 9.17

22. Important ratios to Remember CrossPhenotypicGenotypic BB x bb 4:0 (100% Dom.) 4:0 (all) Bb BB x Bb 4:0 (100% Dom.) 1:1 (50% BB,Bb) Bb x Bb 3:1 (75% Dom, 25% Rec) 1:2:1 (BB, Bb, bb) Bb x bb 1:1(50% Dom, 50% Rec) 1:1 (50% Bb,bb) Two traits (not linked): AaBb x AaBb 9:3:3:1

23. Introductory Questions #5 • A Monohybrid cross and Dihybrid cross always produces Phenotypic rations of 3:1 and 9:3:3:1. What phenotypic ratio will produced from a trihybrid cross? • Solve this trihybrid cross with Pea plants Traits: Seed color, seed shape, height Male Heterozygous all traits Female Heterozygous yellow and tall plant w/ wrinkled seeds • How many offspring would you predict will be Tall with wrinkled, yellow seeds? • How many offspring would have green seeds that are round and tall?

24. IQ #5-Solution 1) (male) (female) 2) YyRrTt x YyrrTt # allele combo in gametes: 8 4 Possible genotypes of the offspring: (a) TT rr YY: ¼ x ½ x 1/4 = 1/32 TT rr Yy: ¼ x ½ x 1/2 = 1/16 A. Answer Tt rr YY: ½ x ½ x ¼ = 1/16 Tt rr Yy : ½ x ½ x ½ = 1/8 = 9/32 B. 3/32 will be predicted that will be tall with round green seeds

25. Introductory Questions #6 • A female heterozygous for Seed shape and color is crossed with a male that is heterozygous for seed shape but homozygous recessive for seed color. How many offspring would you predict (expect) to be yellow and wrinkled if 500 were produced? • If only 50 offspring were yellow and wrinkled how can you tell if your results were only due to chance? What statistical test could you do in order to determine if there;’s a significant difference between what you actually got (50) vs. what you expected? 201 round & yellow 204 round & green 45 wrinkled and green 50 wrinkled & yellow

26. IQ#6: Question #2 (cont’d) • Expected (calculated Values) Expect. ValueObs. value wrinkled & green 62 45 Round & yellow 187 201 Round & green 187 204 Wrinkled & Yellow 62 50

27. IQ #6-Answers • RrYy x Rryy How many will be yellow and wrinkled? (500) Poss. genotypes: Yyrr = 1/2 x 1/4 = 1/8 Answer: 1/8 x 500 = 62 are expected to be yellow, green Do a chi-squared test Chi-Square value: ???? (Homework) Expected Value for each phenotype: 187,187, 62, 62 Chi-squared value: Critical value from table (in lab) 7.82 (0.05 or 95%)

28. Sample Problem using Chi square • Two hybrid Tall plants are crossed. If the F2 generation produced 787 tall plants and 277 short plants. Does this confirm Mendel’s explanation? • What is the expected value? This is your null hypothesis (HO) • Total number of plants: 1064 • 3:1 Phenotypic ratio • Expected value should be: 798 tall and 266 short (75%) (25%)

29. Statistical Tools to Analyze results • Chi-Square: Will tell you how much your data is different from expected (calculated) results. It is Non-Parametric and deals with different catagorical groups vs. Parametric which deals with numbers and which case you would use a T-test instead. Formula:2 = (o – e)2 e 2: what we are solving: o: observed value e: expected (calculated value) 

30. When to use Chi-Squared Test • Can only be used with raw counts (not measurements) • Comparing Experimental & expected (theo.) values • Sample size must be more than 25 to be reliable • Aims to test the null hypothesis (H0) • (H0): the hypothesis that there’s no difference between the data sets • Alternative hypothesis: there is a significant difference • Compare with a critical value table (p values) • To reject the (H0): value must be GREATER than the critical value & favor the alternative hypothesis. • Accepting the null means that there’s no significant difference between the data sets.

31. Calculation of Chi Square Value 2 = (O – E)2 E 2 = (787 – 798)2 + (277 – 266)2 = 0.61 798 266 There are two categories and therefore the degrees of freedom would be 2-1 = 1 . • Look up the critical value for 1 degree of freedom: 3.84 (next slide-always given) • If your value is LARGER than the critical value then you reject the null hypothesis and assume that there is a significant difference between the observed value and the expected. Values are statistically different. • 0.61 is less than 3.84 therefore we accept the null hypothesis and accept that our values are similar enough There’s no significant difference between the observed & expected values. Values are not random.

32. Answers to IQ #5 & #6 IQ #5----1. 27:9:9:9:3:3:3:1 = 64 phenotypes IQ #6----2. 2 = (O – E)2 E 2 = (45 – 62)2 + (201-187)2 + (204 – 187)2 + (50-62)2 = 9.58 62 187 187 62 Critical Value: 4 groups = 3 degrees of freedom= 7.81 9.58 is greater than 7.81 therefore we reject the null hypothesis What does this mean?

33. Accepting or Rejecting your hypothesis? • Accepting the Null (H0) means that: • Test value is less than the critical value • Values are similar enough • there is a not SIGNIFICANT difference between the observed and expected value (p<0.05). More than 95% confidence • Chance alone cannot explain the differences observed. • Rejecting the Null (H0) means that: • Test value is greater than the critical value • Values are very different from each other • Random Chance can cause the results • the observations are significantly different from the expectations. (p>0.05). Evaluate the results. Less than 95% confidence in the values

34. Solving Question #3 Formula:2 = ∑ (o – e)2 e Degrees of FreedomCritical Value 1 3.84 2 5.99 3 7.81 4 9.49 5 11.07

35. Testcrosses Determining an Unknown Genotype

36. Geneticists use the testcross to determine unknown genotypes • The offspring of a testcross often reveal the genotype of an individual when it is unknown TESTCROSS: GENOTYPES B_ bb Two possibilities for the black dog: BB or Bb B B b GAMETES b Bb b Bb bb OFFSPRING Figure 9.6 All black 1 black : 1 chocolate

37. Purebreds and Mutts — A Difference of Heredity These black Labrador puppies are purebred—their parents and grandparents were black Labs with very similar genetic make-ups • Purebreds often suffer from serious genetic defects

38. Their behavior and appearance is more varied as a result of their diverse genetic inheritance • The parents of these puppies were a mixture of different breeds

39. Independent assortment of two genes in the Labrador retriever Blind Blind Black coat, normal visionB_N_ Black coat, blind (PRA)B_nn Chocolate coat, normal visionbbN_ Chocolate coat, blind (PRA)bbnn PHENOTYPES GENOTYPES MATING OF HETEROZYOTES(black, normal vision) BbNn BbNn 9 black coat,normal vision 3 black coat,blind (PRA) 3 chocolate coat,normal vision 1 chocolate coat,blind (PRA) PHENOTYPIC RATIO OF OFFSPRING Figure 9.5B

40. How is Codominance different from Incomplete Dominance?

41. Non-single Gene Genetics(pg. 260-262) Incomplete dominance: -neither pair of alleles are completely expressed when both are present. -Typically, a third phenotype is produced Ex: snapdragons (pink flowers) hypercholesterolemia Codominance: Two alleles are expressed in a heterozygote condition. Ex: Human Blood types

42. Incomplete Dominance

43. Incomplete dominance results in intermediate phenotypes P GENERATION Whiterr • When an offspring’s phenotype—such as flower color— is in between the phenotypes of its parents, it exhibits incomplete dominance Red RR Gametes R r PinkRr F1 GENERATION 1/2 R 1/2 r 1/2 R 1/2 R Eggs Sperm RedRR 1/2 r 1/2 r PinkRr PinkrR F2 GENERATION Whiterr Figure 9.12A

44. Incomplete dominance in human hypercholesterolemia GENOTYPES: HH Homozygousfor ability to makeLDL receptors Hh Heterozygous hh Homozygousfor inability to makeLDL receptors PHENOTYPES: LDL LDLreceptor Cell Normal Mild disease Severe disease Figure 9.12B

45. Many genes have more than two alleles in the population • In a population, multiple alleles often exist for a characteristic • The three alleles for ABO blood type in humans is an example Codominance

46. Codominance-Observed in Blood Types

47. Blood Type Frequencies of different Ethnic Groups

48. Pleiotrophy vs. Polygenic Inheritance

49. Non-single Gene Genetics Pleiotropy: genes with multiple phenotypic effect. Ex: sickle-cell anemia combs in roosters coat color in rabbits Polygenic Inheritance: an additive effect of two or more genes on a single phenotypic character Ex: human skin pigmentation and height

50. A single gene may affect many phenotypic characteristics • A single gene may affect phenotype in many ways • This is called pleiotropy • The allele for sickle-cell disease is an example