2. Midterm 1 Mean = 74.6% > 100: | | > 90: | | | | | | | | | | > 80: | | | | | | | | | | | | | > 70: | | | | | | | | | | | | | > 60: | | | | | | | | | | | | > 50: | | | | | | | | | | | > 40: | Extra Credit, To Date: = 18 pts Please see me if you have questions about arithmetic, or obvious misgrades. If you want to argue a point, I reserve the right to re-grade the entire test.

3. Midterm 2 Lectures, Assignments, Through Chapter 5, Chromosomal Mutations.

4. Linkage Genes located on the same chromosome do not recombine, –unless crossing over occurs, The recombination frequency gives an estimate of the distance between the genes.

5. Recombination Frequencies Genes that are adjacent have a recombination frequency near 0%, Genes that are very far apart on a chromosome have a recombination frequency of 50%, The relative distance between linked genes influences the amount of recombination observed.

6. Linkage Ratio (How do you determine it?) recombinant total progeny GWGwgWgw ???? = Linkage Ratio P GGWW x ggww Testcross F1: GgWw x ggww ExperimentallyExperimentally

7. Linkage Ratio (If Sorting Independently) 50 (Gw) + 50 (gW) 200 (all classes) GWGwgWgw 50 =.5 P GGWW x ggww Testcross F1: GgWw x ggww

8. Linkage Ratio Units % = mu (map units) - or - % = cm (centimorgan)

9. Fly Crosses (white eyes, minature, yellow body) In a white eyes x miniature cross, 900 of the 2,441 progeny were recombinant, yielding a map distance of 36.9 mu, In a separate white eyes x yellow body cross, 11 of 2,205 progeny were recombinant, yielding a map distance of 0.5 mu, When a miniature x yellow body cross was performed, 650 of 1706 flies were recombinant, yielding a map distance of 38 mu.

10. Simple Mapping white eyes x miniature = 36.9 mu, white eyes x yellow body = 0.5 mu, miniature x yellow body = 38 mu, my 38 mu 36.9 mu w 0.5 mu

11. Three Point Testcross Triple Heterozygous (AaBbCc ) x Triple Homozygous Recessive (aabbcc)

12. Three Point Mapping Requirements The genotype of the organism producing the gametes must be heterozygous at all three loci, You have to be able to deduce the genotype of the gamete by looking at the phenotype of the offspring, You must look at enough offspring so that all crossover classes are represented.

13. Representing linked genes... W G D w g d x w g d P Testcross = WwGgDd = wwggdd

14. Representing linked genes... + + + w g d x w g d P Testcross = WwGgDd = wwggdd

15. Phenotypic Classes W- ww G- gg G- gg D- dd D- dd D- dd D- dd W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- wwggdd

16. W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- Crossovers 0 W G D w g d 1 1 2 # 179 52 46 4 22 2 wwggdd1730 1 1 2

17. W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d III

18. W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d I Total = 500 Region I: 46 + 52 + 2 + 4 500 x 100 = 20.8 mu

19. W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d II Total = 500 Region II: 22 + 22 + 2 + 4 500 x 100 = 10.0 mu 20.8 mu

20. W G D w g d W-gg-D wwG-dd 4 2 Recombinants, double crossover Total = 500 10.0 mu20.8 mu 0.1 x 0.208 = 0.0208 6/500 = 0.012 NO GOOD!

21. Interference …the affect a crossing over event has on a second crossing over event in an adjacent region of the chromatid, –(positive) interference: decreases the probability of a second crossing over, most common in eukaryotes, –negative interference: increases the probability of a second crossing over.

22. Gene Order in Three Point Crosses Find either double cross-over phenotype, based on the recombination frequencies, Two parental alleles, and one cross over allele will be present, The cross over allele fits in the middle...

23. # 2001 52 46 589 990 887 600 1786 Which one is the odd one? A C B a c b III A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- aabbcc

24. A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- # 2001 52 46 589 990 887 600 aabbcc1786 Region I A C B a c b I 990 + 887 + 52 + 46 6951 x 100 = 28.4 mu

25. A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- # 2001 52 46 589 990 887 600 aabbcc1786 Region II A C B a c b 28.4 mu 600 + 589 + 52 + 46 6951 x 100 = 18.5 mu II 18.5 mu

26. Master Problems 1, 2 Questions 4.1 - 4.4, 4.6 - 4.16, 4.19 - 4.20

27. Quantitative Traits Optional; Klug and Cummings (at the library), –Chapter 5, pp. 115 - 120, Insights and Solutions #1 (pp. 130), Questions 1-6 (pp. 131-132)

28. Reading Assignments Available as a PDF, online. Will answer question on Monday. Read 15.5

29. Quantitative Traits …traits that show a continuous variation in phenotype over a range, …often result from multiple genes and are further termed polygenic traits.

33. Ribera, 1642 The Club Footed Boy Heart disease Spina bifida Neural tube disorders Diabetes Etc.

35. Environmental Factors Optional: Chapter 15

36. Discontinuous Traits Discontinuous Traits, –exhibit only a few distinct phenotypes and can be described in a qualitative manner, Mendel and others worked with “inbred” lines, Father: DD BB cc aa EE FF GG Mother: dd BB cc aa EE FF GG

37. Continuous Traits “Outbred” transmission, Father: dd BB CC AA EE FF GG Mother: DD bb CC aa EE ff gg Continuous Traits, –Display a spectrum of phenotypes, and must be described in quantitative terms,

39. F3 #1: Cross Two Individuals With 10 Cm Ears. F3 #2: Cross Two Individuals With 17 Cm Ears.

40. Polygenic Traits and Mendel It is possible to provide a Mendelian explanation for continuous variation by considering numbers of genes contributing to a phenotype, –the more genes, the more phenotypic categories, –the more categories, the more the variation seems continuous.

41. Melanin Pigmentation

43. Calculating the Number of Genes If you know the frequency of either extreme trait, then the number of genes (n) can be calculated using the formulae… 1 4 n = ratio of F2 individuals expressing either extreme phenotype.

44. 1/4 show the extreme traits, 14n14n = 141141 n = 1

45. 14n14n = n = 2 1/16 show either extreme traits, 142142 1 16 =

46. Continuous Variation Two or more genes that influence the same phenotype, often in an additive way, –additive allele: contributes a set amount to the phenotype, –non-additive allele: does not contribute to the phenotype. The affect of each allele on the phenotype is relatively small, and roughly equivalent, Substantial variation is observed when multiple genes control a single trait, Must be studied in large populations.

47. Inbred strain #1, mean height = 24 cm, Inbred strain #2, mean height = 24 cm, F1 mean height = 24 cm, F2, 12 - 36 cm range, mean = 24 cm, –4/1000 are 12 cm, smallest class –4/1000 are 36 cm, largest class a. What mode of inheritance? b. How many gene pairs? Quantitative. 1/4 n = 1/250, n = 4 4/1000 = 1/250 are the extreme class, solve 1/4 n = ratio of extreme class 4 2 = 16, 4 3 = 64, 4 4 = 256, etc.

48. Inbred strain #1, mean height = 24 cm, Inbred strain #2, mean height = 24 cm, F1 mean height = 24 cm, F2, 12 - 36 cm range, mean = 24 cm, –4/1000 are 12 cm, smallest class –4/1000 are 36 cm, largest class c. How much does each allele contribute? range: largest (36 cm) - smallest (12 cm) = 24 cm alleles: 4 genes, 8 potential additive alleles 24 cm / 8 additive alleles = 3 cm / additive allelle

49. Inbred strain #1, mean height = 24 cm, Inbred strain #2, mean height = 24 cm, F1 mean height = 24 cm, F2, 12 - 36 cm range, mean = 24 cm, –4/1000 are 12 cm, smallest class –4/1000 are 36 cm, largest class d. Indicate one possible set of P1, F1. P1 AABBccdd x aabbCCDD F1 AaBbCcDd x AaBbCcDd Base Height is 12 cm, so each P1, F1 has 4 additive alleles.