Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

1. Today: Imprinting and LinkageExam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

2. Genotype Phenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple.

3. Lamarck was right? Sort of… Epigenetics: http://www.pbs.org/wgbh/nova/sciencenow/3411/02.html Image from: http://www.sparknotes.com/biology/evolution/lamarck/section2.rhtml

4. Genomic Imprinting • Genomic imprinting is a phenomenon in which expression of a gene depends on whether it is inherited from the male or the female parent • Imprinted genes follow a non-Mendelian pattern of inheritance • Depending on how the genes are “marked”, the offspring expresses either the maternally-inherited or the paternally-inherited allele **Not both

5. A hypothetical example of imprinting similar to Fig 7.10 a B* a B* A* b A=curly hair a=straight hair B=beady eyes b=normal *=methylation A* in males B* in females A* b A*a bB* A*a bB* A*a bB Aa bB* Ab, AB*, ab, aB* A*b, A*B, ab, aB

6. Imprinting and DNA Methylation • Genomic imprinting must involve a marking process • At the molecular level, the imprinting is known to involve differentially methylated regions • They are methylated either in the oocyte or sperm • Not both

7. For most genes, methylation results in inhibition of gene expression • However, this is not always the case

8. Fig 7.11 Changes in methylation during gamete development alter the imprint Haploid female gametes transmit an unmethylated gene Haploid male gametes transmit a methylated gene

9. Thus genomic imprinting is permanent in the somatic cells of an animal • However, the marking of alleles can be altered from generation to generation

10. Tbl 7.2 • To date, imprinting has been identified in dozens of mammalian genes

11. Tbl 7.2

12. Imprinting plays a role in the inheritance of some human diseases: Prader-Willi syndrome (PWS) and Angelman syndrome (AS) • PWS is characterized by: reduced motor function, obesity, mental deficiencies • AS is characterized by: hyperactivity, unusual seizures, repetitive muscle movements, mental deficiencies Usually, PWS and AS involve a small deletion in chromosome 15 • If it is inherited from the mother, it leads to AS • If it is inherited from the father, it leads to PWS

13. AS results from the lack of expression of UBE3A (encodes a protein called EA-6P that transfers small ubiquitin molecules to certain proteins to target their degradation) • The gene is paternally imprinted (silenced) • PWS results (most likely) from the lack of expression of SNRNP (encodes a small nuclear ribonucleoprotein that controls gene splicing necessary for the synthesis of critical proteins in the brain) • The gene is maternally imprinted (silenced)

14. Fig 7.12 The deletion is the same in males and females, but the expression is different depending on who you received the normal version from.

15. The relationship between genes and traits is often complex • Complexities include: • Complex relationships between alleles

16. The relationship between genes and traits is often complex • Complexities include: • Multiple genes controlling one trait

17. Two genes control coat color in mice Fig 4.21

18. Variation in Peas Fig 3.2

19. Fig 2.8 Inheritance of 2 independent genes

20. Approximate position of seed color and shape genes in peas Gene for seed color y Y r R Gene for seed shape Chrom. 1/7 Chrom. 7/7

21. Fig 2.9 There must be a better way…

22. Inheritance can be predicted by probability Section 2.2, pg 30-32

23. Sum rule • The probability that one of two or more mutually exclusive events will occur is the sum of their respective probabilities • Consider the following example in mice • Gene affecting the ears • De = Normal allele • de = Droopy ears • Gene affecting the tail • Ct = Normal allele • ct = Crinkly tail

24. If two heterozygous (Dede Ctct) mice are crossed • Then the predicted ratio of offspring is • 9 with normal ears and normal tails • 3 with normal ears and crinkly tails • 3 with droopy ears and normal tails • 1 with droopy ears and crinkly tail • These four phenotypes are mutually exclusive • A mouse with droopy ears and a normal tail cannot have normal ears and a crinkly tail • Question • What is the probability that an offspring of the above cross will have normal ears and a normal tail or have droopy ears and a crinkly tail?

25. Applying the sum rule • Step 1: Calculate the individual probabilities P(normal ears and a normal tail)= 9 (9 + 3 + 3 + 1) = 9/16 P(droopy ears and crinkly tail)= 1 (9 + 3 + 3 + 1) = 1/16 • Step 2: Add the individual probabilities 9/16 + 1/16 = 10/16 • 10/16 can be converted to 0.625 • Therefore 62.5% of the offspring are predicted to have normal ears and a normal tail or droopy ears and a crinkly tail

26. Product rule • The probability that two or more independent events will occur is equal to the product of their respective probabilities • Note • Independent events are those in which the occurrence of one does not affect the probability of another

27. Consider the disease congenital analgesia • Recessive trait in humans • Affected individuals can distinguish between sensations • However, extreme sensations are not perceived as painful • Two alleles • P = Normal allele • p = Congenital analgesia • Question • Two heterozygous individuals plan to start a family • What is the probability that the couple’s first three children will all have congenital analgesia?

28. Applying the product rule • Step 1: Calculate the individual probabilities • This can be obtained via a Punnett square P(congenital analgesia)= 1/4 • Step 2: Multiply the individual probabilities 1/4 X 1/4 X 1/4 = 1/64 • 1/64 can be converted to 0.016 • Therefore 1.6% of the time, the first three offspring of a heterozygous couple, will all have congenital analgesia

29. Different genes are not always independent Crossing-over Meiosis I (Ind. Assort.) Meiosis II 4 Haploid cells, each unique

30. Fig 5.1 The haploid cells contain the same combination of alleles as the original chromosomes The arrangement of linked alleles has not been altered

31. Fig 5.1 These haploid cells contain a combination of alleles NOT found in the original chromosomes This new combination of alleles is a result of genetic recombination These are termed parental or non-recombinant cells These are termed recombinant cells

32. Linked alleles tend to be inherited together

33. Recombinants are produced by crossing over

34. Crossing over produces new allelic combinations

35. Only 2 chromosomes cross-over, and so the maximum number of recombinants that can be produced is 50%.

36. For linked genes, recombinant frequencies are less than 50 percent

37. Homologous pair of chromosomes

38. Does this show recombination? d/d M2/M2 D/d M1/M2 D/d M2/M2 d/d M1/M2 D/d M1/M2 d/d M2/M2

39. Does this show recombination? d/d M2/M2 D/d M1/M2 D/d M2/M2 d/d M1/M2 D/d M1/M2 d/d M2/M2

40. Longer regions have more crossovers and thus higher recombinant frequencies

41. Some crosses do not give the expected results

42. =25% 42% 41% 9% 8%

43. These two genes are on the same chromosome

45. By comparing recombination frequencies, a linkage map can be constructed

46. By comparing recombination frequencies, a linkage map can be constructed = 17 m.u.

47. Linkage map of Drosophila chromosome 2: This type of map, with mapping units more than 50, can only be put together by making comparisons of linked genes.

48. The probability of crossing over can be used to determine the spatial relationship of different genes

49. similar to Fig 5.3,also see Fig 5.9,and pg 115-117 What is the relationship between these 3 genes? What order and how far apart?