1 / 33

Today…

Genome 351, 14 April 2014, Lecture 5. Today…. Meiosis: how the genetic material is partitioned during the formation of gametes (sperm and eggs) Probability: -the product rule -the sum rule Independent assortment of nonhomologous chromosomes during meiosis.

skip
Download Presentation

Today…

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Genome 351, 14 April 2014, Lecture 5 Today… • Meiosis: how the genetic material is partitioned during the formation of gametes (sperm and eggs) • Probability: • -the product rule • -the sum rule • Independent assortment of nonhomologous chromosomes during meiosis

  2. Meiosis: the formation of gametes DNA Replication • Meiotic Division 1 • Copied chromosomes (sister chromatids) stay joined together at the centromere. • Homologous chromosomes pair up and physically join at sites of recombination • Proteins pull the twohomologs to opposite poles • Meiotic Division 2 • Proteins pull the two sister chromatids to opposite poles • Each gamete gets a copy of only one homolog (usually a maternal-paternal hybrid). DNA Recombination (crossing over)

  3. Mitosis vs. Meiosis 2n 1m 1p 2n 1m 1p DNA Replication 2x1m 2x1p DNA Recombination 2x1p 2x1m 2x 1m/p 2x 1m/p 2x 1p/m 2x 1p/m 2x1m 2x1p 2x 1m/p 2x 1p/m 1m 1m 1p 1p 1m 1p/m 1p 1m 1m 1m/p 2n 1p 1p 1n exact copies half as many chromosomes

  4. Meiotic Division I Crossovers hold the homologues together until all of the chromosomes are attached to the spindle Spindle

  5. Meiotic Division I Crossovers hold the homologues together until all of the chromosomes are attached to the spindle The homologues then separate from one another, exchanging corresponding portions as they do so

  6. Mitosis vs. Meiosis 1m 1p 1m 1p DNA Replication 2x1m 2x1p DNA Recombination 2x1p 2x1m 2x 1m/p 2x 1m/p 2x 1p/m 2x 1p/m 2x1m 2x1p 2x 1m/p 2x 1p/m 1m 1m 1p 1p 1m 1p/m 1p 1m 1m 1m/p 1p 1p exact copies half as many chromosomes

  7. Meiotic Division II The two daughter cells from meiotic division Igo directly into meiotic division II Sister chromatids separate during meiotic division II

  8. One round of DNA synthesis with one cell division Two genetically identical daughters Sister chromatidssegregate Homologs do not line up or separate Homologs do not exchange corresponding segments (no crossing over) Final products are diploid (2n) One round of DNA synthesis with two rounds of cell division Four cells genetically different Sister chromatids segregate only in meiosis II Homologous chromosomes align and separate in meiosis I Homologs exchange corresponding segments (crossing over; recombination) Final products are haploid (1n) Mitosis vs. Meiosis

  9. Probability in genetics

  10. Probability is important in genetics Needed for… • testing hypotheses • mapping disease genes • genetic counseling — need to predict outcomes, then assess how well they match the results — depends on being able to calculate probabilities of specific outcomes — risk assessment Two basic methods… product rule and sum rule

  11. Pedigree of a family segregating phenylketonuria (PKU) What can we infer from the pedigree? Disease can skip generations Affected persons can have unaffected parents Both sexes equally affected PKU is an autosomal recessive disorder

  12. Use of the Product and Sum rules A couple has a first child who tests positive for PKU. What can you infer about their genotypes? PAH+/- PAH+/- 1. What is the probability that their next child will have PKU? ?? PAH-/- 2. What is the chance the next child, if he or she is not affected, will be a carrier? caused by mutations in the gene that encodes phenylalanine hydroxylase (PAH)

  13. Following the fate of the PAH gene in a PAH+/- heterozygote during meiosis PAH- PAH+ DNA Replication 2 copies PAH+ 2 copies PAH- DNA Recombination 2 copies PAH+ 2 copies PAH- PAH+ PAH- PAH+ PAH- PAH+ PAH- PAH+ PAH- PAH- PAH+ PAH- PAH+ ½ of each type PAH+ PAH+ PAH- PAH-

  14. Following the fate of the PAH gene in a PAH+/- heterozygote during meiosis PAH- PAH+ DNA Replication 2 copies PAH+ 2 copies PAH- DNA Recombination 2 copies PAH+ 2 copies PAH- PAH+ PAH+ PAH- PAH- PAH+ PAH+ PAH- PAH- PAH+ PAH+ PAH- PAH- still ½ of each type PAH+ PAH- PAH+ PAH-

  15. Genetic accounting eggs What are the possible genotypes and phenotypes of the children of parents who are both phenylketonuria carriers? ½ PAH+ ½ PAH- PAH+ PAH- PAH+ PAH+ ½ PAH+ sperm PAH+ PAH- PAH- PAH- ½ PAH- 1/4 PAH-/- 1/2 PAH+/- 1/4 PAH+/+ Affected Normal

  16. Product rule PAH+/- PAH+/- eggs 1. What is the probability that their next child will have PKU? ½ PAH+ ½ PAH- PAH-/- PAH+ PAH- PAH+ PAH+ ½ PAH+ Product Rule: The probability of 2 or more independent events occurring simultaneously sperm PAH+ PAH- PAH- PAH- ½ PAH- = the product of the individual probabilities ¼ Probability of PAH-/-? ½ PAH-x ½ PAH-= ¼ PAH-/-

  17. Sum rule 2. What is the chance the next child, if he or she is not affected, will be a carrier? PAH+/- PAH+/- PAH-/- eggs Sum Rule: The probability of an event that can occur in 2 or more ways ½ PAH+ ½ PAH- PAH+ PAH- PAH+ PAH+ ⅓ ½ PAH+ = sum of the separate probabilities sperm PAH+ PAH- PAH- PAH- ½ PAH- ⅓ Probability of PAH+/-? = ⅓ + ⅓ = ⅔

  18. Punnett Square Execution Determine types of gametes from each parent Combine each type of female gamete with each male gamete Advantages of Punnett Square Organized and systematic Gives all possible combinations of genotypes automatically Disadvantages of Punnett Square Slow and labor intensive, especially for complex genotypes (e.g., AaBbCc X AabbCc)

  19. Aa Aa Two events necessary: II-3 must be Aaand they must have aa child A a A a Using the product and sum rules a = no pigment Example: Albinism… What is the probability that III-1 will be albino? aa Aa P(II-3 is Aa) = 2/3 P(Aa x Aa giving aa) = 1/4 P(III-1 is aa) = 2/3 x 1/4 = 1/6

  20. Independent assortment of nonhomologous chromosomes during meiosis What happens to non-homologous chromosomes during meiosis? But attachment of chromosomes to the spindle is random, so this arrangement is equally likely

  21. 1m 1p 1p 1m 2m 2p 2m 2p 1p 1p 1p 1m 1m 1m 1m 1p 1p 1p 2p 2p 2p 2m 2m 2m 2m 2p 2p 2p 1m 1m 2m 2m Independent assortment of nonhomologous chromosomes during meiosis Two equally probable arrangements:

  22. Meiosis and independent assortment of nonhomologous chromosomes can create many different types of gametes Examples: 1 pair of homologous chromosomes gives 2 types of gametes (21 = 2) 2 pairs of homologous chromosomes gives 4 types of gametes (22 = 4) n pairs of homologous chromosomes gives 2n types of gametes 23 pairs of homologous chromosomes gives 223 (8 million) types of gametes

  23. An example of independent assortment • Following the fate of genes on different (nonhomologous) chromosomes • Cystic fibrosis on chromosome 7 • A gene that influences ABO blood types on chromosome 9

  24. Some background on ABO blood groups A red cells = A antigen AB red cells = B antigen O red cells B red cells

  25. The ABO (I) gene There are 3 different versions (alleles) of the I gene: adds A sugar to red cell surface A I alleles IA B adds B sugar to red cell surface IB adds no sugar to red cell surface i

  26. The ABO gene - dominance relationships IA is dominant to i IA/IA or IA/i - A blood type IB is dominant to i IB/IB or IB/i - B blood type i is recessive i/i - O blood type IAis co-dominant with IB IA/IB- AB blood type both phenotypes are seen The I gene lies on chromosome 9q34

  27. CFTR+ CFTR- CFTR- CFTR+ IA i IA i CFTR+ CFTR- CFTR- CFTR+ i IA i IA CFTR+ CFTR+ CFTR+ CFTR- CFTR- CFTR+ CFTR- CFTR- i IA i IA IA i i IA Independent assortment of genes on nonhomologous chromosomes Gametes formed from a CFTR+/- IA/i double heterozygote:

  28. Gametes that arise from a CFTR+/- IA/i double heterozygote: IA CFTR+ 1/4 IA CFTR+ i CFTR- 1/4 22 = 4types i CFTR- CFTR- IA In equal proportions 1/4 CFTR- IA i CFTR+ 1/4 i CFTR+

  29. CFTR+ IA CFTR- i CFTR- IA CFTR+ i CFTR+ IA CFTR- i CFTR+ i CFTR- IA Possible genotypes and phenotypes from a mating of CFTR+/- IA/i double heterozygotes What genotypes give CF and A blood type? Eggs What genotypes give nonCF and A blood type? C+/+ IA/IA C+/+ IA/i C+/- IA/i C+/- IA/IA C-/- i/i C+/- IA/i C+/- i/i C-/- IA/i What genotypes give nonCF and O blood type? Sperm C+/- IA/i C+/- i/i C+/+ IA/- C+/+ i/i What genotype gives CF and O blood type? C-/- IA/IA C+/- IA/IA C-/- IA/i C+/- IA/i

  30. 1/4 1/4 1/4 1/4 ¼ x ¼ = 1/16 Probability of CF & type O? CFTR+ IA CFTR- i CFTR- IA CFTR+ i C+/+ IA/IA CFTR+ IA Mendel’s law of independent assortment: allele pairs separate independently during the formation of gametes (traits are transmitted to offspring independently of one another C+/+ IA/i C+/- IA/i C+/- IA/IA 1/4 9 N, A 3 N, O 3 CF, A 1 CF, O CFTR- i C-/- i/i C+/- IA/i C+/- i/i C-/- IA/i 1/4 1/16th of Children Are CF & Type 0 C+/- i/i CFTR+ i C+/+ i/i C+/- IA/i C+/+ IA/- 1/4 C+/- IA/i C-/- IA/IA CFTR- IA C-/- IA/i C+/- IA/IA 1/4

  31. Probability of CFTR+/-? 1/2 1/4 1/4 1/4 1/4 Probability of CFTR+/- IA/i? CFTR+ IA CFTR- i CFTR- IA CFTR+ i C+/+ IA/IA CFTR+ IA C+/+ IA/i C+/- IA/i C+/- IA/IA 1/4 CFTR- i C-/- i/i C+/- IA/i C+/- i/i C-/- IA/i 1/4 C+/- i/i CFTR+ i C+/+ i/i C+/- IA/i C+/+ IA/- 1/4 C+/- IA/i C-/- IA/IA CFTR- IA C-/- IA/i C+/- IA/IA 1/4

  32. Probability of CFTR+/-? 1/2 1/4 1/4 1/4 1/4 Probability of CFTR+/- IA/i? CFTR+ IA CFTR- i CFTR- IA CFTR+ i C+/+ IA/IA CFTR+ IA C+/+ IA/i C+/- IA/i C+/- IA/IA 1/4 CFTR- i C-/- i/i C+/- IA/i C+/- i/i C-/- IA/i 1/4 C+/- i/i CFTR+ i C+/+ i/i C+/- IA/i C+/+ IA/- 1/4 C+/- IA/i C-/- IA/IA CFTR- IA C-/- IA/i C+/- IA/IA 1/4

  33. Probability of CFTR+/-? 1/2 1/4 1/4 1/4 1/4 Sum rule or product rule? Probability of CFTR+/- IA/i? Probability of IA/i? 1/2 CFTR+ IA CFTR- i CFTR- IA CFTR+ i 1/2 X 1/2 = 1/4 C+/+ IA/IA CFTR+ IA C+/+ IA/i C+/- IA/i C+/- IA/IA 1/4 CFTR- i C-/- i/i C+/- IA/i C+/- i/i C-/- IA/i 1/4 C+/- i/i CFTR+ i C+/+ i/i C+/- IA/i C+/+ IA/- 1/4 C+/- IA/i C-/- IA/IA CFTR- IA C-/- IA/i C+/- IA/IA 1/4

More Related