Lecture 1 2012 slideshare

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Introduction to Genetics,
Prokaryotic & Eukaryotic cells, Chromosome structure, Alleles, Genotype and Phenotype, Mitosis & Meiosis, Genetic Variation, Transmission/ Mendelian Genetics, Dominance, Theoretical Probability, Binomial Expansion, Testcross, Observed Ratios of Progeny, Goodness-of-Fit Chi-Square Test.

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  • Nucleus - defines eukaryotic cells Nuclear membrane - divides cells into two compartments This provides an extra level of protein activity regulation - spatial Proteins that act in the nucleus, such as transcription factors, DNA repair molecules, signal transduction factors have to be efficiently imported into the nucleus, whereas intranuclear molecules such as mRNA, tRNA or even particles like assembled ribosomal subunits have to be exported to the cytoplasm. This traffic occurs entirely through tightly regulated aqueous channels that are formed by the NPC. The key feature of eukarotic cells is their organization into compartments which are separated by membranes. This gives them the high complexity in contrast to the simple organization of prokaryotic cells. The various compartments are connected by complex gating machineries which allow the shuttling of all sorts of cargos. These transport events are of highest priority for the viability and functionality of the individual cells and also for the entire organism. One of the major transport routes is constituted between the nucleus and the cytoplasm. These two compartments are separated via the double membrane nuclear envelope. This transmission electron micrograph on this slide was taken from Samuel Dales and was taken from monkey liver cells. On the right side you can see the cytoplasm and the left side there is the nucleus containing the hetero- and euchromatin. The nuclear envelope is clearly visible. Examples for cargo imported from the cytoplasm into the nucleus includes proteins with nuclear localization signals which have been discovered by my boss about 30 years ago, and hnRNPs, snRNPs and ribosomal proteins. On the other hand the ribosomal subunits are exported after their assembly and also tRNAs, snRNPs and proteins with nuclear export sequences are exported. All transport occurs through large proteinaceous structures called nuclear pore complexes
  • Figure 2.6. Diploid eukaryotic cells have two sets of chromosomes. (a) A set of chromosomes from a female human cell. Each pair of chromosomes is hybridized to a uniquely colored probe, giving it a distinct color. (b) the chromosomes are present in homologous pairs, which consists of chromosomes that are alike in size and structure and carry information for the same characteristics. [Part a: Courtesy of Dr. Thomas Ried and Dr. Evelin Schrock.]
  • Figure 2.7 Each eukaryotic chromosome has a centromere and telomeres.
  • Figure 2.9 The cell cycle consists of interphase and M phase.
  • Figure 2.16 Crossing over produces genetic variation.
  • Figure 2.17 Genetic variation is produced through the random distribution of chromosomes in meiosis.
  • Lecture 1 2012 slideshare

    1. 1. Genetics/lab BIOL 3600 EV1 BIOL 3600 EV2 Winter 2012
    2. 2. Review of the syllabus Instructor Dr. Paula Faria Waziry Text Pierce, Benjamin A 2010. Genetics: A Conceptual Approach 4th Edition, W.H. Freeman, New York, NY (ISBN number = 1429232501) Website All course materials will be posted on Blackboard (when possible) or on SlideShare prior to the lectures. Videos of the lectures will be posted on Blackboard (when possible) or on my Youtube page under playlist: Genetics_Dr. Paula Waziry http:// www.youtube.com/playlist?list=PL3CBA6D82E9B35226&feature=view_all There are several other playlists related to genetics on my Youtube channel. I frequently add related videos to those playlists in order to enrich your learning experience.
    3. 3. Review of the syllabus Grades The final average will be calculated as follows: Four unit tests @ 12.5% each 25% Midterm exam 20% Final exam 20% Quizzes 10% Lab final 10% Lab activities 15% 100% The grading scale :  93% = A, 90-92%=A-, 87-89%=B+, 83-86%=B, 80-82%=B-, 77-79%=C+, 73-76%=C, 70-72%=C-, 60-69%=D, <60%=F - Class grades will be curved if necessary If having problems, address them as early as possible!!
    4. 4. Tips for success 1. Attendance is required at all lectures, labs, and exams - You are responsible for getting any missed info 2. Ask questions in class and participate in discussion 3. Don’t just read and highlight the handouts. Understand and practice questions. 4. You will be responsible for the conceptual material covered in class/ powerpoints. 5. Be considerate: no cell phones, texting, sleeping, snoring…
    5. 5. Chapter 1 Introduction to Genetics <ul><li>People have understood the hereditary nature of traits and have practiced genetics for thousands of years </li></ul><ul><li>The rise in agriculture began when people started to apply genetic principles to the domestication of plants and animals </li></ul>
    6. 6. <ul><li>Pharmaceutical industry: </li></ul><ul><ul><li>Numerous drugs are synthesized by fungi and bacteria. </li></ul></ul><ul><ul><ul><li>Ex: growth hormone, insulin </li></ul></ul></ul><ul><li>In medicine: </li></ul><ul><ul><li>Many diseases and disorders have a hereditary component. </li></ul></ul><ul><ul><ul><li>Ex: hemophilia, sikcle-cell anemia, diabetes, </li></ul></ul></ul><ul><ul><ul><li>muscular distrophy, </li></ul></ul></ul>EMD1: Emerin; Xq28; Recessive EMD2: Lamin A/C; 1q21.2; Dominant EMD3: SYNE1; 6q25; Dominant EMD4: SYNE2; 14q23; Dominant Other: Sporadic & Dominant                                  from A Kornberg MD
    7. 7. 2 years-old Kristian 12 years-old Ashley and 2-weeks old brother Evan Hutchinson-Gilford Progeria
    8. 8. Defect in Lamina A processing Hutchinson-Gilford Progeria Lamin A/C (a) Lamin A (c) DAPI Control Patient
    9. 9. Samuel Dales mRNA Ribosomal subunits tRNA snRNPs Proteins with NES hnRNPs Ribosomal proteins Proteins with NLS snRNPs
    10. 10. TPR (translocated promoter region) Nup98 Nup98 Nup98 The Nuclear Pore Complex
    11. 11. <ul><li>Fusion Transcript </li></ul><ul><li>Nup98-HOXA9 </li></ul><ul><li>Nup98-HOXD13 </li></ul><ul><li>Nup98-PMX1 </li></ul><ul><li>Nup98-DDX10 </li></ul><ul><li>Nup98-RAP1GDS1 </li></ul><ul><li>Nup98-TOP1 </li></ul><ul><li>DEK-Nup214 </li></ul><ul><li>SET-Nup214 </li></ul>Translocation t(7; 11)(p15; p15) t(2; 11)(q31; p15) t(1; 11)(q23; p15) inv(11)(p15; q22) t(4; 11)(q21; p15) t(11; 20)(p15; q11) t(6; 9)(p23; q34) inv(9) Nucleoporin Translocations in Acute Leukemia
    12. 12. <ul><li>Genetic Diversity and evolution </li></ul><ul><li>Living organisms have an important feature in common: </li></ul><ul><li>all use similar genetic systems </li></ul><ul><li>A complete set of genetic instructions for any organism is its genome </li></ul><ul><ul><li>All genomes are encoded in mucleic acid: </li></ul></ul><ul><ul><li>either DNA or RNA </li></ul></ul><ul><ul><ul><ul><li>Suggestive of a common ancestor </li></ul></ul></ul></ul>
    13. 13. <ul><li>Since all organisms have similar genetic systems, the study of one organism’s genes eveals principles that apply to other organimsms </li></ul>GAS motif Accessory TF motifs TATA Accessory TF factors STAT 1 STAT 1 P P
    14. 14. Nucleoporins and Nuclear Traffic Proteins are involved with Mitotic Spindle Checkpoints and Aging Baker DJ, Jeganathan KB, Malureanu L, Perez-Terzic C, Terzic A, van Deursen JM. J Cell Biol. 2006 Feb 13;172(4):529-40. <ul><li>Enabling us to use animal models to study diseases or natural processes </li></ul>
    15. 15. - Influenza virus killed as many as 50 million people in a single year (1918-1919) - Extremely virulent - Fast, nasty, killed all ages - Up until 2006, had no idea why it was so virulent - Had no way to tell if a similar strain was forming - In October 2005, scientists recreated the 1918 strain using various genetic techniques - Obtained samples of the virus from preserved wax specimen and from a Eskimo woman who had died from it - Able to characterize it and predict / prevent future outbreaks?) Influenza A pandemic
    16. 16. Microarray study reveals changes in regulation of genes Upon influenza viral infection Replication-dependent genes Geiss G, J.Virol, May 2001,p4321-4331
    17. 17. Cells expressing low levels of Rae1 are susceptible to influenza infection
    18. 18. <ul><ul><li>Influenza virus targets the mRNA export machinery and the nuclear pore complex. </li></ul></ul>Satterly, Tsai, van Deursen, Nussenzveig, Wang, Faria, Levay, Levy, Fontoura. Proc Natl Acad Sci U S A. 2007 Feb 6;104(6):1853-8.
    19. 19. Genetics What is it? http://www.youtube.com/watch?v=0OnwOKiMVb8&feature=channel_video_title
    20. 20. • Genetics = Study of HEREDITY and VARIATION - Heredity = Passing down of traits from one generation to another - Variation = Differences in inherited characteristics among members of a population
    21. 21. Genetics What is it? • Genetics has many subfields 1. Transmission genetics – Study of heredity - How traits are passed down between generations - Example: My mom has disease, spouse's uncle has same disease  Will our kids will have it? - Is allele dom/rec, is it on X chr or autosome? 2. Molecular genetics – Structure and function of individual genes - Includes study of cancer (cancer genetics), genetic engineering (manipulation of genes), study of chromosome structure (cytogenetics) 3. Population/Evolutionary genetics – Study genetic variation in populations - Includes conservation genetics
    22. 23. History of genetics <ul><li>• Early genetics was more philosophy than science </li></ul><ul><li>- No experimentation </li></ul><ul><li>- Many concepts were incorrect </li></ul><ul><li>- Pangenesis – traits collected from all over body and </li></ul><ul><li>put into sperm/eggs </li></ul><ul><ul><li>- Pre-formationism – little person inside of gametes (homunculus) </li></ul></ul><ul><li>- Blending inheritance – actual mixing of genetic information </li></ul>
    23. 24. “ Blending”
    24. 25. History of genetics • Technological and scientific developments changed genetics in 1800s - Microscopes invented – direct observation of gametes - Darwin and Mendel revolutionize genetics - Evolutionary and transmission genetics begin - Chromosomes observed and discovered to carry genetic information (early 1900s)
    25. 26. Chapter 2 Chromosomes and Cellular Reproduction
    26. 27. 1. Nucleus vs nucleoid - Nucleus – Membrane-enclosed organelle inside of eukaryotic cells that holds the chromsomes - Nucleoid – Region of a prokaryotic cell cytoplasm in which the chromosome resides 2. Chromosome - Single piece of DNA + proteins - Can be linear (eukaryotes) or circular (prokaryotes) - Eukaryotes have many, prokaryotes have one - Divided into many genes
    27. 28. The Nucleus in Context
    28. 29. <ul><li>Origin of the Nucleus: Most likely, from invagination of cell membrane of an ancient bacterium. The interior of the nucleus is topologically equivalent of the cytoplasm. </li></ul><ul><li>The ER is continuous with the nuclear membrane. Topologically equivalent to the extracellular space. </li></ul>Evolution of the Eukaryotic Nucleus
    29. 32. Chromosome distribution is not random Chromosome Territories
    30. 33. Good Review: Meaburn and Misteli, Nature, Jan 25, 2007, vol445, pp379 - 381 Chromosome Territories <ul><li>Chromosomes maintain their individuality </li></ul><ul><ul><li>they have fixed homes  spatially limited volume </li></ul></ul><ul><ul><li>Each chromosome of a specific cell type can be assigned a preferential position relative to the center of the nucleus </li></ul></ul><ul><ul><li>territories have highly branched and interconnected channels </li></ul></ul>
    31. 34. Chromosome Structure <ul><li>Centromere: attachment point for spindle microtubules </li></ul><ul><li>Telomeres: tips of a linear chromosome </li></ul><ul><li>Origins of replication: where the DNA synthesis begins </li></ul>
    32. 35. At times a chromosome Consists of one single chromatid At other times it consists of 2 (sister) chromatids The telomeres are the stable ends of chromosomes The centromere is a constricted region of the chromosome where The kinetochores form and the Spindle microtubules attach
    33. 36. Centromeres can be located at different sites on a chromosome
    34. 37. 3. Gene - Defined segment of a chromosome that provides the instructions to make a single produ ct (a protein) - One chromosome may be subdivided into 1,000 different genes  Chromosomes are divided into genes, which are composed of DNA 4. Diploid vs. haploid - Most eukaryotic organisms have 2 copies of each chromosome in each of their somatic cells - 2 copies = 2n = DIPLOID - Gamete cells (e.g. sperm and eggs) contain only 1 copy - 1 copy = 1n = HAPLOID 1 chromosome insulin ATP synthase actin hemoglobin
    35. 38. • Some more genetic terms (and how they relate to one another): 5. Alleles - Alternate forms of the same gene caused by minor differences in the DNA sequnce 6. Genotype vs phenotype - Genotype – The genetic makeup of an organism (what the genes look like) - Phenotype – The actual observable characteristics that we see - Influenced by the genotype and the environment eye color blue eyes brown eyes
    36. 39. Quick review of Mitosis http://www.youtube.com/watch?v=AhgRhXl7w_g&feature=fvst
    37. 41. Genetic consequences of the cell cycle <ul><li>Producing two cells that are genetically identical to each other and with the cell that gave rise to them. </li></ul><ul><li>Newly formed cells contain a full complement of chromosomes. </li></ul><ul><li>Each newly formed cell contains approximately half (but not necessarily identical) the cytoplasm and organelle content of the original parental cell. </li></ul>
    38. 42. Quick review of Meiosis http://www.youtube.com/watch?v=iCL6d0OwKt8&feature=channel_video_title
    39. 43. Room for genetic variation
    40. 45. Room for genetic vatiation
    41. 47. Genetic Variation is Produced through the random distribution of chromosomes in meiosis <ul><li>Try Animation 2.3 </li></ul><ul><li>file:///E:/media/ch02/Animations/0203_genetic_var_meiosis.html </li></ul>Genetic Variation is Produced through crossing over
    42. 48. Consequences of Meiosis and Genetic Variation <ul><li>Four cells are produced from each original cell. </li></ul><ul><li>Chromosome number in each new cell is reduced by half. The new cells are haploid. </li></ul><ul><li>Newly formed cells from meiosis are genetically different from one another and from the parental cell. </li></ul>
    43. 49. Concept Check <ul><ul><li>crossing over </li></ul></ul><ul><ul><li>contraction of chromosomes </li></ul></ul><ul><ul><li>separation of homologous chromosomes </li></ul></ul><ul><ul><li>separation of chromatids </li></ul></ul>Which of the following events takes place in meiosis II, but not in meiosis I?
    44. 50. Transmission genetics Mendel and beyond
    45. 51. Chapter 3 Basic Principles of Heredity
    46. 52. Transmission genetics Overview • Transmission genetics  Study of how genes/traits are passed down through generations - Wrong hypotheses in history : - Pangenesis: Traits absorbed from all over body  sperm/eggs - Lemarckism: Kids inherit their parents acquired characteristics - Preformationism: Little people inside of gametes (homunculus) - Blending inheritance: Traits of parents blend  passed on • Gregor Mendel – 1850-60s - Experiments with peas disproved all above - Found discrete units of inheritance passed down - Provide instructions for brand new organism - No blending - Unique combination of genes provides unique traits in offspring
    47. 53. Transmission genetics Review of terminology • Genes = Short segments of chromosomes/DNA that provide instructions to make a polypeptide • Alleles – Different forms (variations) of the same gene that result from minor differences in the DNA sequence - Example : Gene controlling melanin • Homozygous vs. heterozygous Homozygous – Organism has 2 copies of the same allele for a given gene Heterozygous – Organism has 2 different alleles for a given gene • Dominant vs. recessive - In a heterozygous individual , both alleles are expressed (assuming no imprinting) - However, one allele often has a greater impact on the final observable trait - Those alleles that mask the effects of other alleles  DOMINANT Those alleles that get masked  RECESSIVE
    48. 54. Simple Mendelian inheritance Monohybrid cross • Simple Mendelian inheritance - Each trait is controlled by one gene, each gene controls one trait - Alleles have a clear dominant-recessive relationship - Individuals have expected phenotype - If examining more than 1 gene  they are on separate chromosomes • Monohybrid cross (1 gene) - Example : Human earlobe shape - Dominant allele  Unattached (E) Recessive allele  Attached (e) - Example mating : Two heterozygotes (Ee x Ee) - Need to predict offspring genotypes/phenotypes - Theory of segregation – Two copies of the gene are segregated and packaged into separate gametes - 1 gets the &quot;E&quot; and the other the &quot;e&quot;  Predict a 3:1 ratio of dom:rec phenotypes - Only a probability – may not see exact ratio e E ee Ee Ee EE e E
    49. 55. Transmission genetics Review of terminology • Genetic shorthand - Dominant allele  Capital letter Recessive allele  Lowercase letter A = normal melanin production AA (homozygous dominant) a = no melanin aa (homozygous recessive) Aa (heterozygous) • Genotype vs. phenotype Genotype – Genetic background of a trait (AA, aa, Aa) Phenotype – Actual observable trait that we see (skin color) - Multiple genotypes can give rise to the same phenotype (e.g. AA and Aa) - Not all individuals with a given genotype will have the expected phenotype • Generation shorthand - P = Parental generation - F 1 = First filial generation (produced from mating two parents) - F 2 = Second generation (produced from mating 2 F1 offspring) Choose letter wisely!
    50. 56. • Why was Mendel so successful? 1) His choice of experimental subject, the pea plant ( Pisum sativum ) is easy to cultivate and grow relatively rapidly ( by 17 th century standards); 2) Peas produce many offspring (seeds); 3) Large variety of traits and traits were genetically pure; 4) He focused on characteristics that have 2 forms; 5) He adopted an experimental approach and used math! 6) He kept careful records of his experiments and was very patient…
    51. 57. Simple Mendelian inheritance Monohybrid and dihybrid crosses • Simple Mendelian inheritance - Each trait is controlled by one gene, each gene controls one trait - Alleles have a clear dominant-recessive relationship - Individuals have expected phenotype - If examining more than 1 gene  they are on separate chromosomes • Monohybrid cross (1 gene) - Example : Human earlobe shape - Dominant allele  Unattached (E) Recessive allele  Attached (e) - Example mating : Two heterozygotes (Ee x Ee) - Need to predict offspring genotypes/phenotypes - Theory of segregation – Two copies of the gene are segregated and packaged into separate gametes - 1 gets the &quot;E&quot; and the other the &quot;e&quot;  Predict a 3:1 ratio of dom:rec phenotypes - Only a probability – may not see exact ratio e E ee Ee Ee EE e E
    52. 58. • Dihybrid cross (2 genes) - Mendel's original experiments : - Done to determine if inheritance of one gene affects inheritance of another - Found that when genes are on different ( nonhomologous ) chromosomes, they move independently during meiosis and don't affect one another  Principle of independent assortment - Example (assuming independent assortment) : - Earlobe shape (gene E ) and hairline shape (F – widow's peak, f – straight) - If mate EeFf x EeFf, what phenotypic ratios in offspring? FIRST SEPARATE THE GENES… Then do separate Punett squares… ee Ee Ee EE e E e E ff Ff Ff FF f F f F
    53. 59. E_F_ = double dom. = 3/4 x 3/4 = 9/16 eeF_ = 1 rec, 1 dom = 1/4 x 3/4 = 3/16 E_ff = 1 dom, 1 rec = 3/4 x 1/4 = 3/16 eeff = double rec. = 1/4 x 1/4 = 1/16 INDEPENDENT TERMS 9:3:3:1 9+3+3+1=16 e E ee Ee Ee EE e E f F ff Ff Ff FF f F
    54. 60. Simple Mendelian inheritance Monohybrid and dihybrid crosses • Dihybrid cross (2 genes) - Example problem : - Imagine you mate a person who has attached earlobes and a smooth hairline with a person who is heterozygous for both genes. What is the probability of them having a child with unattached earlobes and a widow's peak hairline? - Step 1: Convert to letters - Parent 1 = attached, smooth  eeff - Parent 2 = EeFf - ? offspring = E_F_ (2nd letter for each doesn't matter) e e e E f f f F - Step 2: Do two Punett squares - Split up the E's from the F's - Step 3: Do the math - Multiply the individual probabilities - E_F_ = 1/2 x 1/2 = 1/4 ee ee Ee Ee ff ff Ff Ff
    55. 61. Simple Mendelian inheritance Theoretical Probability for Simple Events P(A) = # of outcomes favorable to A Total # of outcomes <ul><li>What is the probability of getting all 3 Heads is you flip 3 coins? </li></ul><ul><li>All possible combinations </li></ul><ul><li>8 outcomes </li></ul><ul><li>Only 1 is favorable </li></ul>1/8 P(A) = e e ee ee Ee Ee e E f f ff ff Ff Ff f F
    56. 62. Simple Mendelian inheritance Theoretical Probability for Simple Events P(A) = # of outcomes favorable to A Total # of outcomes <ul><li>What is the probability of getting all 3 Heads is you flip 3 coins? </li></ul>OR <ul><li>Probability for each coin: 1/2 </li></ul><ul><li>Each coin is “independent” </li></ul>½ X ½ X ½ = 1/8 e e ee ee Ee Ee e E f f ff ff Ff Ff f F
    57. 63. Simple Mendelian inheritance Theoretical Probability for Simple Events <ul><li>What is the probability of getting all 6 Heads is you flip 6 coins? </li></ul><ul><li>Probability for each coin: 1/2 </li></ul><ul><li>Each coin is “independent” </li></ul>½ X ½ X ½ X ½ X ½ X ½ = 1/64 <ul><li>What is the probability of getting either heads or tails in all 6 coins? </li></ul>
    58. 64. Simple Mendelian inheritance Theoretical Probability for Simple Events Probability – tree diagrams <ul><li>All in a bag. Pick one, replace in the bag, pick another </li></ul>7/10 7/10 7/10 3/10 3/10 3/10 (7/10) X (7/10) = 49/100 (7/10) X (3/10) = 21/100 (3/10) X (7/10) = 21/100 (3/10) X (3/10) = 9/100 100/100 B R B B R R B B B R R B R R
    59. 65. Simple Mendelian inheritance Theoretical Probability for Simple Events Probability – tree diagrams <ul><li>All in a bag. Pick one, replace in the bag, pick another </li></ul><ul><li>What is the probability of picking 2 reds </li></ul><ul><li>What is the probability of picking no reds </li></ul><ul><li>What is the probability of picking at least 1 B? </li></ul>(49/100) + (21/100) + (21/100) = 91/100 <ul><li>What is the probability of picking one of each? </li></ul>(21/100) + (21/100) = 42/100 B B B R R B R R (7/10) X (7/10) = 49/100 (7/10) X (3/10) = 21/100 (3/10) X (7/10) = 21/100 (3/10) X (3/10) = 9/100 100/100
    60. 66. The binomial expansion and probability a A aa Aa Aa AA a A Revisiting out Punett square and taking a look at kids with albinism… Heterozygous parents: Chances of having albino child: 1/4 If couple has 3 children, what are the chances for all 3 with albinism: and another albino child: 1/4 And another albino child: 1/4 (1/4) X (1/4) X (1/4) = 1/64 a A aa Aa Aa AA a A a A aa Aa Aa AA a A
    61. 67. The binomial expansion and probability Now… 3 kids: What are the chances of having one child with albinism? (3/4) X (1/4) X (1/4) = 3/64 (9/64) + (9/64) + (9/4) = 27/64 a A aa Aa Aa AA a A non albino 3/4 1/4 a A aa Aa Aa AA a A a A aa Aa Aa AA a A non non albino albino 3/4 3/4 1/4 1/4 albino albino albino non non non non albino 3/4 3/4 3/4 3/4 1/4 1/4 1/4 1/4 Non, non, non Non, non, albino Non, albino, non Non, albino, albino albino, non, non albino, non, albino albino, albino, non albino, albino, albino 64/64 (3/4) X (3/4) X (3/4) = 27/64 (3/4) X (3/4) X (1/4) = 9/64 (3/4) X (1/4) X (3/4) = 9/64 (1/4) X (3/4) X (3/4) = 9/64 (1/4) X (3/4) X (1/4) = 3/64 (1/4) X (1/4) X (3/4) = 3/64 (1/4) X (1/4) X (1/4) = 1/64
    62. 68. The binomial expansion and probability Now… 5 kids: What are the chances of having one child with albinism? Here is where the binomial equation is useful…. (p+q) n Where p = probability of one event = albino q = probability of the other event n = # times the event occurs (p + q) 5 = p 5 + 5p 4 q + 10p 3 q 2 + 10p 2 q 3 + 5pq 4 + q 5 Now… 5 kids: What are the chances of having one child with albinism? = 5pq 4 = 5(1/4)(3/4) 4 = 5(1/4)(81/256) = 405/ 1024 = 0.39 a A aa Aa Aa AA a A a A aa Aa Aa AA a A a A aa Aa Aa AA a A a A aa Aa Aa AA a A a A aa Aa Aa AA a A
    63. 69. The binomial expansion and probability Here is where the Pascal’s triangle is useful…. (p+q) n (p+q) 0 1 (p+q) 1 1 p + 1 q (p+q) 2 1 p 2 + 2 p 1 q 1 + 1 q 2 (p+q) 3 1 p 3 + 3 p 2 q 1 + 3 p 1 q 2 + 1 q 3 (p+q) 4 1 p 4 + 4 p 3 q 1 + 6 p 2 q 2 + 4 p 1 q 3 + 1 q 4 (p+q) 5 1 p 5 + 5 p 4 q 1 + 10 p 3 q 2 + 10 p 2 q 3 + 5 p 1 q 4 + 1 q 5 (p+q) 6 1 p 6 + 6 p 5 q 1 + 15 p 4 q 2 + 20 p 3 q 3 + 15 p 2 q 4 + 6 p 1 q 5 + 1 q 6 a A aa Aa Aa AA a A a A aa Aa Aa AA a A a A aa Aa Aa AA a A a A aa Aa Aa AA a A a A aa Aa Aa AA a A
    64. 70. • An individual has a dominant phenotype  what is the genotype (AA or Aa)? - Do a testcross • Testcross - Take your individual in question and mate them with a homozygous recessive individual (aa) - Predictions : 1) If the individual is AA AA x aa  all offspring should have DOMINANT pheno 2) If the individual is Aa Aa x aa  1/2 should have dom. pheno 1/2 should have rec. pheno.  Routinely done to determine the genotype of an individual a A aa Aa aa Aa a a A A Aa Aa Aa Aa a a
    65. 72. • Observed ratio of progeny may deviate from expected ratios by chance. We expected a 1:1 ratio, but after counting Y ellow and B rown roaches… There were 22 B rown and 18 Y ellow Observed Ratios of Progeny The Goodness-of-Fit Chi-Square Test So… when do we use the Chi-Square Test? When what comes out is not what we expected! To see how well observed values FIT the expected values It indicates the probability that the difference between observed and expected values is due to chance.
    66. 73. <ul><li>• The hypothesis that chance alone is responsible for any deviation </li></ul><ul><li>between observed and expected values is called the null hypothesis . </li></ul><ul><li>Looking at the cats: Black (B) is dominant over Gray (b) </li></ul>Observed Ratios of Progeny The Goodness-of-Fit Chi-Square Test • If we cross 2 heterozygous black (Bb X Bb), we would expect a 3:1 ratio: • Now we have 50 kittens: 30 black and 20 gray b B bb Bb Bb BB b B
    67. 74. Observed Ratios of Progeny The Goodness-of-Fit Chi-Square Test • If we cross 2 heterozygous black (Bb X Bb), we would expect a 3:1 ratio: <ul><li>Observed values: 50 kittens: 30 black and 20 gray </li></ul><ul><li>First get the expected values: </li></ul>Black kittens expected: (3/4) of 50 = 37.5 Grey kittens expected: (1/4) of 50 = 12.5    (observed – expected) 2  expected Chi-Square value =  X 2 =  (30 – 37.5) 2 37.5 + X 2 =  (20 – 12.5) 2 12.5 X 2 = 6 b B bb Bb Bb BB b B
    68. 75. Observed Ratios of Progeny The Goodness-of-Fit Chi-Square Test • Then we figure-out the degrees of freedom = n-1 n = the number of ways that things can vary in the cats’ case: it’s “2 phenotypes” X 2 = 6 • degrees of freedom = 2-1 = 1 • Now we look at the CHI table and see where “6” is For a degree of fredom = 1
    69. 76. The probability of the event due to chance decreases When value is less than 0.05, chance is not responsible for this! Solve Problem 35 at the end of chapter 3

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