MBG*2040 Foundations in Molecular Biology and Genetics Lecture 1-2 Introduction and Review PDF

MBG*2040 Foundations in Molecular Biology and Genetics Dr. Jim Uniacke 1 [email protected] 2  Review: Basic Principles of Heredity and Pedigree Analysis (Chapter 3.1-3.4 and Chapt...

MBG*2040 Foundations in Molecular Biology and Genetics Dr. Jim Uniacke 1 [email protected] 2  Review: Basic Principles of Heredity and Pedigree Analysis (Chapter 3.1-3.4 and Chapter 6.1-6.2) Define, recognize, describe and apply Mendel’s first principle Distinguish genotype and phenotype Solve genetic problems relating to monohybrid crosses Interpret pedigrees – Assess mode of inheritance – Assign genotypes Use the rules of probability to determine genotypic and phenotypic proportions 3  Why Study Genetics? Genetics plays a critical role in understanding disease Genetics improves agricultural practices Genetics used biotechnology industry for drug design, development, & production 4 Zinfandel and Primitivo Carménère: The Jurassic grape 5  Fundamental Concepts Review A gene is the fundamental unit of heredity Genes come in multiple forms called alleles Genotype confers phenotype Genetic information is carried in DNA Genes are located on chromosomes Genetic information is transferred from DNA to RNA to protein 6  Genetic Terminology Term Definition Gene A genetic factor (region of DNA) that helps determine a characteristic Allele One of two or more alternative forms of a gene Locus Specific place on a chromosome occupied by an allele Genotype Set of alleles possessed by an individual organism Heterozygote An individual organism possessing two different alleles at a locus Homozygote An individual organism possessing two of the same alleles at a locus Phenotype or Trait The appearance or manifestation of a character 7  Basic Principles of Heredity Gregor Johann Mendel discovered basic principles of heredity 8 Monohybrid Cross 9 10  Concept Check If an F1 plant depicted in Figure 3.6 is backcrossed to the parent with round seeds, what proportion of the progeny will have wrinkled seeds? (Use a Punnett square.) a. ¾ b. ½ c. ¼ d. 0  Concept Check If an F1 plant depicted in Figure 3.6 is backcrossed to the parent with round seeds, what proportion of the progeny will have wrinkled seeds? (Use a Punnett square.) a. ¾ b. ½ c. ¼ d. 0 Dihybrid Crosses 13 14 15 Mitosis and Meiosis review 17  Human Karyotype Karyotype: G-banding Karyotype: fluorescent probes Human chromosome set: 46 chromosomes. Here we see two copies of the set (diploid) 18  Cell division initiates packaging into chromosomes while doubling the DNA MITOSIS 2N = 46 Paternal Maternal A pair of homologous chromosomes 19  Cell division initiates packaging into chromosomes Sister chromatids while doubling the DNA MITOSIS 2N = 46 Paternal Maternal A pair of homologous chromosomes Paternal Maternal 2N = 46 2N = 46 20 Chapter 3-5 MEIOSIS I 2N = 46 *Segregate the homologous Chromosomes* Paternal Maternal 21  MEIOSIS II Paternal Maternal *Just like mitosis. Separate the sister chromatids Gametes N = 23 N = 23 N = 23 N = 23 22 Person A Person B gametes gametes 2N = 46 N = 23 N = 23 23 Genes are located on chromosomes A- -A a- -a MEIOSIS I 2N = 46 *Separate the homologous Chromosomes* Paternal Maternal 24  MEIOSIS II Paternal Maternal *Just like mitosis. Separate A- -A the sister chromatids a- -a Gametes N = 23 N = 23 N = 23 N = 23 25  Person A Person B gametes gametes 2N = 46 N = 23 a- -a A- a- A- a- N = 23 A- a- A- a- 26  MEIOSIS II Paternal Maternal *Just like mitosis. Separate A- -A the sister chromatids a- -a Chapter 8 *But mistakes happen! Non-disjunction Gametes N = 24 N = 22 N = 23 N = 23 27  Person A Person B gametes gametes 2N = 47 A- a- A- -A a- a- A- a- 28 Genes are located on chromosomes 2N = 46 A- -A a- -a B- -B b- -b MEIOSIS I AaBb *Separate the homologous Chromosomes* Following two chromosome pairs, now you can see how Pat Mat Pat Mat independent assortment produces many variations in gametes Option #1 29  MEIOSIS II *Just like mitosis. Separate the sister chromatids A- -A B- -B a- -a b- -b Option #1 Gametes N = 23 N = 23 N = 23 N = 23 30 Genes are located on chromosomes 2N = 46 A- -A a- -a B- -B b- -b MEIOSIS I *Separate the homologous Chromosomes* Following two chromosome pairs, now you can see how Pat Mat Pat Mat independent assortment produces many variations in gametes Option #2 31  MEIOSIS II *Just like mitosis. Separate the sister chromatids A- -A b- -b a- -a B- -B Option #2 Gametes N = 23 N = 23 N = 23 N = 23 32 A 4 X 4 Punnett square will determine probabilities Person A Person B 2N = 46 N = 23 -b a- -a -B -b -b A- -b a- -b A- a- N = 23 -B -B -B -B A- a- A- a- 33 Genes are located on chromosomes MEIOSIS I BUT A- -A a- -a *Separate the homologous b- -b B- -B Chromosomes* Consider that there are thousands of genes on each chromosome! 2N = 46 Chapter 7 AaBb Linkage Paternal Maternal 34  MEIOSIS II Paternal Maternal *Just like mitosis. Separate A- -A the sister chromatids a- -a b- -b B- -B Chapter 7 Consider that there are thousands of genes on each chromosome! Gametes N = 23 N = 23 N = 23 N = 23 35  Person A Person B gametes gametes 2N = 46 N = 23 a- -a B- -B A- A- a- b- a- b- B- B- N = 23 A- a- A- a- b- B- b- B- 36 Pedigree symbols Generation line Sibship line 37  Pedigree Assumptions If the trait is RARE and autosomal recessive then individuals who marry/mate into the pedigree are not carriers but are… homozygous for the ‘normal’ allele… unless there is evidence to the contrary. 38  Recessive Mutations Often involve loss of gene function: Null/Amorphic Alleles I. A nonfunctional protein is produced OR II. No protein is produced Hypomorphic Alleles I. A poorly functioning protein is produced OR II. Reduced amounts of a normally functioning protein is produced 39  Pedigree Probability Problems Autosomal Recessive Pedigree Assign Genotypes to Each Member…GO! aa Aa AA Aa aa AA Aa Aa Aa AA 1/2Aa 1/2AA 1/2Aa ? Aa Aa aa 1/3AA 2/3Aa 1/2Aa Tutorial #1 1/2AA 1/2AA 1/2Aa, 3/8AA, 1/8aa 42  Concept Check 3 If the probability of being blood-type A is 1/8 and the probability of blood-type O is 1/2, what is the probability of being either blood-type A or O? a. 5/8 b. 1/2 c. 1/8 d. 1/16  Concept Check 3 If the probability of being blood-type A is 1/8 and the probability of blood-type O is 1/2, what is the probability of being either blood-type A or O? a. 5/8 b. 1/2 c. 1/8 d. 1/16  Autosomal Dominant Traits Now let’s examine autosomal dominant traits bb normal Bb affected heterozygote BB affected homozygote Examples: Huntington Disease 45  Dominant Mutations Often Involve Gain/Change of Gene Function Dominant Hypermorphic Alleles Negative phenotypic consequences due to the over- production of a normal protein OR Negative phenotypic consequences due to the production of a protein with increased activity levels Neomorphic Alleles Negative phenotypic consequences due to the presence of an altered protein that has a new function Negative phenotypic consequences when the altered protein interferes with the wildtype protein (Dominant-negative allele) 46  Autosomal Dominant Traits For traits that are rare and dominant, affected individuals are most likely to be heterozygous (Bb) WHY? If the trait is rare in the population, then matings between Bb individuals would be infrequent. In general you will see Bb x bb crosses. Bb x Bb crosses are RARE. 47 Pedigree Probability Problems Autosomal Dominant Pedigree Assign Genotypes to Each Member…GO!! 48  Take Home Problem Albinism is a recessive condition resulting from the inability to produce the dark pigment melanin in skin and hair. A man and woman with normal skin pigmentation want to have two children. The man has one albino parent; the woman has parents with normal pigmentation, but an albino brother. What is the probability their first child will be albino? 49