GEN 3022 1st Edition Lecture 8Outline of Last Lecture I. Introductiona. Simple Mendelian Inheritanceb. Complexity of InheritanceII. Wild-Type Allelesa. Wild-typei. Definitionii. Two explanations for the wild-type phenotypeiii. Exampleb. Genetic polymorphismc. Mutant allelesi. Definitionii. Dominant mutants1. Gain of function2. Dominant negative3. HaploinsufficiencyIII. Incomplete Penetrancea. Definitionb. Ex: PolydactylyThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.c. Degree of PenetranceIV. Expressivitya. Definitionb. FactorsV. Environmental Effectsa. Definition and examplesi. Coat colorii. PKUb. Temperature sensitive mutationsi. Example: Siamese catsVI. Incomplete Dominancea. Definitionb. Example: flower colorVII. Overdominancea. Definitionb. Example: Sickle Cell AnemiaVIII. Multiple Allelesa. Definitionb. Example: blood typec. CodominanceIX. X-linked Genesa. Reviewb. Reciprocal CrossX. Lethal Allelesa. Definitionb. Essential and nonessential genesc. Conditional lethal allelesd. Semilethal allelesXI. Gene interactionsa. Gene interationsb. Epistasisc. NoteXII. ComplementationXIII. Gene Redundancya. Definitionb. Gene knockoutc. ParalogsOutline of Current Lecture I. Incomplete Penetrancea. Two explanationsII. Blood typesa. Example 1III. Types of dominancea. Loss-of-function recessive alleleb. Dominant negative c. Incomplete dominancei. Example 1: loss-of-function alleleii. Example 2: probabilityd. Overdominancee. ComplementationIV. Environmental EffectsV. PenetranceVI. Chromosomesa. Homologous Chromosomeb. Chromatids c. Sister Chromatidsd. Bivalentse. Karyotype VII. X-linked recessive inheritancea. Example 1b. Example 2c. Example 3Current LectureI. Incomplete Penetrancea. Two explanations for incomplete penetrance of a dominant allele are:i. Other genes may affect the phenotype (gene interaction/redundancy)ii. The environment may affect the phenotype (environmental effects)II. Blood Typesa. Example 1: a female who is IBi has offspring with a male who is IAIA. Which blood type(s) would not be possible for their offspringi. Type B and type O because the each offspring has an IA allele in his/her genotype which makes IBi, IBIB, and ii impossibleIII. Types of Dominancea. Loss-of-function recessive allele: an allele affecting flower color in snap dragons shows simple Mendelian dominant/recessive pattern of inheritance. If the recessive allele is a loss-of-function allele, a reasonable explanation for the phenotype of this trait is: i. 50% of the functional protein gives the same phenotype as 100% (this means that a heterozygote will display the dominant trait)b. Dominant negativei. Two proteins interact to form a complex. When a gene that encodes one protein in the complex is mutated, the complex suffers a substantial loss of activity. This type of mutation is classified as: dominant negativec. Incomplete Dominancei. Example 1: loss-of-function allele1. A trait exhibits incomplete dominance. If one of the two alleles is a loss-of-function allele, a reasonable molecular explanation would be: 50% of the functional protein is not enough to give the same phenotype as 100% (a heterozygote will give an intermediate trait that is neither dominant nor recessive)ii. Example 2: probability 1. It has been demonstrated that flower color in four-o’-clocks showsincomplete dominance. When two pink-flowered four-o’-clocks are crossed to each other, what are the following probabilities of the offspring?a. A plant will be red-flowered: 25% (genotype RR)b. A plant will be either white or pink: 75% (genotype Rr or rr)d. Overdominancei. An example of overdomiance would be two lines of true breeding tomatoes that are susceptible to verticillium wilt (a fungal infection) produce offspring that are resistant to verticillium wilte. Complementationi. Two different strains of plant exhibit a recessive phenotype of white flowers. When crossed, they produce offspring with wild-type flowers. This phenomenon is called complementation and it indicates that the recessive alleles are in two different genesIV. Environmental Effectsa. You are trying to develop true breeding blue hydrangeas, but the plants in one corner of your plot consistently turn pink regardless of your mating strategy. Thecause of this would be environmental effects because it the phenotype of the flowers is resulting from the physical location of the flowers in the plot. V. Penetrancea. What does is mean that a disease is 80% penetrant?i. This means that only 80% of the people who carry the dominant allele display the phenotype. It is important to specify that it is the dominant allele that is not expressed. VI. Chromosomesa. Homologous Chromosomes: the two copies (pair) of the same chromosome in a diploid cell. One is inherited from the mother and the other from the father. They are similar but are not necessarily identical because they can contain different alleles.b. Chromatids: these are the chromosomes after replication. One chromatid has thesame amount of DNA as a single chromosome. c. Sister Chromatids: after the chromosomes have replicated, the two chromatids remain attached at the centromere and are called sister chromatids. Sister chromatids have twice the amount of DNA as a single chromosome. d. Bivalent: formed when both pairs of sister chromatids associate with each other. Bivalents have four times the amount of DNA as a single chromosomee. Karyotype: an organized representation of the chromosomes within a cellVII. X-linked recessive inheritancea. Red-green color blindness is inherited as an x-linked recessive trait. i. Example 1: What is the probability that a woman with phenotypically normal parents and a color-blind brother will have a color-blind son with a man who is not color blind? (they have not yet had any children)1. Because the woman’s brother is color-blind but her father is not, then the mother must have been a heterozygote. She therefore has a 50% probability of inheriting the affected allele. The probability that she would pass this on to a child would be 50% and the probability that she will have a son is 50%, so the overall probability is 0.5 x 0.5 x 0.5 = 12.5%.ii. Example 2: What is the probability that a phenotypically normal woman, who has already had one color blind son with a man who is not color blind, will have a color blind son with the same man?1. Because she has