BIO207_TermTest 1.pdf

BIO207H5S Introductory Genetics Winter 2024 Lecture 1 Preeti Karwal, Ph.D. BIO207H5S LEC0101 Introductory Genetics Course Outline - Winter 2024 Instructor: Preeti Karwal E-mail Address: [email protected] Class Time and Location Mon, 01:00 PM - 02:00 PM, IB 110 Tue, 11:00 AM - 01:00 PM, IB 110 Office Hours Schedule and Location Monday 11:00 AM - 12:00 PM, DV3093CD Tuesday 1:00 PM - 2:30 PM, DV3093CD Textbooks: 1. 2. Pierce, Benjamin A. Genetics: A Conceptual Approach 7th ed. - W.H. Freeman & company, New York. Daniel L. Hartl. Genetics: Analysis of Genes and Genomes 9th Edition - Jones & Bartlett Publishers. The content covered in the lecture slides, classroom discussion and tutorials is the main source of information. Points of Contact for your Tutorials and Grading of Assessments TUT0106 and TUT0108: Eniolaye Balogun (Admin TA) E-mail Address [email protected] TUT0110, TUT0113, TUT0115 and TUT0117: Asawari Albal E-mail Address [email protected] TUT0105 and TUT0107 : Neha Purakan E-mail Address [email protected] TUT0101 and TUT0103: Nicholas Boehler E-mail Address [email protected] TUT0109 and TUT0111 : Farhan Majumder E-mail Address [email protected] TUT0102 and TUT0104 : Timothy McLean E-mail Address [email protected] TUT0114 and TUT0116 : Yijie (Jacky) Zhu E-mail Address [email protected] TUT0112: Farheen Mohammed E-mail Address [email protected] Grading TA: Cameron Sinclair E-mail Address [email protected] Tutorial Quizzes - There will be weekly tutorial quizzes. The Lowest quiz mark will be dropped. Participation in all tutorial quizzes is highly recommended as NO makeup quizzes will be taken in any circumstances. Final Exam - The final Exam is CUMULATIVE, it covers all the content covered during lectures and tutorials. The Final exam, like the term tests, will likely include multiple choice, short answer and few long answer subjective type questions. (more details will be shared on Quercus before the final exam). Term Tests The Term Tests are NOT cumulative and will cover all material covered before or in between two term tests. There will be 3 term tests and the two highest-scored term tests will count toward your final grade. Participation in all three tests is highly recommended as NO retests or makeup tests will be taken in any circumstances. If you miss a term test, use the Absence Declaration tool on ACORN to declare an absence and contact your instructor. In case of two missed tests due to unforeseen circumstances, a weighted average based on the redistribution of grades for other similar assessments will be considered subject to evaluation of individual case based on the timely submission of appropriate documents for proof of your reason for absence. For this academic consideration, the student will be required to submit appropriate documentary proof of their reason for absence e.g., medical documentation to Eniolaye Balogun, the admin TA within two days of their absence in the term test. This redistribution of grades, if at all considered will apply to only one of the two essential term tests, not both. Genetics Genetics is the study of heredity, how inherited variation is encoded, replicated, and expressed, and how it evolves over time. This definition encompasses the three major subdivisions of genetics: transmission genetics, molecular genetics, and population genetics. 1.6 Genetics can be divided into three major subdisciplines 1.7 Model genetic organisms are species with features that make them useful for genetic analysis TABLE 1.1 Early concepts of heredity Concept Proposed Correct or Incorrect Pangenesis Genetic information travels from different parts of the body to reproductive organs. Incorrect Inheritance of acquired characteristics Acquired traits become incorporated into hereditary information. Incorrect Preformationism Miniature organism resides in sex cells, and all traits are Table 1.1 Early concepts of heredity inherited from one parent. Incorrect Blending inheritance Genes blend and mix. Incorrect Germ-plasm theory All cells contain a complete set of genetic information. Correct Cell theory All life is composed of cells, Correct and cells arise only from cells. Mendelian inheritance Traits are inherited in accord with defined principles. Correct Father of Genetics Gregor Johann Mendel 1822–1884 Why do you think Mendel was successful? Mendel’s Skills ▪ He adopted an experimental approach unlike many earlier investigators who simply described the results of crosses. He formulated hypotheses based on his initial observations and then conducted additional crosses to test his hypotheses. ▪ He always interpreted his results by using mathematics. He kept careful records of the numbers of progeny possessing each trait and computed ratios of the different traits. ▪ He was adept at seeing patterns in detail and was patient and thorough, conducting his experiments for 10 years before attempting to write up his results in 1866. Advantages from the Model Organism and the Experimental Setup ▪ Garden pea plants complete entire generation in a single growing season (annual plants) and produce many offspring. ▪ Garden pea plants are known to differ in detectably distinct characters. e.g., seed colour, seed shape, flower colour, pod colour, texture. ▪ Pea plants are self fertilizing plants – True breeding lines. Also artificial cross-pollination was possible. ▪ The loci for any two characters studied simultaneousy in a dihybrid cross were located far apart in most cases – Independent assortment. ▪ In Mendel’s peas, the genotype and not the environment, largely determined the characters. Some more facts about Mendel…… Mendel did these experiments about 20 years before microscopy advances allowed identification of chromosomes and their role in inheritance. He published his work in 1866 but his work went unappreciated for about 34 years. Hugo de Vries, Erich Tschermark and Carl Correns independently confirmed Mendel’s conclusions around year 1900. The theory that genes are located on chromosomes (the chromosomal theory of heredity) was developed in the early 1900s by Sutton and Boveri. 3.5 (1) Mendel conducted monohybrid crosses Monohybrid cross Mendel began by conducting monohybrid crosses, the crosses between parents that differed in a single characteristic. Principle of Segregation Mendel drew several conclusions from the results of his monohybrid crosses 1. He reasoned that although the F1 plants display the phenotype of only one parent, they inherit genetic factors from both parents as both parental phenotypes are seen in F2 generation. Each plant must therefore possess two genetic factors encoding a characteristic. 2. The two alleles in each plant separate when gametes are formed, and one allele goes into each gamete. 3. The traits that appeared unchanged in the F1 heterozygous offspring Mendel termed as dominant, and those that disappeared in the F1 heterozygous offspring were named as recessive. The Concept or Law of Dominance states that when two different alleles are present in a genotype, only the trait encoded by one of them—the dominant allele—is observed in the F1 phenotype. 4. The two alleles of an individual plant separate with equal probability into the gametes. All these four conclusions formed the basis of Principle of segregation (Mendel’s first law). “This principle states that each individual possesses two alleles (diploid organism) for any particular trait, one inherited from the maternal parent and one from the paternal parent. These alleles segregate from each other at the time of gamete formation.” Mendel further confirmed the Principle of Segregation by allowing the F2 plants to self-fertilize and produce an F3 generation. TABLE 3.1 Summary of important genetic terms Term Definition Gene An inherited factor (encoded in the 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 Table 3.1 Homozygote An individual organism possessing two of the same alleles at a locus Heterozygote An individual organism possessing two different alleles at a locus Characteristic or character An attribute or feature possessed by an organism Phenotype or trait The appearance or manifestation of a characteristic Molecular basis of Mendelian law of Segregation 3.7 Segregation results from the separation of homologous chromosomes in meiosis TABLE 3.2 Comparison of the principles of segregation and independent assortment Principle Observation State of Meiosis* Segregation (Mendel’s first law) 1. Each individual organism possesses two alleles encoding a trait. Before meiosis 2. Alleles separate when gametes are formed. Anaphase I Table 3.2 3. Alleles separate in equal proportions. Anaphase I Alleles at different loci separate independently. Anaphase I Independent assortment (Mendel’s second law) *Assumes that no crossing over occurs. If crossing over takes place, then segregation and independent assortment may also occur in anaphase II of meiosis. Punnet Square can be used to calculate the results of genetic crosses The Punnett square is a shorthand method of predicting the genotypic and phenotypic ratios of progeny from a genetic cross. This method was developed by the English geneticist Reginald C. Punnett in 1917. A Punnett square is constructed by drawing a grid, listing the gametes produced by one parent along the upper edge, and listing the gametes produced by the other parent down the left side. By crossing two varieties of pea plants that differed in height, Mendel established that tall (T) was dominant to short (t). He tested his theory concerning the inheritance of dominant traits by crossing an F1 tall plant that was heterozygous (Tt) with the short homozygous parental variety (tt). Backcross The type of cross, between an F1 genotype and either of the parental genotypes, is called a backcross. Backcrosses are usually done in artificial selection to have organisms with desirable traits. A type of backcross is testcross. Testcross The cross in which one individual of unknown genotype is crossed with another individual with a homozygous recessive genotype for the trait in question. Application: A testcross helps to test or reveal the genotype of the first individual. A Phenotypic ratio of 1:1 in F1 generation resulting from a test cross is characteristic of a heterozygous genotype of the individual tested. Another method for determining the outcome of a genetic cross is to use the rules of probability, as Mendel did with his crosses. Probability refers to the likelihood of the occurrence of a particular event. It is the number of times that a particular event takes place, divided by the number of all possible outcomes. Probability values range from 0 to 1. Two rules of probability are useful for predicting the ratios of offspring from genetic crosses: 1. Addition or Sum rule P (A or B) = P (A) + P (B) 1. Multiplication or Product rule P (A and B) = P (A) X P (B) Genotype TT or Tt or tT Possible gametes Probability Phenotype ¼+¼+¼=¾ tall Dihybrid cross In addition to his work on monohybrid crosses, Mendel performed crosses in which the varieties of peas differed in two characteristics, called dihybrid crosses. 3.11c Mendel’s dihybrid crosses revealed the principle of independent assortment Principle of Independent Assortment Mendel carried out a number of dihybrid crosses for pairs of characteristics and always obtained a 9 : 3 : 3 : 1 ratio in the F2. 9 : 3 : 3 : 1 ratio makes perfect sense in regard to the principle of segregation and the concept of dominance. Mendel recognized in the results of dihybrid crosses, the principle of independent assortment (Mendel’s second law). “This principle states that alleles at different loci separate independently of one another.” The principle of independent assortment is an extension of the principle of segregation as it states that when the two alleles separate at a locus, their separation is independent of the separation of alleles at other loci. Molecular basis of Mendelian law of Independent Assortment 3.12 The principle of independent assortment results from the independent separation of chromosomes in anaphase I of meiosis BIO207H5S Introductory Genetics Winter 2024 Lecture 2 Preeti Karwal, Ph.D. Forked line/Branch Diagram for Dihybrid Cross : applying Probability rules Forked line/Branch Diagram for Trihybrid Cross : applying Probability rules Round seeds, yellow endosperm, gray seed coat X Wrinkled seeds, green endosperm, white seed coat RRYYCC X rryycc F1 : RrYyCc Selfing F2 : R_Y_C_ : R_Y_cc : R_yyC_ : rrY_C_ : R_yycc : rryyC_ : rrY_cc : rryycc Phenotype Ratio : Try calculating yourself!! Problem Solving for calculation of probability of obtaining offspring…… Aa Bb cc Dd Ee × Aa Bb Cc dd Ee Calculate the probability of obtaining offspring with the genotype aa bb cc dd ee The probability of an offspring from this cross having genotype aa bb cc dd ee : ¼ × ¼ × ½ × ½ × ¼ = 1/256 The Chi-Square (𝛘2) Goodness-of-Fit Test The ratios of genotypes and phenotypes actually observed among the progeny, however, may deviate by chance from the expected ratio as per Mendelian principles of segregation, independent assortment, and dominance. This test indicates the likelihood of chance that could produce the deviation between the expected and the observed values. Scientific Question: Is there a difference between observed and expected ratio? Null hypothesis: There is no significant difference between the observed and the expected ratio. Degrees of freedom (d. o. f) = No. of categories (n) - 1 TABLE 3.7 Critical values of the 2 distribution The critical value, P(0.05) at d.o.f = 1 is 3.841. P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 3 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357 7.779 9.488 11.143 13.277 14.860 5 0.412 0.831 1.610 4.351 9.236 11.070 12.832 15.086 16.750 6 0.676 1.237 2.204 5.348 10.645 12.592 14.449 16.812 18.548 7 0.989 1.690 2.833 6.346 12.017 14.067 16.013 18.475 20.278 8 1.344 2.180 3.490 7.344 13.362 15.507 17.535 20.090 21.955 9 1.735 2.700 4.168 8.343 14.684 16.919 19.023 21.666 23.589 10 2.156 3.247 4.865 9.342 15.987 18.307 20.483 23.209 25.188 11 2.603 3.816 5.578 10.341 17.275 19.675 21.920 24.725 26.757 12 3.074 4.404 6.304 11.340 18.549 21.026 23.337 26.217 28.300 13 3.565 5.009 7.042 12.340 19.812 22.362 24.736 27.688 29.819 14 4.075 5.629 7.790 13.339 21.064 23.685 26.119 29.141 31.319 15 4.601 6.262 8.547 14.339 22.307 24.996 27.488 30.578 32.801 P, probability; df, degrees of freedom. *Most scientists assume that, when P < 0.05, a significant difference exists between the observed and the expected values in a chi-square test. Since 𝛘2 = 0.46 is less than 3.841, the null hypothesis fails to be rejected or in other words, the null hypothesis is accepted. In other words, since the value of 𝛘2 = 0.46 lies between P(0.1) and P(0.5), the probability of deviation occurring by chance lies between 10% - 50% and hence null hypothesis is accepted. So, we conclude that there is no significant difference between the observed and the expected ratio. Chi Square Analysis Problem solving for a monohybrid cross…… Using Binomial Equation to calculate Probability Binomial expansion helps solve complex genetics problems related to probability. It helps calculate all possibilities for a given set of two unordered events. If there are 2 events with alternate independent events having probabilities p and q, then in n number of trials, the probabilities of various combinations of events is given by : = (p + q)n where p + q = 1 e.g., the two categories can be Normal and Diseased, WT and Mutant, Tall and Short. Binomial probability that exactly x events belonging to a category occur in n trials = n = no. of trials x = no. of events in one category n – x = no. of events in another category p = Probability of occurrence of x events out of n trials q = Probability of occurrence of other events (n-x) out of n trials where n! = 1 x 2 x 3 x 4 ….n Probability calculation using Binomial Expression….. In a family of six children, what is the probability that atleast four will be girls? G = Girl B = Boy Probability of having a G = ½ Probability of having a B = ½ Event Binomial Expression Probability 4G, 2B 6!/(4! 2!) x (½)4 X (½)2 15/64 5G, 1B Try calculating yourself!! 6G, 0B Total prob. = 22/64 BIO207H5S Introductory Genetics Winter 2024 Lecture 3 Preeti Karwal, Ph.D. Chromosomal theory of Inheritance Early in the twentieth century, Walter Sutton and Theodor Boveri independently noted that the behavior of chromosomes during meiosis is identical to the behavior of genes during gamete formation described by Mendel. Similarities: 1. Genes and chromosomes exist in pairs, and 2. Members of a gene pair and members of a chromosome pair separate from each other during gamete formation. Based on these similarities, Sutton and Boveri independently proposed that genes are carried on chromosomes and formulated the chromosomal theory of inheritance. “The theory states that inherited traits are controlled by genes residing on chromosomes faithfully transmitted through gametes, maintaining genetic continuity from generation to generation.” Much later work by Thomas H. Morgan, Alfred H. Sturtevant and Calvin Bridges on sex linked inheritance and the discovery of nondisjunction in Drosophila established that the Sutton’s and Boveri’s theory was correct. Extensions of Mendelian Inheritance While alleles are transmitted from parent to offspring according to Mendelian principles, the observed phenotypic ratios may differ from those resulting from standard monohybrid, dihybrid, and trihybrid crosses. These exceptions could be because of following reasons: 1. Sometimes genes fail to display the clear-cut dominant/recessive relationship observed by Mendel. 2. More than one gene may influence the phenotype of a single characteristic. 3. Genes may be present on sex chromosomes instead of autosomes, so reciprocal cross may not give same outcomes. 4. Phenotypes may often be the combined result of both genetics and the environment. 5. Extranuclear inheritance, resulting from the expression of genes present in the DNA found in mitochondria and chloroplasts. Incomplete Dominance One Homozygous genotype, A1A1, produces red pigment, resulting in red flowers. Another homozygous genotype, A2A2, produces no pigment, resulting in white flowers. The phenotype of the heterozygote determines the type of dominance. When a cross between parents with contrasting traits generates offspring with an intermediate phenotype, this phenomenon is called incomplete dominance. For example, if plants such as four-o’clocks or snapdragons with red flowers are crossed with white-flowered plants, the offspring have pink flowers. Some red pigment is produced in the F1 intermediate pink-colored flowers. Therefore, neither red nor white flower color is dominant. This situation is known as incomplete, or partial, dominance. In complete dominance, the genotypic ratio (1:2:1) of the F2 generation is identical to that of Mendel’s monohybrid cross. However, because neither allele is dominant, the phenotypic ratio is identical to the genotypic ratio, unlike Mendelian monohybrid cross ratio. Plausible biochemical mechanism or molecular basis of incomplete dominance: Insufficient expression of the gene or gene dosage may result in incomplete dominance. For example, human biochemical disorder Tay–Sachs disease, in which homozygous recessive individuals are severely affected with a fatal lipid storage disorder, there is almost no activity of the enzyme hexosaminidase A in those with the disease. Heterozygotes, with only a single copy of the mutant gene, are phenotypically normal but express only about 50 percent of the enzyme activity found in homozygous normal individuals. Codominance In codominance, the phenotype of the heterozygote is not intermediate between the phenotypes of the homozygotes; rather, the heterozygote simultaneously expresses the phenotypes of both homozygotes. For example, MN blood types of humans. The MN blood-group locus encodes one of the types of antigens on the surface of red blood cells. At the MN locus, there are two alleles: the LM allele, which encodes the M antigen; and the LN allele, which encodes the N antigen. - Homozygotes with genotype LMLM express the M antigen on the surface of their red blood cells and have the M blood type. - Homozygotes with genotype LNLN express the N antigen and have the N blood type. - Heterozygotes with genotype LMLN exhibit codominance and express both the M and the N antigens; they have blood type MN. Again, the typical genotypic and phenotypic ratio for codominance is 1:2:1. TABLE 5.1 Differences between complete dominance, incomplete dominance, and codominance Type of Dominance Definition Complete dominance Phenotype of the heterozygote is the same as the phenotype of one of the homozygotes. TABLE 5.1 Differences between complete dominance, incomplete dominance, and codominance Incomplete dominance Phenotype of the heterozygote is intermediate (falls within the range) between the phenotypes of the two homozygotes. Codominance Phenotype of the heterozygote includes the phenotypes of both homozygotes. Multiple Alleles For some loci, more than two alleles are present within a group of organisms—the locus has multiple alleles. Although there may be more than two alleles present within a group of organisms, the genotype of each individual diploid organism still consists of only two alleles. With multiple alleles, a greater variety of genotypes and phenotypes are possible. For example, fur or coat colour in rabbits. Order of Dominance: C > Cch > Ch > c C – provides full coat colour Cch – partial defect in pigmentation Ch – temperature sensitive conditional mutant allele that provides pigmentation only in the colder parts of body c – prevents pigmentation Himalayan Rabbit ABO blood types (Multiple alleles showing codominance) 5.8 Expression of the ABO antigens depends on alleles at the H locus (no balloons) Lethal Alleles An allele that causes the death of an organism during an early stage of development often before birth so that certain genotypes can never be observed. Lethal alleles may be recessive, dominant, or conditional in nature. Recessive lethal mutation: Mutations resulting in the synthesis of a gene product that is nonfunctional can often be tolerated in the heterozygous state. So one wild-type allele may be sufficient to produce enough of the essential product to allow survival. However, the homozygous recessive individuals do not survive, so they cause lethality in recessive form. These mutations affect genes that are essential to life – ‘Essential Genes’. Dominant lethal mutation: Mutations that cause ectopic production or overexpression of a toxic product. This allele is lethal both in the homozygote and the heterozygote. Dominant lethal alleles are very rare because of negative selection by nature and also because they can only be transmitted if the lethality phenotype occurs after reproductive age. The phenotypic ratio of 2 : 1 produced in a cross results from a recessive lethal allele. Conditional lethal allele e.g., the disorder called Huntington’s disease in humans is due to such an allele (H), where the onset of the disease and eventual lethality in heterozygotes (Hh) is delayed. Age of onset : 30 – 50 Temperature sensitive alleles that kill under only a particular temperature. BIO207H5S Introductory Genetics Winter 2024 Lecture 4 Preeti Karwal, Ph.D. Gene Interactions Gene-gene interactions take place when genes at multiple loci determine a single phenotype. This does not mean, however, that two or more genes, or their products, necessarily interact physically with one another. Rather, the cellular function of numerous gene products contributes to a common process or pathway and so the development of a common phenotype. Types: 1. 2. 3. 4. Epistasis Complementary gene action Duplicate gene action Supplementary/Additive genes Since inheritance patterns based on the gene-gene interactions listed above will involve at least two genes governing a trait, so they generally manifest as alterations in dihybrid cross ratio (9:3:3:1). Epistasis Epistasis refers to a phenomenon where the expression of one gene or gene pair masks or modifies the expression of another gene or gene pair. e.g. a mutation in wingless gene (epistatic), responsible for the wing formation in Drosophila masks the mutations in crossveinless gene (hypostatic) that controls wing characteristics. e.g., inheritance of coat color in mice True breeding black mice X True breeding albino mice aaCC AAcc F1: All Agouti Mice AaCc F2: Agouti : Black: Albino 9:3:4 Molecular Basis: Normal wild-type coat color is agouti. Agouti (A allele) is dominant to black (a allele) hair. AA – agouti , aa – black Expression of A gene (A/a) require C gene product in dominant form. F2 results: Genotype Ratio Phenotype A_C_ 9 Agouti A_cc 3 Albino aaC_ 3 Black aacc 1 Albino Phenotypic ratio - 9 : 3 : 4 Conclusion: Comparing both albino mice, it is clear that mutation in C gene (cc genotype) masks the mutation in A gene (aa genotype), so gene C is epistatic to A gene in recessive form → RECESSIVE EPISTASIS. Inheritance of flower colour: Molecular basis: Gene A Precursor → Gene B Pigment 1 → (Red) Pigment 2 (Blue) F2 results: Genotype Ratio Phenotype A_B_ 9 Blue A_bb 3 Red aaB_ 3 White aabb 1 White Phenotypic ratio - 9 : 3 : 4 Conclusion: The inheritance of flower colour shows recessive epistasis. Gene A is epistatic to gene B in recessive form. Dominant epistasis vs Recessive epistasis: Dominant epistasis in fruit colour in summer squash: W allele in dominant form gives white colour regardless of genotype at second locus Y. In the absence of W allele in dominant form (or ww background), Y allele in dominant form gives yellow colour, while yy gives green colour. Genotype Ratio Phenotype W_Y_ 9 White W_yy 3 White wwY_ 3 Yellow wwyy 1 Green Phenotypic ratio - 12 : 3 : 1 Conclusion: The inheritance of summer squash colour shows dominant epistasis. Gene W is epistatic to gene Y in dominant form. Occasionally, a ratio of 13:3 is also observed in certain cases of epistasis. Complementary gene interaction (Duplicate Recessive Epistasis) It is demonstrated in a cross between two true-breeding strains of white-flowered sweet peas. Phenotypic ratio – 9 : 7 Conclusion: The presence of at least one dominant allele of each of two gene pairs, A and B is essential for flowers to be purple. So, the two genes complement each other. All other genotype combinations yield white flowers because the homozygous condition of either recessive allele masks the expression of the dominant allele at the other locus. Complementation Test This test helps to determine whether two mutations associated with a specific phenotype represent two different forms of the same gene (alleles) or are variations of two different genes. No Complementation Recessive mutations, a and b occur at the same locus (Allelic mutations) Complementation Recessive mutations, a and b occur at the different locus (Non-allelic mutations) When two mutations occur in different genes, they are said to be complementary, because the heterozygote condition rescues the function otherwise lost in the homozygous recessive state. The alternative name cis-trans test describes the two central components of the test. The terms cis and trans refer to the relationship of the two mutations, with cis used to describe mutations occurring on the same chromosome and trans used to describe mutations occurring on different chromosomes. Duplicate Gene Action It is demonstrated in a cross for kernel colour in wheat: F2 results: Genotype Ratio Phenotype A_B_ 9 Coloured A_bb 3 Coloured aaB_ 3 Coloured aabb 1 No colour Gene A/Gene B Precursor → Phenotypic ratio – 15 : 1 Conclusion: The biosynthesis of red pigment near the surface of wheat seeds involves many genes, two of which we label A and B. Normal, red colouration of the wheat seeds is still maintained if function of either of these genes is lost in a homozygous mutant. It can be because the two loci’s gene products have the same (redundant) functions within the same biological pathway and so can replace each other. Pigment Supplimentary genes (Additive/Cumulative effect) Novel phenotypes often result from the interactions of two genes, as in the case of the comb shape in chickens. Two genes affect the comb characters such as shape and color in chicken. Rose gene: R or r Pea gene: P or p Rose gene, if present in R_ (RR or Rr) will produce a "rose type" comb– but ONLY if Pea gene is present in pp condition. Pea gene, if present in P_ (PP or Pp) will produce a "pea type" comb-but ONLY if Rose gene is present in rr condition. If both alleles are present in double recessive condition, (rrpp), the wild type, single comb results. Walnut comb, a novel phenotype, is produced when the genotype has at least one dominant allele of each gene (R_P_). RRpp X rrPP Rose Pea F1 : RrPp Walnut Selfing F2 results: Variable Expressivity The gene expression and the resultant phenotype are often modified through the interaction between an individual’s particular genotype and the external environment. Because of this interaction, there could be a variability in the expressivity of a given gene in the individual’s phenotype. Expressivity reflects the range of expression of the mutant genotype. When the degree of phenotypic expression in dominant or homozygous recessive form varies from one individual to another, the gene is said to have Variable Expressivity. e.g. Flies homozygous for the recessive mutant eyeless gene yield phenotypes that range from the presence of normal eyes to a partial reduction in size to the complete absence of one or both eyes. Incomplete Penetrance The percentage of individuals who show at least some degree of expression of a mutant genotype defines the penetrance of the mutation. e.g., If 15 percent of mutant flies show the wild-type appearance, the mutant gene is said to have a penetrance of 85 percent. It is incompletely penetrant. Incomplete Penetrance refers to a condition, when a trait is not manifested in the detectable phenotype despite the presence of gene. It is an extreme case of variable expressivity. e.g., Polydactyly exhibits both incomplete penetrance and variable expressivity. Presence/Absence of extra digits : Penetrance/Incomplete Penetrance Presence of extra digits in all limbs or only hands or only feet or only one limb - Variable Expressivity Incomplete Penetrance vs Variable Expressivity Some of the factors responsible for incomplete penetrance and variable expressivity may be: ▪ ▪ ▪ ▪ ▪ Effect of external environment e.g. role of diet on cholesterol level in familial hypercholesterolemia. Mosaicism for X linked characters due to X inactivation (Barr body) in females Gene-Gene interactions e.g., epistasis Effect of sex hormones e.g., Sex influenced traits Late onset diseases may show age related penetrance. TABLE 5.2 Modified dihybrid phenotypic ratios due to gene interaction Genotype A_ B_ A_ bb aa B_ aa bb 9:3:3:1 9 3 3 1 9:3:4 9 3 Ratio* 12 : 3 : 1 9:7 9 9:6:1 9 Seed shape and seed color in peas Recessive epistasis Coat color in Labrador retrievers 3 Dominant epistasis Color in squash Duplicate recessive epistasis Albinism in snails 1 Duplicate interaction — 1 Duplicate dominant epistasis — Dominant and recessive epistasis — 1 TABLE 5.2 Modified dihybrid phenotypic ratios due to gene interaction 7 6 15 : 1 13 : 3 None 4 12 15 13 3 Type of Interaction Example Discussed in Chapter *Each ratio is produced by a dihybrid cross (Aa Bb × Aa Bb). Shaded bars represent combinations of genotypes that give the same phenotype. Pleiotropy The term pleiotropy is derived from the Greek words pleio, which means "many," and tropic, which means "affecting.“ Genes that affect multiple, apparently unrelated, phenotypes are called pleiotropic genes. A mutation in a pleiotropic gene may cause a disease with a wide range of symptoms. For example, Phenylketonuria (PKU), a metabolic disorder in humans caused by a deficiency of the enzyme phenylalanine hydroxylase, which is necessary to convert the essential amino acid phenylalanine to tyrosine. A defect in the single gene that codes for this enzyme results in multiple phenotypes, including mental retardation, eczema, and pigment defects that make affected individuals lighter skinned. Pleiotropy https://www.nature.com/scitable /topicpage/pleiotropy-one-genecan-affect-multiple-traits-569/ Pleiotropy should not be confused with: Gene gene interactions like epistasis, where an interaction between two or more genes governs a trait or Polygenic traits, in which multiple genes converge to result in a single phenotype e.g., skin color phenotype. Gene 1 Gene 2 Gene 3 Trait BIO207H5S Introductory Genetics Winter 2024 Lecture 5 Preeti Karwal, Ph.D. Topics covered: 1. Sex determination 2. Sex-linked inheritance XX-XY System In humans and all other placental mammals, females inherit an X chromosome from each parent, whereas males always inherit their X chromosome from their mother and their Y chromosome from their father. Consequently, all of the somatic cells in human females contain two X chromosomes, and all of the somatic cells in human males (also called hemizygous) contain one X and one Y chromosome. Males (heterogametic sex) produce X and Y gametes, and females (homogametic sex) produce only X gametes. In this system, referred to as the XX-XY system, the maleness is determined by sperm cells that carry the Y chromosome. Y chromosome has ~75 genes compared to 900–1400 genes on the X. Presence of Y chromosome determines maleness During early development, every human embryo is potentially hermaphroditic for the first few weeks of gestation. By the fifth week of gestation, gonadal tissues arise as a pair of gonadal (genital) ridges associated with each embryonic kidney. At this stage, its gonadal phenotype is sexually indifferent or neutral, so the male or female reproductive structures cannot be distinguished. The cortex of this neutral gonadal tissue is capable of developing into an ovary, while the medulla may develop into a testis. In addition, two sets of undifferentiated ducts called the Wolffian and Müllerian ducts exist in each embryo. Wolffian ducts differentiate into other organs of the male reproductive tract, while Müllerian ducts differentiate into structures of the female reproductive tract. If cells of the gonads have an XY constitution, the development of the medulla into a testis is initiated around the seventh week. However, in the absence of the Y chromosome, no male development occurs, the cortex of the gonadal tissue subsequently forms ovarian tissue. Testis determining factor, a transcription factor causes the undifferentiated gonadal tissue of the embryo to form testes. Pseudoautosomal Inheritance Present on both ends of the X and Y chromosome are so-called pseudoautosomal regions (PARs) that share homology and synapse and recombine with each other during meiosis. The presence of such a pairing region is critical to segregation of the X and Y chromosomes during male gametogenesis. The remainder of the chromosome, about 95 percent of it, does not synapse or recombine with the X chromosome. Since this region is located on sex chromosomes but has two copies in both males and females like autosomal regions, it is called pseudoautosomal region. One of the X chromosomes undergoes inactivation in the somatic cells of females for dosage compensation (explained on next slide) but the PARs lying on X chromosome escape this inactivation and continue to be expressed even from the Barr body. Less or more than two copies of PARs result in genetic consequences in case of sex chromosome aneuploidies like Turner’s Syndrome (XO) – one copy of PAR and Klinefelter’s Syndrome (XXY) – three copies of PAR, respectively. Dosage Compensation Murray Barr and Ewart Bertram demonstrated a genetic mechanism in mammals that compensates for X chromosome dosage disparities. They observed a darkly staining body in the interphase cells of female cats that was absent in similar cells of males. In humans, this body can be easily demonstrated in all somatic female cells e.g., those derived from the buccal mucosa (cheek cells) or in fibroblasts (undifferentiated connective tissue cells), but not in similar male cells. This highly condensed structure called Barr body or Sex chromatin lies against the nuclear membrane and is comprised of heterochromatic inactivated X chromosome. By inactivating one of the two X chromosomes in the cells of females, the dosage of genetic information that can be expressed in males and females becomes equivalent. Single active principle: Regardless of how many X chromosomes a somatic cell possesses, all but one of them appear to be inactivated and can be seen as Barr bodies. Therefore, the number of Barr bodies follows an N - 1 rule, where N is the total number of X chromosomes present. Female Male Lyon Hypothesis Lyon postulated that the inactivation of X chromosomes occurs randomly in somatic cells at a point early in embryonic development, most likely sometime during the blastocyst 4- to 8-cell stage of development. Once inactivation has occurred, all descendant cells have the same X chromosome inactivated as their initial progenitor cell. This explanation, which has come to be called the Lyon hypothesis, was based on observations of mosaic patterns that occur in the black and yellow-orange patches of female tortoise shell and calico cats. Such X-linked coat color patterns do not occur in male cats because all their cells contain the single maternal X chromosome and are therefore hemizygous for only one X-linked coat-color allele. Female calico cat In this figure, XO confers orange colour Xo confers black colour https://www.bio.miami.edu/dana/dox/calico.html In this figure, XB confers orange colour Xb confers black colour https://www.khanacademy.org/science/biology/classical-genetics/sexlinkage-non-nuclear-chromosomal-mutations/a/x-inactivation Mechanism of Dosage Compensation A region of the mammalian X chromosome located in the p arm in humans, is called the X-inactivation center (Xic), and its genetic expression occurs only on the X chromosome that is inactivated. One of these, X-inactive specific transcript (Xist), is known to be a critical gene for X-inactivation. A long non-coding RNA that is transcribed from the XIST gene spreads over and coats the X chromosome bearing the gene that produced it. https://link.springer.com/article/10.1007/s00439-011-1027-4 XX-XO System In XX-XO system found in crickets, grasshoppers, and some other insects, sperm cells that lack an X chromosome (referred to as O) determine maleness. Females carry two X chromosomes (XX) and only produce gametes with X chromosomes – Homogametic sex X X XX Males carry only one X chromosome (XO) and produce some gametes with X chromosomes and some gametes with no sex chromosomes at all – Heterogametic sex X O XO ZZ-ZW System In the ZZ-ZW sex determination system found in birds, snakes, some amphibians, fish and insects, females carry the mismatched chromosome pair (ZW) and males carry the identical pair (ZZ). Females produce gametes with Z chromosome or with W chromosome – Heterogametic sex Males carry two X chromosomes (ZZ) and only produce gametes with Z chromosomes – Homogametic sex Haplodiploidy It is found in insects in the order Hymenoptera including bees, ants and wasps. Sex is based on the number of chromosomes found per cell. There are no sex chromosomes. Males develop from unfertilized eggs – Haploid Females develop from fertilized eggs - Diploid TABLE 4.1 Some common sex-determining systems System Mechanism XX-XO Females XX XX-XY Females XX Heterogametic Sex Organisms Males X Male Some grasshoppers and other insects Males XY Male Many insects, fishes, amphibians, reptiles; mammals, including humans Table 4.1 Some common sexdetermining Males ZZsystems Female ZZ-ZW Females ZW Butterflies, birds; some reptiles and amphibians Genic sex determination No distinct sex chromosomes Sex determined by genes on undifferentiated chromosomes Varies Some plants, fungi, protozoans, and fishes Environmental sex determination Sex determined by environmental factors None Some invertebrates, turtles, alligators Sex determination in Drosophila melanogaster Y chromosome is not involved in sex determination in Drosophila melanogaster. Instead, Bridges proposed that the X chromosomes and autosomes together play a critical role in sex determination. The Chromosomes of Drosophila melanogaster The diploid chromosome complements of a male and a female Drosophila melanogaster. Sex-linkage One of the first cases of sex-linkage was documented by Thomas H. Morgan around 1920 during his studies of the white mutation in the eyes of Drosophila. Morgan got curious after observing a white eyed male fly in contrast to red eyed flies in his Drosophila lab cultures. The normal wildtype red eye color is dominant to white. Drawings (A) of a male and a female fruit fly, Drosophila melanogaster. The photographs (B) show the eyes of a wildtype red-eyed male and a mutant white-eyed male. Illustrations © Carolina Biological Supply Company. Used with permission. Photographs courtesy of E. R. Lozovsky. F1 : Red eyed female : Red eyed male :: 1:1 F2 : Red eyed female : Red eyed male : white eyed male :: 2:1:1 F1: Red eyed female : White eyed male :: 1:1 F2: Red eyed female : Red eyed male : white eyed male : white eyed female : 1:1:1:1 Morgan’s Hypothesis: White eye gene is a X-linked characteristic. Morgan was able to correlate these observations with the differences found in the sex-chromosome composition between male and female Drosophila. He hypothesized that the recessive allele for white eyes is found on the X chromosome, but its corresponding locus is absent from the Y chromosome. Females thus have two available gene sites, one on each X chromosome, whereas males have only one available gene site on their single X chromosome. One result of X-linkage is the crisscross pattern of inheritance, whereby phenotypic traits controlled by recessive X-linked genes are passed from homozygous mothers to all sons. Reciprocal Cross The definitive method to test for sex-linkage is conducting reciprocal crosses. Reciprocal crosses - It means to cross a male and a female that have different phenotypes, and then conduct a second set of crosses, in which the phenotypes are reversed relative to the sex of the parents in the first cross. e.g., a female of a certain genotype A is first crossed with a male of genotype B. Then, in the reciprocal cross, a female of genotype B is crossed with a male of genotype A. If the gene or trait is autosomal, it will not matter which parent has which phenotype; all of the offspring will show the dominant phenotype for F1 for any monohybrid cross. P1 Tall X P2 Dwarf F1 All Tall P1 Tall X P2 Dwarf F1 All Tall If the gene or trait is sex-linked, the offspring in F1 or F2 obtained in a reciprocal cross will be different. P1 Red eyed X P2 White eyed F1 All red eyed P1 X P2 White eyed Red eyed F1 Red eyed : White eyed 1 : 1 Non-disjunction as a proof for Chromosomal theory of inheritance Flies with unexpected phenotypes continued to appear in Morgan’s crosses at a frequency more than the expected mutation rate. Red eyed males X White eyed females F1 results: 95% - red eyed females and white eyed males (expected) 2.5% of male offspring were red eyed (unexpected) 2.5% of female offspring were white eyed (unexpected) Calvin Bridges, one of Morgan’s students investigated the genetic basis of these exceptions. He hypothesized that these white eyed females and red eyed males result from non-disjunction of chromosomes during gamete formation. To confirm this hypothesis, Bridges observed these flies with unusual phenotypes cytogenetically (under microscope) and indeed found that the appearance of rare phenotypes is associated with particular chromosomes. Therefore, this observation gave definite proof for chromosomal theory of inheritance.

BIO207_TermTest 1.pdf

Related

Transcript

Tags

Upgrade to continue