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AGRY 32000 Exam 2 Study Guide

by: Gayatri

AGRY 32000 Exam 2 Study Guide AGRY 32000

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Includes all the learning objectives from Lectures 20-33
Dr. Brenda Owens
Study Guide
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This 7 page Study Guide was uploaded by Gayatri on Sunday April 10, 2016. The Study Guide belongs to AGRY 32000 at Purdue University taught by Dr. Brenda Owens in Spring 2016. Since its upload, it has received 224 views. For similar materials see Genetics in Agriculture and Forestry at Purdue University.

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Date Created: 04/10/16
AGRY 32000 Exam 2 Study Guide Lecture 20: Mendelian Genetics 1 • To understand the importance of Mendel’s work in genetics o Mendel - Austrian monk that worked on pea-crosses to find out how inheritance worked o His work was not initially appreciated because it disproved the blending theory, which was the popular belief in the 19 century § Blending theory meant that parental traits were mixed together and not conserved in offspring • To understand the elegant and scientific design of Mendel’s experiments o Used pea plants because they were easy to glow and to pollinate in a controlled way § Self fertilization as well as cross fertilization was possible o Pea plants he used were all from pure-breeding lines (self pollinated plants that produced the same traits in every generation), and used various traits (phenotypes) to conduct proper genetic analysis § Round or wrinkled seeds § Yellow or green seeds § Purple or white flowers o Experiment 1: Round x wrinkled cross à F1: all round à F2: 3 round 1 wrinkled § One phenotype disappeared in F1, reappeared in F2 with 3:1 ratio § Round phenotype was dominant over wrinkled o Experiment 2: RR x Rr à all round (1:1 RR/Rr) o Experiment 3: Rr x rr à half round half wrinkled (1:1 Rr/rr) • To understand the experimental observations that led to Mendel’s Law of Equal Segregation o Based on Experiments 1-3, Mendel discovered the following: § Equal separation of alleles occurred during gamete formation § A gamete contained only one form of the alleles § At fertilization, gamete fusion was random • To understand how the following concepts were discovered and their meaning: o Genotype: pair of alleles carried by individuals (RR, Rr, rr) o Phenotype: appearance of genotypes (round, wrinkled) o Alleles: alternative forms of a gene (R, r) o Dominant: a phenotype that appears in F1 of two pure breeding lines o Recessive: a phenotype that disappears in F1 and reappears in F2 of heterozygous cross Lecture 21:Mendelian Genetics - Chromosomal Basis of Single Gene Inheritance • To understand the structural components of chromosomes that function during segregation th o Chromosomes and cell division were observed in the late 19 century by 3 other scientists, Mendel was right! o DNA is a double helix o Homologous chromosomes are two members of a chromosome pair in a diploid nucleus, one is from mom and one from dad o Ploidy: # of homologous sets of chromosomes in a cell § Haploid (n): organism with one complete genome § Diploid (2n): organism with two complete genomes o Parts of a chromosome § Chromatid: one of two that are attached together to form a chromosome (get separated during anaphase of mitosis) § Kinetochore: where spindle fibers attach § Kinetochore microtubules: push and pull to line up chromosomes at metaphase • To understand the mechanical similarities and differences between mitosis and meiosis o Mitosis: somatic (body cells) divide to increase their number § 1 diploid cell (2n) à 2 diploid cells (2n) § 1 haploid cell (n) à 2 haploid cells (n) § Steps of mitosis in a diploid • Interphase (2n) • Prophase (4n) • Metaphase (4n) • Anaphase (4n) • Telophase (2n) • Daughter cells (2n) o Meiosis: specialized diploid cells (germ cells/meiocytes) divide to produce sex cells (sperm and egg) § 1 diploid cell (2n) à 4 haploid cells (n) § Steps of meosis in a diploid (same as meiosis x 2 but without replication twice) • Interphase (2n) • Prophase (4n) • Metaphase (4n) • Anaphase (4n) • Telophase (2n) • Daughter cells (2n) Following one daughter cell… • Prophase II (2n) • Metaphase II (2n) • Anaphase II (2n) • Telophase II (1n) • Daughter cells (1n) • To be able to follow the fate of parental alleles during meiosis o Members of each pairs of alleles in a gene separate equally when gametes are produced • To understand how multiple alleles can be identified and used in genetic experiments o Discovery of genes can be done through observing segregation ratios § Flower color: +/+ (red) x alb/alb (white) à F1: +/alb à ¼ +/+, ½ +/alb, ¼ alb/alb • Red is dominant over white § Fruit fly wing: +/+ (long) x +/sh (short wing) à F1 ½ +/+, ½ +/sh • Short is dominant over long § To determine if an individual is homozygous or heterozygous, cross with rr • If all are red, then it is RR • If some are red and some white, then it is Rr Lectures 22 & 23:Sex Determination, Sex-Linked Inheritance, and Pedigree Analysis • To know differences between sex chromosomes and autosomes o Sex chromosomes: chromosomes that are connected to sex determination § 1 from each parent o Autosomal chromosomes: chromosomes that are unrelated to sex determination § 22 from each parent • To understand how the XY-based sex determination works in human o Sex is determined by number of X or presence of Y § Females have X/X (Abnormals can have X/null) § Males have X/Y (Abnormals can have X/X/Y or X/X/X/Y) • Y encodes testis determining factor (SRY) which is a dominant activator in maleness pathway • To understand the sex-linked inheritance that was first discovered in the fruit fly – Drosophila (by Morgan) o Eye color was controlled by a single X-linked gene, Y has no influence because it does not carry gene § w+ = wild type, dominant (red) § w = mutant, recessive (white) o Cross #1: Female w+/w+ (wild-type red fly) x Male w/w (mutant white fly) à F1: all red à F2: 3 red : 1 white § F1 are all red because only X with w+ exists § F2 has two red females and one white male and one red male because w+ and w exist o Cross #2: Female w/w (mutant white fly) x Male w+/w+ (wild-type red fly) à F1: 1 red : 1 white à F2: 2 red : 2 white § F1 has half and half split because X with w+ and w exists § F2 has half and half split again, with white male and female and red male and female because of more w than w+ • To be able to deduce a type of allele based on human pedigrees o 4 types of disorders: 1. Autosomal recessive: progeny with unaffected parents can have it, males and females are equally affected 2. Autosomal dominant: disorder is seen in every generation, males and females are equally affected 3. X-linked recessive: sons of affected males are not affected, all daughters of affected males are carriers, more males affected 4. X-linked dominant: daughters of all affected males are affected, affected females pass it on to half of their offspring • To be able to calculate simple risk factors for transmitting disease-causing alleles based on human pedigrees o Mutiplication of probabilities of each parent (and grandparents)= probability of offspring Lecture 24: Independent Assortment • To know the genotypes and phenotypes of gametes that are produced by parental, F1, and F2 individuals in a dihybrid cross o Mendel used round (R) and wrinkled (r) seeds along with yellow (Y) and green (y) seeds o RRyy x rrYY à F1: RrYy (round yellow) à F2: 9/16 round yellow, 3/16 round green and wrinkled yellow, and 1/16 wrinkled green • To understand the chromosomal basis of independent assortment (Mendel’s 2 law) nd o Different allele pairs assort independently during gamete formation • To be able to predict the expected phenotypic and genotypic ratios of a dihybrid cross and understand their probability of occurrence o For 2 pairs of alleles use Punnett square o For more than 2, use branch diagram o Product rule: the probability of independent events occurring together is the product of their individual probabilities o Sum rule: the probability of either of two naturally exclusive events is the sum of their probability • To be able to use the chi-square test to statistically evaluate phenotypic or genotypic ratios o Chi-square test: determines the probability of obtaining observed proportions by chance under a specific hypothesis § Sum of ((observed-expected) /expected) § Df = # of categories – 1 § P < value, interpreted as “less than value % of your hypothesis is true” Lecture 25: Non-nuclear Inheritance • To know about the existence of organelles and genomes besides those within the nucleus o Organelles exist in the cytoplasm of cells that carry their own genomes, and these genomes are called non-nuclear genomes. Two types of organelles exist: § Mitochondria: power house of cell, makes ATP (17 kb in humans) § Chloroplast: unique to plants, used in photosynthesis (121 kb in liverwort) o Organelle DNA is single stranded and circular (similar to plasmid) • To understand the concept of cytoplasmic segregation and non nuclear inheritance o Non-nuclear inheritance: displays a special type of inheritance known as uniparental inheritance § Progeny inherit organelle genes exclusively from the mother (materal inheritance) § This is because in organelle genes, only the mother’s eggs contribute to the bulk of the cytoplasm, and father’s sperm do not (Independent of Mendelian patterns of inheritance) o Cytoplasmic segregation: a population of distinct organelle chromosomes often show segregation of two types into the daughter cells at cell division § E.g. variegated leaves of four o’clock plant caused by mutation in chloroplast DNA § A à green, a à white § Depending on how cytoplasm is divided up when two daughter cells are made, the phenotype (expression) is affected § A cell could have all A’s all a’s, or even/uneven distribution of A and a (Segregation refers to a cell having either A or a only) • To recognize the non-mendelian patterns of inheritance for traits controlled by non nuclear genes o In non-nuclear inheritance, all kids acquire genes from mother only o All F1 kids, regardless of sex, are affected because F mom is o Only females in F1 pass it on to their kids, F1 males do not o Only F2 kids that have F1 moms are affected o F1 dads do not pass it on to their kids Lectures 26 – 28: Genetic Linkage, Recombination, and Mapping • To understand the concept of linkage, how it was discovered and how to test for it o Linkage of loci was discovered using Drosophila as a model organism o Recombinant genotypes were made from crossing over of chromosomes § Occurred between two non-sister chromatids when pairs of homologous chromosomes were lined up during metaphase o In diploids, recombination is detected with a testcross of heterozygote and recessive tester: cross F1 with aabb and see if recombinant phenotypes show up o Independent assortment produces 50% recombinants, written as separated by semicolon (;) o Linked genes produce less than 50% recombinants § Cis linked = two dominant or wild type alleles on same homolog (AB/ab or ++/ab) § Trans linked = two dominant or wild type alleles on different homologs (aB/Ab or +b/a+) § Unknown linkage is given by dot (.) • To understand how recombination frequency is measured and its use in the generation of genetic maps o Recombinants are the products of meiosis with allele combinations different from those of the haploid cells formed in meiotic diploid o Recombination frequency between two loci (RF) = recombinant chromosomes/total chromosomes x 100% (units are centimorgan (cM) or map unit (m.u.) o In diploids, recombination is detected with a testcross of heterozygote and recessive tester • To understand the concept of genetic mapping and its importance o Longer regions have more crossovers and more recombos (higher RF) • To know how to conduct and interpret a 3-point test cross o 3 point testcross is used to determine linkage, gene order, and map distance o Double crossovers can occur between two linked genes, need to be counted twice (usually the smallest two numbers) § To find double crossover chance, multiply individual cross over percentages o Interference refers to one crossover inhibiting crossover in an adjacent region, calculated with the following formula: § I = 1 – (observed frequency/expected frequency) Lecture 29: Gene Interaction • To understand the concept of multiple alleles and how to determine the dominance relationship between them o Interactions between alleles of a single gene can be dominant or recessive, 5 types of interactions exist: § Complete dominance: haplosufficient – one dose is enough for wild type • Null mutation = one that results in complete absence of function for the gene • In this case, mutations of haplosufficient genes are recessive § Complete recessiveness: haplosufficient – one dose is enough for mutant § Incomplete dominance: heterozygote shows intermediate phenotype • E.g. RR = Red, rr = white, Rr = pink § Codominance: distinct expression of two phenotypes in heterozygote (NOT intermediate mix of the two) • E.g. blood typing; I /I or I /i = A type, I /I or I /i = B type, I /I = AB type, i/i = O type § Recessive lethal alleles: recessive allele causes death of organism • E.g. Mouse coat; A/A = brown, A /A = yellow (yellow heterozygote is dominant over brown) Y Y • A / A = yellow, but does not survive (yellow homozygote is lethal), cross outcome of yellow heterozygote self cross is 2:1 yellow to brown (since one dies) • To be yellow, one A is needed (dominant), but to be lethal two A ’s are needed (recessive) • Pleotropic allele = an allele that influences more than one trait • To understand how biosynthetic pathways were discovered in Neurospora o Fungus, or bread mold: mutant needs requires a nutrient to be able to grow, wild type can make it itself o Three different mutations associated with the nutrient pathway (ability to produce arginine), each was a single gene mutation § Arg1, Arg2, and Arg 3 cannot make arginine, need it as a supplement to grow on media § Arg1 grows w/ Ornithine (1 step is made up for with it) § Arg2 grows w/ Citrulline (can do 1 step but not 2 step, Citru. makes up for it) § Arg3 grows only w/ Arginine (Can’t do 3 step, needs Arg) (Pathway: Or. à(enzyme) à Citrul. à (enzyme) à Arg.) Lecture 30: Genetic Interaction II • To know how to determine if a common phenotype is caused by a mutation in the same gene or mutations in different genes o Done with a complementation test: a genetic test to determine if two mutants with a similar phenotype have a mutation in the same gene o Can only be applied to recessive mutations • To learn how to perform complementation test o In a haploid organism (fungus), the complementation test can be performed by haploid cell fusion § Fusion results in a heterokaryon: the haploid nuclei from the different strains occupy one cell § E.g. Arg1 and Agr 2 fusion à in absence of arginine, growth occurs (if no growth, the mutation is from the same gene) o In diploid organisms, cross two mutants together: a/a;B/b X A/A;b/b/ è F1 with A/a;B/b § Example 2: testing to see if three alleles are on 1, 2, or 3 genes • Wild type: blue, dominant • mutant I: white $, recessive • mutant II: white £, recessive • mutant III: white ¥, recessive • Results: white $ X white £ à F1, all white means that they’re both on the same gene • § Self F1 to get F2 with the following results: • A/-;B/- (9/16) à functional • A/-;b/b (3/16) à nonfunctional • a/a;B/- (3/16) à nonfunctional • a/a;B/- (1/16) à nonfunctional (9:7 ratio) • Results of complementation test o A 9:7 F2 ratio suggests interacting genes in the same pathway; o Absence of either gene function leads to absence of the end product of the pathway, meaning the mutation is of the same gene Lecture 31: Genetic Interaction III • To understand the concept of epistasis o Epistasis occurs when the double mutant shows one phenotype but not the other § E.g. flower inheritance; • Wild type = blue (w+/w+;m+/m+) • Mutant 1 = white (w/w;m+/m+) • Mutant 2 = pink (w+/w+;m/m) • w+/w;m/m+ self cross gives the following results o (w+/-;m+/-) à blue (9) o (w+/-;m/m) à pink (3) o (w/w;m+/-) à white (3) AND (w/w;m/m)à white (1) (white is epistatic to pink, pink is hypostatic) • To know how epistasis affects the inheritance of phenotypes o Dominant epistasis: dominant phenotype overrides all others, model: § Precursor à W enzyme à colorless OR precursor à w enzyme à light pink à d enzyme à red § Gives 12:3:1 ratio o Model of recessive epistasis: § Precursor (colorless) à w+ enzyme à pink à m+ enzyme à blue § Gives 9:3:4 ratio • Duplicate gene interaction: two genes are not linked and have the same function o Phenotypic ratio is 15:1 • To understand the concept of pentrance and expressivity of mutant phenotypes o Pentrance = the fraction of individuals of a defined genotype that express the predicted phenotype § E.g. 100 people carry dominant mutation, 97 affected à 97% pentrance o Expressivity = degree of phenotypic expression in individuals that are genotypically identical at a locus § E.g. piebald spotting in beagles § Caused by environment-gene interaction or other genes including QTLs, quantitative trait loci Lectures 32-33: Large scale chromosomal changes • To master basic terms of chromosome variants and know the types of chromosome # variants in different organisms o Euploids: organisms with multiple copies of the basic genome o Monoploid: (n), member of a diploid species that has one chromosome (NOT same as haploid) o Polyploid: organism that has more than two sets of chromosomes § Diploid (2n), Triploid (3n), Tetraploid (4n), etc § Autopolyploid: multiple chromosome sets from same species § Allopolyploid: chromosome sets from multiple species • To understand how polyploids come to be and effect of chromosome variants on meiosis o Triploids (3n) are usually autopolyploids formed from tetraploid (4n) X diploid (2n) § Sterile or highly infertile because it has incomplete sets of chromosomes, intermediate between haploid and diploid # à aneuploid gametes) o Tetraploids (4n) have 4 chromosome sets from the same species § Normal occurrence makes 2 diploid cells from mitosis § Presence of colchicine leads to one tetraploid (4n = 8) because microtubule polymerization and therefore separation into two cells after anaphase is disrupted § Meiosis in a tetraploid can make functional gametes, three possibilities • Two bivalents • One quadrivalent • One trivalent and one univalent o Allopolyploids = hybrids of two species that contain more than one genomes § The two chromosomes are too varied and cannot pair, resulting in sterile hybrid (monoploid-like) § To deal with this, spontaneous doubling occurs, which allows for pairing between homologs of the same genome (2n X 2n à 4n = 36, even though only 2 and 2 pair) • E.g. hexaploid wheat (6n = 42) with A, B, D diploid genomes o Aneuploid = individual whose chromosome number differs from the wild type by part of a chromosome set


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