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Exam 3 Notes!

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by: Rachael Couch

Exam 3 Notes! Biol 2311

Rachael Couch
GPA 3.9

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I am a TA for this class this year. Study my notes and you WILL do well!
Introduction to Biology
John Burr
Study Guide
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This 14 page Study Guide was uploaded by Rachael Couch on Thursday November 12, 2015. The Study Guide belongs to Biol 2311 at University of Texas at Dallas taught by John Burr in Fall 2014. Since its upload, it has received 45 views.


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Date Created: 11/12/15
Chapter 13: Genetics Evolution  The fossil record during Darwin’s time showed evidence for ancient life but no one could explain how this evolution occurred  Lamark was the first to try to explain evolution o Hypothesized inheritance of acquired characteristics from giraffes acquiring long necks by stretching up to eat leaves off of tall trees and then passing it to their offspring o False in his assumption of acquiring desirable traits by work like how a body builder acquires muscles with exercise  Darwin came up with natural selection, that those who had desirable traits would survive to reproduce and pass though desirable traits to their offspring o With giraffes, natural selection leads to a species of giraffes that have variation in neck lengths but the average neck length of the population is longer over time due to natural selection o The selective value of a longer neck for giraffes is actually because male giraffes compete for females with their necks and the longest necks win and get the females and reproduce more causing that trait to carry on  Natural selection occurs when 3 conditions are met o Variation - Individuals within a population vary in their characteristics o Heritability - Trait can be passed on to offspring. o Differential Reproduction - Certain traits cause a better chance of reproduction and survival Sickle Cell Disease  Example of natural selection  Children with sickle cell disease have inherited a mutant hemoglobin gene from both parents  The children who inherit the sickle cell gene are protected from malaria infections  So in areas where malaria is endemic, the bad gene has selective value and a large number of people have the sickle cell gene Genetics  During Darwin’s time, not much was known about genetics. People thought that all traits in children were a blend of traits from their parents. But this inheritance of blended traits doesn’t work with natural selection because a population would blending traits would eventually give rise to uniform individuals  Mendel discovered that traits behaved as if they are particulate (unmixable)  Mendel found that he could explain results of his pea breeding experiments if he assumed that o “Genes” (he didn’t know what they were at the time) came in pairs (alleles) o That one allele can be dominant over another o That each gamete could contain only one of the two alleles of its parent “genes” (Which he called “segregation” of alleles)  This is now known to be because of meiosis o And that when he considered more than one gene, each gene assorted independently to the gametes (independent assortment)  Also because of meiosis  With Mendel’s discovery of the gene, the physical basis for Darwin’s theory of evolution by natural selection could begin to be understood  Mendel studied various phenotypic (observable) traits in garden peas  First experiment included crossing a round seeded father and a wrinkle seeded mother which resulted in a round seed progeny (F1 generation) which conclusively debunked the “blending” hypothesis o If the blending hypothesis were true, you would expect to see a seed that was not round like the father or wrinkled like the mother, but somewhere in between.  Second experiment to answer if the results were due to a maternal or paternal influence. Switched and crossed a wrinkled father with a rounded mother seed. All progeny were again round seeded. This concluded that the round seed trait is dominant to the wrinkled seed trait.  By Mendel’s “principle of segregation”, the gametes produced by each plant contain just one copy of the gene which can then combine randomly with each other during fertilization  Mendel’s “true breeding lines” are what we call homozygous. Both alleles of the gene (on the two homologous chromosomes) are the same o In the cross of a homozygous dominant with a homozygous recessive, all F1 progeny are heterozygous with the dominant phenotype  Draw out the Punnett square of AA x aa o When this F1 generation is crossed with itself (heterozygous x hetero) it results in a 3:1 ratio of phenotype which is consistent with Mendel’s principle of segregation  Draw out the Punnett square of AA x Aa  The Principle of Segregation can be represented by a Punnett square  Mendel then experimented with seeds differing in two traits o Each parent homozygous for different colors and different seed shapes o All offspring heterozygous for both with dominant phenotypes for both  Self-pollination of these F1 plants (all hetero for both traits) led to a 9:3:3:1 ratio  9-dominant for both traits  3- dominant A recessive B  3- recessive A dominant B and  1- recessive for both traits o Draw out the Punnett square for 2 traits AaBb x AaBb  The 9:3:3:1 was predicted by and confirms Mendel’s principle of independent assortment  Mendel then did a test cross of a seed heterozygous for shape and color with a homozygous recessive (on both traits) parent and observed equal proportions of phenotypes in the offspring o A test cross is a cross of any phenotype with a homozygous recessive.  Mendel’s ideas were not worked out until the early 1900s by Sutton and Boveri who worked out what was happening in meiosis.  Sutton and Boveri realized that is chromosomes contained the genetic material then their behavior in meiosis would exactly explain Mendel’s observations Extending the chromosome theory  Thomas Hunt Morgan (early 1900s) chose the fruit fly as his study object o Fruit fly has no preexisting phenotypic variants like the plant has round and wrinkled seeds  It took Morgan a year of breeding flies and examining thousands before finding a mutant fly with white eyes o Used notation w for the dominant allele and w for the recessive allele o Note that this is a different method of notation o Compare previous notation  W = dominant, w = recessive  w+ = dominant w = recessive  Ran basically same experiments as Mendel and found similar results except that the females were always red-eyed and the only white-eyed mutants were males o Crossed red eyed females with white eyed females and was able to breed white-eyed females o Did a reciprocal cross (like Mendel) and shows that the white- eyed trait does depend on the sex of the donor, only boys get it if the mother has it o This was explained by the discovery made by Nettie Stevens at about the same time that sex is governed by “X” and “Y” chromosomes X-Linked Recessive Traits  X & Y are called sex chromosomes, the rest are called autosomes  In Drosophila (the fruit flies) the gene for red or white eye color is on the X chromosome  Females can either have no mutated X chromosomes, have the disease (2 mutated Xs), or be a carrier (one mutated X and one normal X) o Females can be XX (unaffected/normal), XX+ (carrier), or X+X+ (affected)  Boys cannot be carriers, they either have it (mutated X) or don’t (normal X) o Males can be XY (unaffected/normal) or X+Y (affected)  A father with the trait will transmit the mutant allele to all daughters (takes their only X) but no sons (only takes their Y) Sex Chromosomes  X chromosome contains 1400+ genes and mutant X alleles are linked to 59+ diseases  Y chromosomes contains only a few genes and no human diseases are known to be Y-linked o The only known Y-linked trait is hairy ear rims Hemophilia  Recessive X-linked mutation  The wild type allele codes for an enzyme used in blood clotting; boys with the defective allele have blood that fails to properly clot  Famous example: intermarrying royal families of Europe Red-Green colorblindness  X-linked recessive trait  One in 12 men have red or green color blindness while only one in 250 women do  The gene for synthesizing red-sensitive opsin and green-sensitive opsin are adjacent to each other  Some individuals can have 2 or 3 genes for green-sensitive opsin  Green genes can be lost by errors in crossing over Congenital Generalized Hypertrichosis  Rare X-linked dominant trait  Hair covers face and upper body Non-Disjunction of the Sex Chromosomes  The sex chromosome equivalent to Down’s syndrome where there are 3 copies of chromosome 21  XX non-disjunction o XXX female- normal; slightly lower IQ; slightly taller; frequency unknown because phenotype is so normal o X0 female- Turner syndrome- short stature, webbed neck, immature sex organs that do not undergo changes during puberty; low-normal intelligence o XXY male- Kleinfelter syndrome- low fertility many female body characteristics, lowered IQ, 1:500 male births o OY- nonviable  YY non-disjunction o XYY- fertile, normal appearance, 20x more frequent in males in penal and mental institutions but most XYY males do not develop patterns of antisocial behaviors Barr bodies  One might expect an XO female to be same as XX female since one of the X in normal XX is silenced but not all genes are silenced in Barr bodies (the “silenced” Xs)  Many of the genes that escape inactivation are in the regions of the X chromosome that contains genes also present on the Y chromosome (unlike the majority of the X chromosome which are not also present on the Y chromosome). These regions are called pseudoautosomal regions as individuals of either sex will receive two copies of every gene in these regions (like a regular autosome) unlike the majority of genes on the sex chromosomes in which they receive only one copy.  Since individuals of either sex will receive two copies of every gene in a pseudoautosomal region, no dosage compensation is needed for females, so it is postulated that these regions of DNA have evolved mechanisms to escape X-inactivation  Shortly after fertilization, all the cells in a female embryo randomly pick one of their two X-chromosomes and “shut it off” by converting it to a highly compacted form that is the Barr body  In an XXX female, there are 2 Barr bodies and one functional X chromosome  The entire chromosome is not inactivated (if they were, XO’s and XXY’s would be phenotypically normal)  Females are “mosaics” because of X inactivation and because it occurs early in embryological development o The black and orange patches of Calico cats are an example of this X chromosome mosaicism in females, female calico cats have black and orange alleles of an X-linked gene o X-linked genetic diseases can be mosaic is females meaning that they can be red-green color blind in one eye Morgan’s discovery of X-linked inheritance was important evidence in support of the hypothesis that chromosomes are the carriers of the genes Mendel discovered  The Morgan group (Morgan and his students) realized that chromosomes might indeed be a linear array of Mendel’s genes  The group then realized that if genes are located one after another along the length of a chromosomes, then two genes close to each other on a chromosome would be “linked”- unable to independently assort to the gametes in the way that traits governed by genes on separate chromosomes would  The first pair of linked genes that they discovered was on the X chromosome. They were 1) the gene whose alleles determined the traits of red or white eyes and 2) a second gene that governs body color [either wild type gray (dominant) of yellow (recessive)]  To examine the linkage between eye color and body color, they bred a homozygous white-eyed gray bodies female (wy+/wy+) and red-eyed yellow bodied male (w+y/Y) o w+ = red/ w = white o y + = red/ y = yellow  The F1 red eyed gray bodied females (wy+/w+y) (= heterozygous for both traits) were then mated with their sibling white-eyed gray bodied males (wy+/Y)  Rationale: if the genes are located one after another along the length of a chromosome, then only two different kinds of gametes could be formed by the F1 females (a game with wy+ X-chromosome and a game with w+y X-chromosome) which would should that the traits could not switch location. If the w/w+ alleles were on different homologous pairs of chromosomes than the y/y+, the mother could form 4 different kinds of gametes wy/w+y/wy+/w+y+ because they’re not connected so they can form any combination  Furthermore: if they’re linked then only 4 different kinds of flies would be found. Only two types of males would be found. Half would be wy+/Y and the other half would be w+y/Y o This was not what was observed, rather, a small fraction (1.4%) of the male progeny were unexpected, novel phenotypes: wy/Y and w+y+/Y o These novel phenotypes were due to the unexpected genotypes wy and w+y+ X chromosomes o Morgan realized that the recombinant genotypes on the X chromosomes were being generated when crossing over occurred during prophase of Meiosis.  After Morgan’s discovery that genes are linearly arrayed on chromosomes and that crossing over leads to recombinant chromosomes, an undergrad student in his lab (Sturtevant) then realized that the frequency of crossing over would depend on the physical distance between genetic loci on a chromosome. o The recombination frequency (%) = the map units (now called centimorgans) that the genes are apart   Genes that are closer together are less likely to recombine than genes very far apart Experimental crosses of F1 Progeny after Mendel’s work in 1900 often did not result in the 9:3:3:1 ratio. Investigation of these results led to further study of different aspects of inheritance that can alter this ratio. Incomplete Dominance  Heterozygous has a different phenotype because the dominant allele is not completely dominant.  Ex: RR flower is red, rr flower is white, but Rr is pink (having only one dominant copy of the gene results in only half as much pigment as it takes to make a red flower Co-dominance (multiple alleles)  Multiple alleles at play rather than just the dominant allele and recessive allele  Ex: ABO blood types – 3 rather than 2 alleles o The A, B, and O blood groups are defined by the type of oligosaccharide present on the surface proteins and lipids of the blood cells in our bodies  Being blood group O means that all your blood cells have the O-antigen on their surface and you make antibodies against the A-antigen and B-antigen  Being blood group A means that all your blood cells have the A-antigen on their surface and you make antibodies against the B-antigen  Being blood group B means that all your blood cells have the B-antigen on their surface and you make antibodies against the A-antigen  In blood group A and B, other non-blood cells still express the O antigen  If you are blood group AB, your cells have the A-antigen and B-antigen and you do not make antibodies against the A, B, or O antigen o The I allele codes for an enzyme that attaches a sugar to the end of the O-antigen structure to create the A-antigen structure while the I structure codes for a different enzyme that attaches a different sugar to the end of the O-antigen structure to create the B-antigen structure o The i allele is a defective gene that doesn’t code for any enzymes, thus if a person is homozygous ii, the O-antigen structure remains A Bodified. o A heterozygous I I individual will make both enzymes and thus both antigens o The gene that codes for the different enzymes (I I ) or no enzyme at all (i) is located on the long arm of chromosome 9. o No recipient has antibodies against type O cells (universal donor) o AB is the universal acceptor Blood Type Summary Phenotype Genotype Antibodies Can accept blood (antigens) types… Present Blood Group O ii None O Blood Group A I I or I i A O, A Blood Group B I I or I i B O, B A B Blood group AB I I A and B O, A, B, AB Environmental Effect on Gene Expression  Phenylketonuria (PKU) is a disease found in individuals who are homozygous for a defective allele for the PKU gene on chromosome 12 o The non-defective PKU gene encodes an enzyme Phenylalanine Hydroxylase (PAH) that converts phenylalanine to tyrosine. If both alleles are defective, phenylalanine and a related breakdown product (phenylpyruvic acid) accumulate in their bodies o The accumulation of these molecules interferes with developments of the nervous system and causes mental retardation o Genetic screening permits the identification of PKU babies at birth and placing the babies on a low-phenylalanine diet permits completely normal development  Some mutant genes encode proteins that are heat-sensitive o The ch allele in Siamese cats encodes a heat-sensitive enzyme involved in the synthesis of melanin (a dark brown pigment) o The ch version of the enzyme is inactive at temperatures at 33C o The hair cells in the skin of the main body and head are above 33C, the enzyme  is inactive and the hair is white or blonde. o At the extremities, the skin temp is cooler so the enzyme is active, melanin is  produced and the hair is dark brown  Less commonly, some genes are cold­sensitive o The arctic fox has a mutant gene encoding a cold sensitive protein that governs  hair pigment production. In the winter, it’s too cold for the enzyme to be active so the fox is white. In the summer, it isn’t too cold for the enzyme, so the protein is  active and hair color is brown Epistasis: Interactions with other genes  Cross of brown bell pepper with yellow red pepper creates a red bell pepper. o Cross of the F1 red pepper with itself let to production of 9 red, 3 yellow, 3 brown, and 1 green  Bell pepper color is determined by two genes. One gene (R) encodes an enzyme required for production of red pigment. Another gene (Y) governs the presence/absence of green in the form of controlling chlorophyll synthesis. o R= red/ r= yellow o Y= absence of green (-chlorophyll)/ y = presence of green (+chlorophyll)  Furthermore, orange peppers are produced when red pigment is made in reduced amounts. A third gene that also produces an enzyme that functions in the synthesis of red pigment can also come into play. o If this third gene makes a defective enzyme, production of red pigment is reduced, even when the R gene is wild type (dominant). o So two different plants can have the same genotype in terms of gene 1 and 2 (both RRYY) but one will make red peppers and one will make orange peppers. This is called epistasis. The third gene is said to be epistatic to R. Quantitative traits  Mendel worked with discrete traits (yes or no qualities/green or yellow) but many traits aren’t discrete  For example, height, weight, and skin color fall on a scale of continuous  These type of traits are called quantitative traits  Ex: wheat kernel color. Homozygous white crossed with homozygous red results in a blended F1. The F2 generation (from a self-pollinated F1 generation) exhibits in a normal distribution of colors that range from white to red  Nilsson-Ehle proposed a model to explain the “normal distribution” range of colors produced by the F2 generation in wheat kernels. In his model three genes contribute to gene color. Each gene has two allelic forms. o This model is correct and produces 64 different genotypes in a trihybrid cross in a 1:6:15:20:15:6:1 ratio (a normal distribution)  Quantitative traits are produced by the independent actions of many genes in which each locus contributes a small amount to the value of the phenotype Chapter 15: Genes Garrod (1902)  As early as 1902, Archibald Garrod proposed that some diseases might be caused by a  hereditary defect in the ability to synthesize a particular enzyme in a metabolic pathway.  In the disease alkaptonuria, patients excrete large quantities of homogentisic acid in  their urine (turning it black).  o Garrod proposed that the enzyme that normally converts homogentisic acid to a  different acid is missing in these patients o He called these genetic defects inborn errors of metabolism Beadle & Tatum (1941)  1941­Beadle and Tatum stated their conclusion that genes code for proteins  In particular, they proposed that genes code for the enzymes that catalyze the steps in a  metabolic pathway  Their general approach to research was to make a gene defective and study the effect on  the now­mutant organism o They studied common bread mold Neurospora crassa mutants that were defective in the synthesis of the amino acid arginine o They identified 3 classes of Arg mutants based on their ability (or lack thereof) to  grow on minimal medium supplemented with one of the three intermediates in  arginine biosynthesis  From this, they deduced that each class of mutant was unable to carry out  a different one of the three steps in the arginine pathway (3­step synthesis)  This was because each classed lacked the enzyme that catalyzed that step  Therefore, they could conclude that “Gene A” (the defect gene in class 1  arg mutants) coded for “Enzyme A” (the enzyme that the arg 1 mutants  lacked), while “Gene B” (the defect gene in class 2 arg mutants) coded for “Enzyme B” (the enzyme that the class 2 arg mutants were missing).. etc.  From this, they concluded that each gene in an organism is responsible for  making a different enzyme  This was often referred to as the “one gene­one enzyme”  hypothesis o Later broadened in the late 1940s to the “one gene­one  protein” hypothesis Nucleotide Sequence  Amino Acid Sequence  After Watson & Crick discovered the DNA structure in 1953, the question about genes  became “How does the sequence of nucleotides in a DNA molecule direct the assembly  of a sequence of amino acids in a protein?”  Early idea: Amino acids directly interact with the nucleotides in the DNA molecule  Crick proposed instead that the sequence of nucleotide bases in DNA act as some sort of  code (i.e. DNA was only an information­storage molecule)  1958­ Arthur Pardee (working with Jacob & Monod) performed an experiment that used  features of bacterial mating to study the regulation of gene expression o This was called the “PaJaMo” experiment and it provided the key to  understanding the role of ribosome and mRNA in protein synthesis  Brenner and Crick first realized that an implication of PaJaMo was that there must be a  special kind of short lived RNA called “messenger RNA” that serves as a link between  the DNA in the nucleus of a cell and the protein­synthesizing ribosomes (stable, long­ lived RNA) in the cytoplasm o Shortly thereafter, Hurwitz & Furth purified mRNA molecules and then went on  to isolate the enzyme RNA polymerase that synthesizes mRNA molecules off a  DNA template  At this point, it seemed likely that the sequence of bases in DNA was serving as a code  for a sequence of amino acids and that mRNA molecules carried the code from the DNA  to the ribosomes. The next step was to decipher the code Deciphering Nucleotide Code  George Gamow argued a priori that each amino acid is coded for by three nucleotides in  the DNA o His argument was based off of the fact that (at the time) there were 20 known  amino acids in proteins. If each of the four different bases coded for one amino  acid, then there would only be 4 amino acids. If it was a 2 base code, there could  be only 16 (4 ). But with a 3 base code, there could be 64 different amino acids  (4 )  This hypothesis was proved experimentally by Crick and Brenner in 1961 by treating  cells with proflavin which mimics a nucleotide pair and resulted in errors in DNA  synthesis that lead to insertion of deletion of a nucleotide o Crick and Brenner were the first to realize that it caused insertion or deletion of a  nucleotide and that this, in turn, could cause a reading frame shift  Consequently, a nonsense sequence of amino acids would arise following  the frame­shift mutation and the protein would be completely inactive  A compensatory nucleotide subtraction or addition could then bring the  reading frame back to its original order. If this occurred soon enough after  the original mutation, the protein would be active  Crick & Brenner tested their ideas by taking advantage of a genetic system worked out by Benzer with a gene called the “rII locus” of the bacteriophage T4. o The great advantage of this system was that many billions of individual phage  progeny can be observed and even very rare mutations and rare recombinational  events can be detected  Crick & Brenner realized that if the code was a triplet of nucleotides, then a mutant gene  with a single shift event could be corrected if recombined with other mutants, each with a frame shift in the same direction, which is what they observed  the triple mutants  recovered the ability to encode a protein The next discovery: Which bases code for which amino acids?  Nirenberg & Matthaei discovered that UUU is phenylalanine o Set up cell­free protein synthesizing system o Synthesized an RNA molecule that was just a string of Uridine residues (“polyU”) o The poly­U in their protein­synthesizing system formed a string of phenylalanine  Similarly, CCCCC… encoded protein Pro­Pro­Pro­Pro… so  o CCC codes for Proline (Pro) o AAA codes for Lysine (Lys) and… o GGG codes for Glycine (Gly)  This opened the door to a series of similar experiments   Final solution of the code: Nirenberg and Leder found that one could use synthetic  mRNA with just one trial trinucleotide codon at a time to ribosomes and the ribosomes  would then bind the correct tRNA, charged with its radiolabeled amino acid  The code was essentially all worked out in the next 5 years  The code is degenerate meaning that there is more than one codon for most amino acids  (ex: CCC = proline, CCU, CCA, or CCG also = proline)  3 codons are “nonsense” or “stop” codon and one codon (AUG) is a start codon and  codes for methionine o All proteins begin with the amino acid methionine Using the genetic code to predict an amino acid sequence The DNA sequence… transcribed as…. …and translated as… The Central Dogma  So, by the end of the 1960s, what was termed by Crick as “the central dogma” was well  established: DNA codes for mRNA, and mRNA in turn codes for protein.  The term transcription came to be used for the process of copying mRNA off of the  DNA template  The term translation came to be used for the process of converting the information on  mRNA into protein (using ribosomes and tRNA)  DNA sequences define the genotype, proteins create the phenotype


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All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.