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Bio 222 Week 2 notes

by: Gabrielle Woolery

Bio 222 Week 2 notes BIOL 222

Gabrielle Woolery
Penn State
GPA 3.65

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These are more in-depth notes from the second week of class
Paul Babitzke
Class Notes
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This 24 page Class Notes was uploaded by Gabrielle Woolery on Thursday September 1, 2016. The Class Notes belongs to BIOL 222 at Pennsylvania State University taught by Paul Babitzke in Fall 2016. Since its upload, it has received 54 views. For similar materials see Genetics in Biology at Pennsylvania State University.


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Date Created: 09/01/16
Bio 222 Week 2 Notes Chapter 3: Basic Principle of Heredity Monohybrid Cross  Analyzes only one characteristic/gene  All characteristics are separate from one another Dihybrid Cross  Analyzes two characteristics/genes simultaneously o Can be linked or unlinked  Usually has four possible phenotypes o Phenotypes 1 and 2 are dominant  RRYY or RrYy o Phenotype 1 is dominant and phenotype 2 is recessive  RRyy or Rryy o Phenotype 1 is recessive phenotype 2 is dominant  rrYY or rrYy o Both phenotypes are recessive  Rryy Note this lecture is discussing unlinked genes P (parental generation)  Remember these are pure breed F1 (comes from parental cross)  Heterozygous for both gene pairs so they self pollinate and result in F2 F2 (comes from F1 selfing) Ex: R and r are two alleles for one gene for the shape of peas R=round peas r=wrinkles Y and y are two alleles for one gene for the color of peas Y=yellow y=green These genes are unlinked because the genes are on different chromosomes P RRYY X rryy (crossed) = F1 RrYy (self pollinate) = F2 Makes a 9:3:3:1 phenotypic ratio These are what each number stands for respectively: 9= round and yellow peas (homozygous and heterozygous dominant) 3=round and green (homozygous and heterozygous for shape, homozygous for color) 3=wrinkled and yellow (homozygous for wrinkled, homo and hetero for yellow) 1= wrinkled and green peas(both genotypes are recessive) How to fill out the dihybrid cross Punnett Square 1. List gametes from parents on one side 2. List gametes from parents on other side 3. Fill in genetic combinations in the squares 4. Determine genetic ratios (phenotypic or genotypic) by counting squares Independent Assortment  Genes assort independently from one another o Ex: eye color is separately determined from hair color Mendel’s Second Law  law of independent assortment  during gamete formation the segregation of alleles of one gene pair is independent of the segregation alleles from another gene pair o Different gene pairs containing there alleles are not linked, gene pairs on different chromosomes assort independently during meiosis Branch Diagram  For F3 generation ¾ A__ : ¾ B__ and ¼ bb: 9/16 A__B__ and 3/16 A__bb ¼ aa: ¾ B__ and ¼ bb: 3/16 aaB__ and 1/16 aabb Say Aa is a gene for hair color A is brown a is blonde Bb is a gene for eye color B is brown b is blue This branch diagram shows that for the first thing written in both rows is the probability that someone will get brown or blonde hair color. The next column in both rows shows the percentage of those that will have brown eye color (homozygous or heterozygous) and blue hair color (homozygous recessive). The next column in the first row shows the probability of people who will show the dominant phenotype of having both brown eyes and hair and also those who will have brown hair and blue eyes. The last column in the second row shows those who will end up having blonde hair and brown eyes and those who show both recessive phenotypes for having blonde hair and blue eyes. So if you think about it the first column discusses the phenotypes for hair color and the second column shows the phenotypes for eye color. The last column shows the probabilities of those phenotypes occurring together. Overall this diagram portrays how the 9:3:3:1 ratio is established. How to make the branch diagram 1. List the probabilities for one even happening (getting the dominant phenotype for one gene and the recessive phenotype for that same gene) 2. List probabilities of second event next to those of the next (these are the last to columns and this is when you include the probabilities of the second gene) 3. Next list the probabilities of those genes alleles occurring together (blonde hair and blue eyes, brown hair blue eyes, etc..) and use the product rule to determine these genetic ratios refer to branch diagram above for steps 1 and 2 you can draw to separate monohybrid crosses and then figure out the probabilities for step 3 To calculate genotypic frequency EX: F1 AaBbCcDdEe X AaBbCcDdEe 1. Determine the probability of getting each allele for each separate gene  AA o (1/2A)(1/2A)=1/4  Each parent has a 50% chance of passing down an A  Remember to use multiplication rue  bb o (1/2b)(1/2b)=1/4  Each parent has a 50% chance of passing down a b  Cc o (1/2C)(1/c)+(1/2C)(1/2c)=1/2  This is different because this is a heterozygote. Each parent has 2 alleles they can pass down so it has to be taken into consideration the probability of each of the alleles each parent can pass down.  Both parents have a 50% chance of passing down both alleles  Since there are two alleles being taken into consideration the addition rule and multiplication rule are used Now by applying these examples and knowledge you can find probabilities for trihybrid, tetrahybrid crosses and so on Using these parents F1 AaBbCcDdEe X AaBbCcDdEe Finding the probability of getting AAbbCcDdee Respectively (1/4)(1/4)(1/2)(1/2)(1/4)=1/256 P=1/256 The ¼ are always homozygotes and the ½ is the probabilities of the heterozygotes To calculate phenotypic frequency F1 AaBbCcDdEe X AaBbCcDdEe Probability of getting A__, bb, C__, D__, ee  A__ (dominant phenotype) o ¾ because its dominant  bb o ¼ chances of getting a recessive is less because it needs to be homozygous  C__ o ¾  ee o ¼ Respectively: (3/4) (1/4) (3/4) (1/4)= 27/1024 Probability of getting these phenotypes is 27/1024 Human Pedigree Analysis Only ones you need to know are  Unaffected person  Obligate carrier  Family  Consanguinity Autosomal Recessive Disorders  Mendalian inheritance of an autosomal recessive disorder is revealed by the appearance of the phenotype in male and female progeny of unaffected individuals  The gene is recessive so the only way someone would exhibit the disorder or trait is if the genotype is homozygous recessive  A person who is heterozygous for the disorder is a carrier  If both parents are homozygous recessive for the disorder all the children will have the disorder  It tends to skip generations Autosomal dominant disorders  Mendalian inheritance of an autosomal dominant disorder show affected males and females in each generation and also show affected males and females transmitting the condition to sons and daughter in equal proportions  If the family size is large enough you can see the disorder in every generation  Dominant disorders can never be a carrier because if you have the dominant trait you definitely get the disorder Chapter 4: Extensions and Modifications of Basic Principles Sex Determination XX-XO  Females have 2 X chromosomes (XX): homogametic sex  Males have 1 X chromosome (XO) heterogametic sex  During Meiosis in the male ½ of the gametes get an X giving a 1:1 sexual frequency XX-XY  Human situation  Females 2X chromosomes (XX): homogametic sex  Males have 1X and 1Y chromosome (XY): heterogametic sex ZZ-ZW  Females have 1 Z and 1 W chromosome (ZW) heterogametic sex  Males have 2 Z chromosomes (ZZ) homogametic sex Genic Sex Determination  No sex chromosomes. Specific genes on the autosomes determine the sex  Genes on various autosomes in the genome. Genes spread out in a variety  Ex: plants and protozoans Environmental Sex Determination  Environmental factors determine the sex  Ex: temperature Drosophila  X: a ratio dictates the sex THIS IS TRIVIA Humans  The presence of the Y chromosome dictates maleness o SRY gene is on the y chromosome, this gene is what establishes maleness. If SRY is deleted the person can be female SRY Gene  Male determining gene in humans on Y chromosome  Females develop if this gene is missing from the Y in an XY individual  Males develop if this gene is trans located to an X in an XX individual Sex Linked Inheritance  Inheritance pattern of genes on sex chromosomes  Sex chromosomes carry many genes unrelated to male and female development X-linked Recessive Inheritance  Many more males than females have it  Mutation is on a gene on the X chromosome o Females need the mutation on both X chromosomes o Males need the mutation on only one X chromosome because there genotype is XY  Ex gene for colorblindness is on the X chromosome Experiment 1 X+ is dominant for red eye color Xw is recessive for white eye color Remember X is what determines eye color not Y Parental Generation (P)  X+X+ (red eye female) X w Y (white eyed male) F1 Generation  All females are red: X+Xw because X+ is dominant over Xw  All males are also red X+Y because males pass on the Y chromosome and the female was homozygous dominant so the males can only get an X+  X+Xw times X+Y gives F2 generation F2 Generation  All females get red eyes because X+ is passed from male and X+ is dominant  50% of males have white eyes because it’s a 50% chance females pass the Xw or X+ genotype  Phenotype frequency for these flies is 3:1 Experiment 3  Reciprocal cross of experiment 1 X+ is dominant for red eye color Xw is recessive for white eye color Parental Generation XwXw (white female) times X+Y (red male) F1 Generation  Red Female XwX+ because they definitely get the X+ from the male  White Male XwY because male passes down the Y and female is homozygous recessive so it has to get Xw  These cross F2 1:1:1:1 Red Female: White Female: Red Male: White Male (X+Xw) (XwXw) (X+Y) (XwY)  Because female give 50% of one of the alleles and the male gives the recessive type but the females still give 50%. Think in a way that the female overall determines what happens here. Experiment 2 : Test Cross F1 Red Female (X+Xw) X white male Get the same result from experiment for F2 as you did in the parental generation (refer to that above) X-Linked Recessive  Many more males than females show the recessive phenotype  If it is rare almost all affected people are male o If rare none of the offspring of an affected male are affected but all daughters are carriers o Female Carriers will pass condition to ½ her sons and ½ her daughters will be carriers o Recessive will skip generations X-linked Dominant  Affected males pass condition to all daughters never to sons because sons get Y from dads not X  Affected females pass condition to ½ of sons and ½ of daughters Y-linked Inheritance  Other than the SRY gene found on the Y chromosome in males very few human phenotypes are known to be Y- linked Dosage Compensation in mammals: X Inactivation  Female mammals inherit 2 X chromosomes  Early in development one of the 2 X chromosomes is randomly inactivated in each cell  The same X chromosome remains inactive in all of the daughter cells of each progenitor Barr Body  The inactive X chromosome Pseudominance  If the functional X has a recessive allele the recessive phenotype is expressed. Since the inactivation process is random all female mammals are genotypes corresponding to inactivation alternatives. Chapter 4: Extensions and Modifications Dominance  Phenotype of the heterozygote is the same as one of the homozygotes Codominance  Phenotype of heterozygote includes phenotypes from both homozygotes  When two alleles are both shown  Ex: A fish being colored blue and red Incomplete Dominance  Phenotype of the heterozygote is intermediate between the two homozygotes  white and red flowers make a pink flower  The pink flower illustrate incomplete dominance Human Hemoglobin  HbA o One allele that codes for normal blood cells o It is completely dominant with respect to anemia o Incomplete dominance to sickles  HbS o Abnormal “sickle” shaped RBCs= anemia = fatal o HbA is dominant to it Examples  HbAHbA o this is for normal blood cells  HbSHbs o Abnormal sickle shaped RBCs, anemia  HbAHbS o Since there is HbA it is dominant to getting anemia but since HbS is present at low oxygen levels sickles can form o So no anemia but can get sickle shaped cells Pleiotropic Allele  An allele that affects more than one characteristic  HbS is an allele that is an example of this Penetrance and Expressivity Penetrance  Percentage of individuals having a particular genotype that expresses the expected phenotype  Ex: B is an allele for brown hair. BB genotype is expected to have the phenotype of brown hair. So, this is the percent of individuals from a selected group, population, etc who actually show brown hair with this genotype Incomplete Penetrance  A genotype that does not always produce the expected phenotype  Ex: Human polydactyly: it is a dominant allele but not everyone with the allele has extra digits Expressivity  Degree to which a phenotype is expressed Incomplete Penetrance and Expressivity are caused by other genes and environmental factors Lethal Alleles  Alleles of a gene that cause death  Death may occur during early embryonic development, after birth, or later in life Mouse Coat Color  Yellow coat is a variant  Dominant yellow is lethal and the mice die in utero before birth (YY) Ex: Parental Generation Yellow X Yellow F1 2:1 Yellow: non yellow ¼= YY (die in utero) Normally ratios should add to four but this one adds to three because one of the mice were not born which suggest that the mice presenting the yellow phenotype have to be heterozygous because if they were homozygous they would die before they were born Manx Cat  Has no tail when it is heterozygous  When it is homozygous it is lethal  Factor C clotting mutation These two examples of lethality (Mouse and cat) are pleiotropic o Pleotropic: Allele that affects more than one characteristic Multiple Alleles  A gene that has more than two alleles  EX: ABO blood groups Blood Groups  IA, IB, i o IAi and IBi are completely dominant to IAi and Ibi and codominant in IAIB o A and B are codominant Universal donor Univers al Accept AB blood has no antibodies because it can accept both antigen types hence why this is the universal acceptor Pepper Color 9:3:3:1 ratio Two genes affect one characteristic Y+, y and C+, c affects the color of a pepper Parental Generation Y+Y+ C+C+ makes a pepper red  Homozygous Dominant for both gene pairs yycc makes a pepper cream  Homozygous recessive for both gene pairs F1 Generation Y+yC+c  Heterozygous for both gene pairs F2 Generation Y+__C+__  Having one dominant allele for both gene pairs makes a red pepper Y+__ cc  Having one dominant Y and a recessive cc makes the color peach yyC+__  Having recessive yy and one dominant C makes the color orange yycc  Having homozygous recessive for both gene pairs makes the color cream Epistasis  When the effect of one gene masks the effect of another gene at a different locus Recessive Epistasis  Need to be homozygous recessive for one of the gene pairs to hide the expression  Ex: Labrador Retriever Labrador Retriever Bb: pigment production Ee: Pigment disposition B: black color B: brown color Parental Generation BBEE (black) X bbee (yellow)  Yellow is pigment disposition  Recessive ee affects all coloration  Dominant EE or Ee has no impact on color only look at Bb alleles to see the color F1 Generation BbEe(black) X BbEe  Both turn out to be heterozygotes so you still only look at the Bb alleles because there is no recessive ee to dominate the colors F2 Generation Gets a 9:3:3:1 ratio  B_E_ = 9 black  bbE_= 3 brown  B_ee= 3 yellow  bbee= 1 yellow Recessive Epistasis Requires ee Dominant Epistasis  Only need one dominant allele for epistasis  Ex: Summer squash Summer Squash W: the dominant allele for epistasis prevents conversion of compound A to compound B to make the squash a WHITE COLOR Parental Generation WWYY (white) X wwyy(green) F1 WwYy (white) X WwYy (selfing) F2 W_Y_: 9 white W_yy: 3 white wwY_: 3 yellow wwyy:1 green W is epistatic to Y and y References <Dr. Paul Babitzke > “Chapter 4àExtensions and Modifications of Basic Principles (9-2-16)” BIO 222. 2016.” The Pennsylvania State University, Pa.


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