×
Log in to StudySoup
Get Full Access to OleMiss - Bisc 336 - Study Guide - Midterm
Join StudySoup for FREE
Get Full Access to OleMiss - Bisc 336 - Study Guide - Midterm

Already have an account? Login here
×
Reset your password

OLEMISS / Biology / BIOL 336 / Who determined that dna was the carrier of genetic info?

Who determined that dna was the carrier of genetic info?

Who determined that dna was the carrier of genetic info?

Description

School: University of Mississippi
Department: Biology
Course: Genetics
Professor: Ryan garrick
Term: Fall 2016
Tags: Genetics
Cost: 50
Name: Genetics Test 1 Study Guide
Description: This is a compilation of class notes with a mixture of matching and fill in the blank. I recommend trying to do as much as possible without looking back at your notes so you truly know what you need to focus on!
Uploaded: 09/12/2016
38 Pages 31 Views 3 Unlocks
Reviews


LECTURE 1


Who determined that dna was the carrier of genetic info?



Recap on Vocab:  

________ – the study of heredity and variation

Heredity – ___________________________________ 

Variation – differences between individuals  

Understand which major discoveries underpin recent developments in genetics 

1600-1850

_______ – (epigenesis) egg differentiates into adult structures over time - Aka development

Pasteur – (spontaneous generation) living organisms do not derive from non living components  We also discuss several other topics like Syncopation is a note just of what?

_______ – (evolution) _______ ____ ____________, with _______ _________ as the  driver of change  


What are the different forms or variants of a gene?



- Variation in a population  

- Competition for _______ 

- Advantages over others  

1600-1850

________ and “unit factors” – predictable patterns of inheritance: random  segregation, assortment  

• Chromosomal theory  

– _____________________________________________________ 

- process of ________ and _______ 

• Transmission Genetics – _____________________________________________ - chromosome and pedigree analysis  We also discuss several other topics like What is the study of taxonomy?

Q: Who determined that DNA was the carrier of genetic info? What was  believed previously?

A: Avery at Al. – People believed previously that proteins were the carriers of  genetic info


What is the structure and function of genomes?



Q: What about Watson and Crick?  

A: Watson and Crick discovered that DNA was organized as a double helix  with a phosphate backbone and nucleotide pairing rules (___ goes with T, and ___ goes with G)  

Describe relationships between components of the central dogma; contrast DNA, RNA, and proteins.  

THE CENTRAL DOGMA

 DNA ______________

____________

 ________________ Translation

(Matching Vocab)

A. Allele

B. Genotype

C. Gene

D. Genome

E. Phenotype

F.

G. 1. __________ how a person’s genotype interacts with environmental  factors (observable/physical characteristics)

H. 2. __________ section of DNA that codes for RNA to make proteins,  which in turn make us We also discuss several other topics like Who discovered penicillin?

I. 3. __________ an entire haploid set of chromosomes aka what it takes to build an entire organism (“blueprint”) all of the chromosomes in the  nucleus of a cell  

J. 4. __________ simple combination of the alleles at a locus K. 5. __________ different forms or variants of a gene  

L.

M. PROTEINS 

N. Definition: the end products of ________ _____________

O.

P. Remember: All _______________ are proteins but not all proteins are  _______________!

Q.

R. Q: What is Sickle Cell Anemia?  

S. A: Sickle Cell Anemia is a disorder that arises from a mutant gene  which produces a mutant protein that alters phenotype. The mutation  is __________________, which is a single nucleotide change that leads to  a different amino acid, and a ß-globin protein

T.

 U. Recall the series of advances that were possible following the development of recombinant DNA.  

 V. RECOMBINANT DNA 

W. Q: Explain DNA cloning.

X. A: _________________ cut DNA at specific nucleotide sites, then  ___________ (plasmids) take up the DNA fragments from the restriction  enzymes. After that, ___________________, such as bacteria, take up the  vector. As the bacteria colony grows, clones are made. Don't forget about the age old question of Who is phineas gage?

Y.

Z. Q: What are “Model Organisms” and their characteristics?  AA. A: Model organisms are a diverse set of organisms that can be  well understood from a scientific standpoint based on these  characteristics:  

1. ______________________

AB.

2. _______________________________ Don't forget about the age old question of How many types of diabetes are there?

AC.

3. ______________________________________

AD.

 AE. BIOTECHNOLOGY 

AF.

AG. ___________________________

- crop plant resistance (to insect herbivores, viruses, etc), nutrient  enhancement, milk production, etc..  

- controversy:  Don't forget about the age old question of Organisms made up of membrane-bound units are called what?

o engineered traits might “escape” into wild relatives

o patents on genes/living organisms is not OK

AH.

AI. ___________________________

- rapid screening of disease-associated alleles  

- determine risk of developing (or having a child with) a genetic disorder - develop individual-based therapy, based on observed expression of  genes (e.g., in cancer cells).  

AJ.

AK. NEW FIELDS 

AL.

A. Genomics

B. Proteomics  

C. Bioinformatics  

AM.

AN. 1. __________ identify proteins present in a cell at a given  time/location; modifications and interactions  

AO. 2. __________ filtering and analyses of very large genetic  datasets; software development  

AP.3. __________ structure and function of genomes (all of the DNA carried  in an organism), often comparative  

AQ.

AR. LECTURE 2

AS. Know the terminology (cells, chromosomes, etc.) 

AT. Cell Structure and Genetic Function

A. Nucleus

B. Organelles

C. Cytoplasm

D. Cytoskeleton

AU.

AV.1. __________ colloidal matrix

AW. 2. __________ mitochondria, ER, centriole, chloroplast AX. 3. __________ houses DNA

AY. 4. __________ lattice of tubules and filaments  

AZ.

BA. In the Nucleus

A. Chromatin

B. Chromosomes

C. Nucleolus

BB.

BC. 1. ___________ where ribosomal RNA (rRNA) is made and where  ribosomal subunits are initially built  

BD. 2. ___________ loosely or uncoiled DNA that is unattached from  histone; present when cells are NOT dividing  

BE. 3. ___________ condensed (super coiled) DNA with discrete  structures, able to be seen just before and during mitosis and meiosis,  otherwise they are more loosely packed  

BF.

BG. In the Cytoplasm

A. Smooth Endoplasmic Reticulum (SER)

B. Rough Endoplasmic Reticulum (RER)

C. Mitochondria

D. Chloroplasts

E. Centrioles

BH.

BI. 1. __________ maternally inherited organelle in charge of oxidative  respiration (make ATP); its chromosomes live outside of the nucleus  (no heterozygous or homozygous copies of genes because they are  haploid/diploid; “Powerhouse of cell”

BJ. 2. __________ sites of biochemical reactions such as fatty acid and  phospholipid synthesis  

BK. 3. __________ involved in photosynthesis in all plants, algae, and  some protozoa (but not all cells)

BL. 4. __________ site of biochemical reactions such as translation and covered in ribosomes  

BM. 5. __________ used in mitosis and meiosis – in charge of the  organization of spindle fibers which attach to and move chromosomes  BN.

BO. Explaining Chromsomes

BP.• When present as sister chromatids (2 chromatids), chromosomes  have a _____________ used for __________

BQ. • _________ – DNA that may or may not be transcribed or  translated  

BR.

BS.

BT.

BU.

BV.

BW.

BX.

BY.

BZ.

CA.

CB.

CC.

CD. 

CE.

CF.Homologous

Chromosomes

CG. • All members of a species have the same number of  ____________ per somatic cell

CH. • Occur in pairs, 1 from the maternal and 1 from the paternal  parent; matching _______, ____________, and __________________ CI. • Sex chromosomes (although dissimilar) still behave as ___________.  CJ.

CK. Karyotype

CL. • Humans:  

- Diploid (2n) # = _________

- Haploid (n) # = __________

CM.

 CN. Understand the function of mitosis  

CO. Mitosis

CP.• genetic continuity between generations of cells

CQ. • Parent and daughter cells have __________________ CR. Function of mitosis: replacing skin and hair cells; growing  CS.

CT. Describe steps involved in mitosis 

CU.

CV. Regular Function:  

CW. • ________ – cells performing normal function (ex: blood cells  carrying O2)

CX. • ________ – non-dividing cells withdrawn from the cycle heading  towards duplication; continue carrying out normal function CY.• ________ – where DNA replication takes place

CZ. • ________ – similar to G1; cell is still performing normally but  form of chromosomes has been altered.  

DA.

DB. Mitotic Division (PMAT)

DC. Q: What happens in Mitosis, Prophase?

DD. A: ___________ move to pole and organize microtubules and  spindle fibers, _________________ breaks, and ________________ are visible as pairs of genetically identical sister chromatids  

DE.

DF. Q: What happens in Mitosis, Metaphase?

DG. A: Chromosomes migrate to the central axis and then align along the metaphase plate

- Sister chromatids pulled apart in the ___________ direction  DH.

DI.Q: What happens in Mitosis, Anaphase?

DJ.A: Sister chromatids separate and migrate to opposite poles, separated sisters = ______________. Daughter cells will get ___________ sets of  chromosomes

DK.

DL. Q: What happens in Mitosis, Telophase?

DM. A: Cytoplasm starts to divide (cytokinesis), chromosomes uncoil  back to diffuse chromatin, __________________ reforms, and  _________________ disappear.  

DN. Order of Regular function and Mitosis:  

DO. Interphase

1. _________

2. G0  

3. _________

4. G2

DP. Mitosis

1. Prophase

2. ___________

3. ___________

4. ___________

DQ.

DR. LECTURE 3

DS. Mitosis Recap

1. Starts with _____________ (2n =4)

2. __________ – main events: chromosomes condense, nuclear envelope  degenerates, chromosomes move to opposite sides and lay down  spindle fibers  

3. __________ – 4 chromosomes each consisting of a pair of sister  chromatids line up in middle

4. __________ and __________ – creates 2 new diploid (___) daughter cells  DT.

 DU. Describe Steps involved in Meiosis  

DV.

DW. Function of Meiosis:

DX. • Continuity between generations of cells

DY.• Reduces number of chromosomes by ½: diploid –> haploid  DZ. • Passes genetic material from parent to offspring

EA. • Fusing ½ of each parent DNA allows them to reform to create  the next generation’s makeup

EB.

EC. • Meiotic cell cycle involves ONE ______________________ and TWO  _________________

- PMAT I and PMAT II

ED.

A. Prophase I

B. Metaphase I

C. Anaphase I

D. Telophase I

E. Prophase II

F. Metaphase II

G. Anaphase II

H. Telophase II

EE.

EF.

EG.

1. ___________ – Chromosomes mostly visible, homologous chromosomes  pair up (synapse) together, ________________ from physical contact  between chromatid of one chromosome with the chromatid of another  chromosome allows for crossing over

- synapsed homologs have how many chromatids? ________ 2. ___________ – chromosomes migrate and align in middle; equational  division starts

3. ___________ – tetrads pulled apart (one pair of _____________ to each  pole), poles now have random mix of maternal and paternal sister  chromatids that have been previously modified by crossing over

4. ___________ – starts with 2 separate daughter cells from meiosis I;  chromosomes condense, centrioles move to opposite poles, spindle  fibers laid down; no pairing of homologous chromosomes so no  ____________; chromosomes in each cell have already been duplicated  (are now a pair of sister chromatids)  

5. ___________ – 1st division (___________) ends here; nuclear membrane  forms around chromosome sets at each pole, cytoplasm begins to  pinch; interphase occurs briefly after without DNA replication

6. ___________ – chromosomes migrate and align in middle; reductional  division starts  

7. ___________ – 2nd division (___________) ends here; cytoplasm is divided,  nuclear membrane reforms; ultimately 4 haploid cells (gametes), each  with different genetic makeup, are produced from single meiotic cycle  

8. ___________ – dyads pulled apart (one __________ to each pole); poles  now have random mix of chromatids from each chromosome; each full  (hapoloid) set of the genome  

EH.

 EI.Contrast Mitosis vs. Meiosis

EJ. ** See class handout **

EK.

EL. LECTURE 4

EM. Contrast spermatogenesis v. oogenesis  

EN. Spermatogenesis  

EO. • General procedure of meiosis  

EP.

EQ. Male Gamete Production

ER. • Starts with ______________________________ (spermatogonium) ES. • Becomes an enlarged primary spermatocyte, which undergoes  1st (______________) meiotic division  

ET.• _____________________ undergoes 2nd meiotic division EU. • each make 2 haploid spermatids that develop into motile  sperm = total of 4 _________ sperm  

EV.

EW. Oogenesis  

EX. Female Gamete Production

EY.• Starts from ____________, which enlarges to become the primary  oocyte 

- primary oocyte then splits during ____________

EZ.

FA.• Daughter cells of the primary oocyte receive equal _________________,  but unequal _____________ (cellular volume)

- the cell that gets the least is called either the 1st or 2nd_______________  and does not continue through the process of oogenesis FB. • The cell that gets more cytoplasm is either called the  _____________ (after 1st division) or ______________ (after 2nd division) FC. • Oocytes that get shorted cytoplasm do not become eggs FD. • Just ___________________ (egg) produced per meiotic cycle  FE.

FF. Sperm v. Egg Production

FG. • Initially start off similar

– diploid parent cell undergoes growth and maturation

FH. • End of first division (reductional):

- first polar body in oogenesis

- 2 daughter cells as secondary spermatocytes in spermatogenesis  FI. • A single ovum produced at end of oogenesis vs. 4 haploid sperm at  end of spermatogenesis  

- haploid gametes produced are genetically different from parent cell  FJ.

FK. Spontaneous DNA Mutations

FL.Usually caused by rare erros during DNA replication, at low, on-going  “background rates” like a slow ticking clock

FM. • Most mutations are _________ (such as in non-coding DNA so  there is no phenotypic effect)

FN. • Most mutations in coding regions are detrimental, only some  are beneficial  

- most common outcome: production of an allelic variant that does not  function as well as original DNA  

FO.

FP. Q: What are the different Mutations in Protein-Coding genes?  FQ. A: __________ – change in a codon and in its amino acid –>  changes amino acid/protein and ____________ – change in a codon that  creates a premature STOP codon so protein synthesis is terminated  FR.

FS. Q: What could create new allelic variants that did not previously  exist?

FT.A: Crossing over between parts of homologous chromosomes in  ______________ because it creates “mosaic-like” chromsomes FU.

FV.Q: What is independent (random) assortment?

FW. A: This is the ultimate outcome of meiosis because each haploid  gamete is not a clone of its diploid parent.  

FX.

FY.Q: Why is it important to know about sickle-cell anemia when learning  about genetics?

FZ. A: Sickle cell anemia is caused by a single DNA substitution,  which causes a mutant protein to be produced, which in turn alters the  phenotype of an individual, effecting his or her red blood cells.  GA.

GB. Q: What is the difference between a population and a gene pool  GC. A: ______________ – group of individuals living together that have  capacity to interbreed and can also compete for limited resources;  ______________ – total number of allelic variants of a gene that exist  within a particular group of individuals  

GD.

GE. LECTURE 5

 GF. Understand Mendel’s 4 Postulates 

GG. Q: Who was Gregor Mendel?

GH. A: A monk who experimented with pea plants while working in a  monastery garden – discovered units of inheritance by observing  similarities in ______________ between one generation to the next  without any knowledge of _________________________  

GI.

GJ.Q: What traits did he observe were encoded in the same gene?  GK. A: flower color, position of flower, seed color, seed shape, pod  shape, pod color, and stem length. Pea plants are a great model  organism because they are easy to cross, fast growing, and have  observable traits  

GL.

GM. Q: What does “true breeding” mean?

GN. A: This means that the individuals were homozygous for a gene  GO.

GP. Monohybrid Cross

GQ. • Mating between individuals that differ in ________ trait  GR. • P1 – parentals – true breeding green or true breeding yellow  (RR and rr)

- F1 – 1st generation offspring –> ______ % yellow (all Rr)

- F2 – 1st generation crossed with each other –> ______ % yellow, _______  % green (RR, rr, and Rr)

o _________ ratio

o Mendel’s Conclusion: proposed existence of “___________”  – things that pass unchanged from one generation to the next,  and determine the phenotype that is expressed

GS.

GT. Mendel’s First 3 postulates

GU. 1. ________ of a gene: occur in pairs in an individual; only applies  to diploid organisms (including peas)

GV. 2. ________________ : one phenotype dominates over the other  (not true of all genes)

GW. 3. ________________ : during meiosis, chromosomes on which  alleles reside separate and go to different gametes randomly  GX.

 GY. Use a Punnett Square for Mono- and Di-hybrid crosses GZ.

HA. Q: If a cross between P1 resulted in an F1 generation of all  heterozygotes, what will the F2 generation look like genotypically and  phenotypically?

HB. A: genotypic ratio = 1:2:1 and phenotypic ratio = 3:1 (See  Punnett Squares Below)

HC.

HD.

HE. Punnett Squares

Practice

F1

C. d

HF.

P1

Cro

ss

HG.

D

A.  Cr

HH.

D

os

s

B. D

E. __ _

F. __

_

HI.d

HJ. ____

_

_____D. DHK.

G. d

H. __

_

I. __

_

HL.

d

HM.

_____

HN.

___

HO.

HP. Dihybrid Cross

• Simultaneously follow inheritance of 2 phenotypic characters 

• Generates an expected phenotypic ratio 9:3:3:1 

HQ.

HR. yellow, round X green, wrinkled

HS.

HT.

HU.

HV.

HW. F1

HX. All yellow, round

HY.

HZ.

IA.

IB. F1 X F1

IC. yellow, round X yellow round 

ID.

IE.

IF.

IG.

IH. 9/16 yellow, round 3/16 green, round

II.3/16 yellow, wrinkled 1/16 green, wrinkled 

IJ.

IK.

IL.

• As in monohybrid cross, more genotypes than phenotypes 

• Here, 9 genotypes underlie the 4 phenotypes 

­ GGWW, GGWw, GgWW, GgWw, GGww, Ggww, ggWW, ggWw, ggww • 9:3:3:1 ratio requires that 2 traits are fully independent 

• fig 3­7

IM.

IN.

IO.

 IP. Extend Principles of Independent Asortment into three-trait crosses, suing the forked line method  

IQ. Forked Line Method (Phenotypes)

IR.

IS.

IT. round (3/4) yellow, round IU. 3/4 X 3/4 = 0.57 (9/16) IV. yellow

IW.¾

IX. wrinkled (1/4)  yellow, wrinkled

IY. 3/4 X 1/4 = 0.19 (3/16) IZ.

JA.

JB.

JC. round (3/4)   green, round JD. 1/4 X 3/4 = 0.19 (3/16) JE. green 

JF. 1/4

JG. wrinkled (1/4)     green, wrinkled JH. 1/4 X 1/4 =  0.06 (1/16)

JI.

JJ.

JK. LECTURE 6

JL. Mendel’s 4th Postulate

JM.• ______________: different phenotype traits are inherited independently of one another JN.• The 2 alleles of one gene have an equal chance of segregating with either 2 alleles at  another gene

JO.• Need a min. of 2 different genes, each with 2 alleles 

JP.

JQ.What about a trihybrid cross? 

JR.• identical processes of segregation and independent assortment apply to 3 (or more)  traits 

JS. • Punnet Sqare with 64 boxes (crikey!), or forked line method 

(http://www.youtube.com/watch?v=nBTX2h2GYYw) 

JT. • Based on laws of probability and relationships between expected phenotypic ratios  JU.

JV. Expectations v. Observations

JW. • We can calculate expected phenotypic ratios from Mendel’s principals  (segregation, independent assortment) 

JX. • …and we can make observations of phenotypic ratios from experimental crosses JY.• Next, we need a way to assess the “fit” between observed v. expected ratios (i.e,  equivalent or not?)

JZ.

KA. Recap: Mendel’s Postulates  

KB. 1. Alleles of a gene occur in __________.  

KC. 2. _________/__________ : one phenotype dominates KD. 3. _________________: during meiosis, alleles of a gene separate  and go to different gametes randomly  

KE. 4. ____________________ : different phenotypic traits are inherited  independently of one another

KF.- rule or inference only possible following experiments that follow the  inheritance of 2 different phenotypes separately such as pod color and  pod shape  

KG.

KH. • If we want to assess whether Mendelian rules of inheritance are true, we need a sufficient number of observations to really assess that  - why larger sample sizes converge on true answer better than small  sample sizes  

KI.

 KJ. Perform Chi-Squared Analyses, Determine P-values, and interpret them with respect to a null hypothesis  

KK. Testing Mendel’s Principles

KL. • Assuming dominant/recessive, we can test for  ___________________ and _______________ _____________________.  KM. • A dihybrid cross needs a larger sample size than monohybrid  because there are more combinations of phenotypes  

KN.

KO. Q: What is your null hypothesis when performing chi-square  analyses  

KP.A: The null hypothesis would be whatever ratios you would get under  Mendelian principles (Monohybrid: 3:1 genotype; Dihybrid: 9:3:3:1  genotype)

- you either ____________ your null hypothesis or _______ ____ __________  (never accept a null hypothesis for Mendelian X^2 Analyses)  KQ.

KR. Q: What is the Chi-Square equation?  

KS. A: X^2 = ∑((o – e)^2 / e) where ____ represents observed and  ______ represents expected  

KT.

KU. Q: How do you find your degrees of freedom?  

KV. A: Count the number of classes (ex: round, wrinkled, etc.) and  subtract by 1 (n – 1)  

KW.

KX. Q: How do you use a chi square look up table?  

KY.A: After you determine your chi-square number and your degrees of  freedom number, look at the given table (FIG 3-10) and find which row  represents your degrees of freedom. If your df=1, look on the first row,  etc. Next, find where your X^2 value would fall on that row. If the P  value above that is _____________ 5%, or 0.05, reject the null hypothesis because this means that there is less than 5% chance of obtaining  observed data when the null hypothesis is true (aka we start to think  that there may be other factors such as crossing over and independent assortment messing up the data)  

KZ.

LA. Q: You mean to accept the null hypothesis when your p value is  > 5% (p > 0.05)

LB. A: NO!! This is when you would “Fail to reject” the null hypothesis LC.

LD.Using a X^2 test “look up graph” 

LE.• Chi square of 0.53

• locate area on X axis where 0.53 occurs

LF. • Df = 1

• diagonal lines represent degrees of freedom values

LG.• Find 0.53 on x­axis, draw line up to df 1 diagonal line

LH.• Cut across to y axis; provides estimate of p­value 

LI. FIG 3­10

LJ.

LK.Reasons for Rejecting a Null Hypothesis (P<0.05)

LL. • One or more of the underlying assumptions do not hold: 

LM. 1) dominance/recessiveness, 2) segregation, and 3) independent  assortment  

LN.

LO.

LP. LECTURE 7

LQ. Use Pedigrees to Infer Mode of Inheritance of a Trait LR. Pedigrees

LS. • __________________ – is the phenotype of interest dominant or  recessive? Is it an autosomal or X-linked chromosome?  

- Recorded histories of family trees

- Circles are females, squares are males, diamond for unkown sex - Shade shape if individual is expressing a phenotype (ex: albino) - Parents are unrelated – ________ line

- Related parents – _________ line

- Numbers – represent _______________  

LT.

LU. Pedigree Analysis

LV. • Key point – ususally does not provide same level of certainty as  carefully designed _______________________

LW. • Larger pedigrees allow _________ inferences to be made by  providing greater confidence in __________ of ______________.  LX. • Multiple pedigrees _______________________.  

LY.

LZ. Modification of Mendelian Ratios Ch. 4

MA. After Mendel:  

MB. Research started to focus on traits that do not follow the simple  Mendelian model  

MC. • Although the basic mechanism of inheritance is the same,  other assumptions did not always hold

- Some traits are influenced by more than one gene

- Can also be influenced by the environment  

- Maybe dominant/recessive doesn’t have same effect such as when  occurring in the

MD. heterozygous form  

ME.

MF. • _______________ – the (often dominant) allele carried by the  most common phenotype in a population (common and usually  dominant)

MG. • ___________ is the source of new genetic variants (mutant  alleles) and are more rare

MH.

MI.

MJ.More Genotype Notation

MK. • Capital letter alleles don’t have to represent dominance only;  they could also represent a ________________ allele (e.g. as D+) ML. • Also don’t always have to represent with one alleles (e.g., the  Wr allele) which can also be separated by a slash (e.g. Wr+/Wg MM.

MN.

1. Single-Locus Phenomena

MO. - single locus/gene and one gene is controlling a specific trait  MP. - departures from Mendelian genotypic ratios  

MQ.

MR. Incomplete Dominance  

MS. • Offspring’s phenotype is an ______________________ of the  parents (think calico cats)

MT. • In a monohybrid cross, each of 3 genotypes (homozygous  dominant, heterozygous, homozygous recessive) yields a  ___________________________.  

MU. • Expected phenotypic and genotypic ratios = ____________ MV.

MW. Co-Dominance

MX. • Offspring’s phenotype is different form the 2 parents,  ___________________

- Joint expression of both alleles in ___________  

- Expected genotypic and phenotypic ratios in monohybrid cross = 1:2:1 MY.

MZ. Multiple Alleles  

NA. • Strictly a population-level phenomenon

NB. • Ex: Human ABO blood group – controlled by a single autosomal  locus with 3 alleles (I^A, I^B, and I^O)

- 3 alleles = ________ possible diploid genotypes

NC.

ND. • These 6 genotypes yield 4 alternative phenoytpes  - A&B is _______ dominant to O but _______________ to each other  1 AO or AA – _______ is expressed  

2 AB – both expressed

3 BO or BB – _______ is expressed  

4 OO – neither A nor B expressed

NE.

NF.Recessive Lethal Alleles

NG. • Selective disadvantage (or advantage) of a given allele causes deviation from  Mendelian ratios

NH.      • Some ___________________ can be tolerated by     ______________,     but not  _____________ 

NI. • ____________________: individuals that are homozygous for the recessive allele have a short life expectancy

• sometimes a single copy of the recessive allele is enough to cause problems  NJ.

NK. Agouti has complex inheritance

NL.• This locus affects two traits (hair color and embryonic development);  dominance/recessiveness switches 

• The yellow allele (AY) is a lethal recessive (AY/AY is never seen – embryos die before birth) 1 homozygotes with that genotype die very young

2

• AY is dominant with respect to coat color (AY/A survives but has an unusual yellow coat) • The dual action of this locus on survival and coat color cause __________________ from  Mendelian ratios 

1 ­ we know from a series of crosses that if we cross a pair of agouti mice (AA X AA), all  offspring will show the agouti color, and therefore they all survive. 

2

NM. yellow (AAY) X yellow (AAY) = _____ yellow and _____ agouti, the survivors (the  other 1/3 die due to detrimental homozygote genotype so you do not account for these in ratios) 

NN.

NO. agouti (AA) X yellow (AAY) = ½ agouti and ½ yellow (all survive) NP.

NQ.

NR. Dominant Lethal Alleles

NS. • ____________________: degeneration of nerve cells in the brain, caused by a run of “CAG” repeat, chromosome 4. 

NT.• Usually late onset (around 40 years) in heterozygotes

NU. • Alleles can persist in a population when carrier individuals survive long enough to  reproduce.

NV.

NW. Q: So, Why has Huntington’s diseas not been weeded out of the genepool?  NX. A: because of those who are heterozygous for the trait have a later onset and can  mate before showing signs or symptoms. 

NY.

NZ.Genes on Sex Chromosomes

OA.      • __________: compared to autosomal genes, those on the X­chromosome have  unique inheritance 

OB. • ___________________ only has ONE copy (e.g. XY males, in mammals) OC.

OD. X­linkage and Drosophila White eyes

OE.

OF. Sex­Limited Traits

OG. • Autosomal genes that act differently in sexes

OH. • e.g., feathers of male and female chickens, where a certain phenotype found only  in one of the sexes

• Hormones determine expression 

OI.

OJ.Sex­Influenced Traits

OK. • Autosomal genes that act differently in sexes, but the phenotype is not limited to  only one of the sexes

J.

K.

L. G

e

n

ot

y

p

e

M.

N.

O. F

e

m

a

l

e

P.

Q.

R. M

a

l

e

S.

T.BB

U.

V. B

a

l

d

W.

X. B

a

l

d

Y.

Z. B

b

AA.

AB.

Not

AC.

AD.

Bald

AE.

AF.b

b

AG.

AH.

Not

AI.

AJ. N

o

t

Principles 

PF.

PG. LECTURE 8

PH.

OL.ex: baldness 

OM.

ON.

OO.

OP.

OQ.

OR.

OS.

OT.

OU.

OV.

OW.

OX.

OY.

OZ.

PA.

PB.

PC.

PD. • Mitochondria and Chloroplasts have their  own protein­coding genes, so can affect  phenotype 

PE.• Since they are haploid and uniparentally inherited, ___________ follow basic Mendelian 

2. Locus-by-locus interactions

PI.

PJ. Physical Linkage

- A form of ___________________________

- Two genes that occur closely on the same chromosome will have  alleles hthat exhibit ____________ because they tend to be inherited as a single unit

- Probability of crossing over to separate the 2 is possible but not very  likely  

PK.

PL.Locus-By-Locus Interactions

PM. • Many phenotypic traits are controlled by more than one gene PN. • interactions among genes and their products can be direct:  - In some cases, alleles at one locus __________ the expression of alleles  at another (epistasis)

- __________________ are masked by an epistatic locus  

PO.

PP.Epistasis: The Bombay Phenotype

PQ. • The single locus that determines human ABO blood type can be impacted by a second locus, ___________  

- the recessive allele at ____________ causes individuals that are AA, BB,  or AB to express the O blood phenotype  

- ABO blood group are the _________________ alleles and that locus is the  ___________ locus  

PR. Ch. 5 –> Sex Determination and Sex Chromosomes  PS.

PT. Sexual Differentiation  

• sex chromosomes are the location of many (but not all) sex  determining genes

• mechanisms of sex-determination are _______________ among  eukaryotes (not just ‘XY’ vs. ‘XX’)

• differences between males (♂) and females (♀) (sexual dimorphism)  are separated into 2 classes:

PU. 1. ___________: gonads only

PV. 2. _____________: other traits related to gender

PW.

PX. Homo- vs. Heterogametic  

• the ________________ (XY) has a pair of dissimilar sex chromosomes (vs.  ___________________(XX))

• human males make _____________________ (some with X, others with Y)  that determine sex of offspring

PY.

• ______________________: failure of the X chromosomes to segregate  properly during meiosis

PZ.

QA. Q: What (usually) determines the sex of a human embryo? QB. A: presence v. absence of the Y chromosome

QC.

QD. ____________________________: paring and synapse/  recombination; rest of the Y (male-specific region) has no  counterparts on the X

QE.

QF. Q: What acts as a “switch” for expression of other genes?  QG. A: ___________: (where other 74 genes are) contains genes, incl.  sex-determining region Y (SRY)* present in all (male) mammals

- SRY codes for the testis-determining factor (TDF) protein, a ‘switch’ for  expression of other genes

- SRY has a large effect by either facilitating or preventing transcription  or translation of other genes

QH.

QI.Dosage Compensation

• females have the potential to produce 2x as much protein encoded by X-linked genes than males

• _______________________ mechanisms balance out the expression of X linked genes

- _____________ : inactivated X in diploid cells of females, visible during  mitotic interphase

QJ.

QK. • Mechanism in silencing the other X chromosome is usually by  _______________

- that X chromosome is never going to express any of its genes - this way, each gender has an equal amount of genes  - Early on both chromosomes contribute to gene expression until one is  silenced  

QL.

QM. Lyon Hypothesis  

• one X chromosome per ___________ cell is randomly ___________ during  embryonic development

• ________________ have the same X inactivated as their parent cell, but  not all parent cells are the same

QN.

QO. Temperature Dependent sex Determination  QP. • Not like that across tree of life – sex may not even be  genetically determined  

QQ. - Most common in reptiles that incupation temp sets the path  whehter indivudual will be male or female  

QR.

LECTURE 1

Recap on Vocab:  

________ – the study of heredity and variation

Heredity – ___________________________________ 

Variation – differences between individuals  

Understand which major discoveries underpin recent developments in genetics 

1600-1850

_______ – (epigenesis) egg differentiates into adult structures over time - Aka development

Pasteur – (spontaneous generation) living organisms do not derive from non living components  

_______ – (evolution) _______ ____ ____________, with _______ _________ as the  driver of change  

- Variation in a population  

- Competition for _______ 

- Advantages over others  

1600-1850

________ and “unit factors” – predictable patterns of inheritance: random  segregation, assortment  

• Chromosomal theory  

– _____________________________________________________ 

- process of ________ and _______ 

• Transmission Genetics – _____________________________________________ - chromosome and pedigree analysis  

Q: Who determined that DNA was the carrier of genetic info? What was  believed previously?

A: Avery at Al. – People believed previously that proteins were the carriers of  genetic info

Q: What about Watson and Crick?  

A: Watson and Crick discovered that DNA was organized as a double helix  with a phosphate backbone and nucleotide pairing rules (___ goes with T, and ___ goes with G)  

Describe relationships between components of the central dogma; contrast DNA, RNA, and proteins.  

THE CENTRAL DOGMA

 DNA ______________

____________

 ________________ Translation

(Matching Vocab)

A. Allele

B. Genotype

C. Gene

D. Genome

E. Phenotype

F.

G. 1. __________ how a person’s genotype interacts with environmental  factors (observable/physical characteristics)

H. 2. __________ section of DNA that codes for RNA to make proteins,  which in turn make us

I. 3. __________ an entire haploid set of chromosomes aka what it takes to build an entire organism (“blueprint”) all of the chromosomes in the  nucleus of a cell  

J. 4. __________ simple combination of the alleles at a locus K. 5. __________ different forms or variants of a gene  

L.

M. PROTEINS 

N. Definition: the end products of ________ _____________

O.

P. Remember: All _______________ are proteins but not all proteins are  _______________!

Q.

R. Q: What is Sickle Cell Anemia?  

S. A: Sickle Cell Anemia is a disorder that arises from a mutant gene  which produces a mutant protein that alters phenotype. The mutation  is __________________, which is a single nucleotide change that leads to  a different amino acid, and a ß-globin protein

T.

 U. Recall the series of advances that were possible following the development of recombinant DNA.  

 V. RECOMBINANT DNA 

W. Q: Explain DNA cloning.

X. A: _________________ cut DNA at specific nucleotide sites, then  ___________ (plasmids) take up the DNA fragments from the restriction  enzymes. After that, ___________________, such as bacteria, take up the  vector. As the bacteria colony grows, clones are made.

Y.

Z. Q: What are “Model Organisms” and their characteristics?  AA. A: Model organisms are a diverse set of organisms that can be  well understood from a scientific standpoint based on these  characteristics:  

1. ______________________

AB.

2. _______________________________

AC.

3. ______________________________________

AD.

 AE. BIOTECHNOLOGY 

AF.

AG. ___________________________

- crop plant resistance (to insect herbivores, viruses, etc), nutrient  enhancement, milk production, etc..  

- controversy:  

o engineered traits might “escape” into wild relatives

o patents on genes/living organisms is not OK

AH.

AI. ___________________________

- rapid screening of disease-associated alleles  

- determine risk of developing (or having a child with) a genetic disorder - develop individual-based therapy, based on observed expression of  genes (e.g., in cancer cells).  

AJ.

AK. NEW FIELDS 

AL.

A. Genomics

B. Proteomics  

C. Bioinformatics  

AM.

AN. 1. __________ identify proteins present in a cell at a given  time/location; modifications and interactions  

AO. 2. __________ filtering and analyses of very large genetic  datasets; software development  

AP.3. __________ structure and function of genomes (all of the DNA carried  in an organism), often comparative  

AQ.

AR. LECTURE 2

AS. Know the terminology (cells, chromosomes, etc.) 

AT. Cell Structure and Genetic Function

A. Nucleus

B. Organelles

C. Cytoplasm

D. Cytoskeleton

AU.

AV.1. __________ colloidal matrix

AW. 2. __________ mitochondria, ER, centriole, chloroplast AX. 3. __________ houses DNA

AY. 4. __________ lattice of tubules and filaments  

AZ.

BA. In the Nucleus

A. Chromatin

B. Chromosomes

C. Nucleolus

BB.

BC. 1. ___________ where ribosomal RNA (rRNA) is made and where  ribosomal subunits are initially built  

BD. 2. ___________ loosely or uncoiled DNA that is unattached from  histone; present when cells are NOT dividing  

BE. 3. ___________ condensed (super coiled) DNA with discrete  structures, able to be seen just before and during mitosis and meiosis,  otherwise they are more loosely packed  

BF.

BG. In the Cytoplasm

A. Smooth Endoplasmic Reticulum (SER)

B. Rough Endoplasmic Reticulum (RER)

C. Mitochondria

D. Chloroplasts

E. Centrioles

BH.

BI. 1. __________ maternally inherited organelle in charge of oxidative  respiration (make ATP); its chromosomes live outside of the nucleus  (no heterozygous or homozygous copies of genes because they are  haploid/diploid; “Powerhouse of cell”

BJ. 2. __________ sites of biochemical reactions such as fatty acid and  phospholipid synthesis  

BK. 3. __________ involved in photosynthesis in all plants, algae, and  some protozoa (but not all cells)

BL. 4. __________ site of biochemical reactions such as translation and covered in ribosomes  

BM. 5. __________ used in mitosis and meiosis – in charge of the  organization of spindle fibers which attach to and move chromosomes  BN.

BO. Explaining Chromsomes

BP.• When present as sister chromatids (2 chromatids), chromosomes  have a _____________ used for __________

BQ. • _________ – DNA that may or may not be transcribed or  translated  

BR.

BS.

BT.

BU.

BV.

BW.

BX.

BY.

BZ.

CA.

CB.

CC.

CD. 

CE.

CF.Homologous

Chromosomes

CG. • All members of a species have the same number of  ____________ per somatic cell

CH. • Occur in pairs, 1 from the maternal and 1 from the paternal  parent; matching _______, ____________, and __________________ CI. • Sex chromosomes (although dissimilar) still behave as ___________.  CJ.

CK. Karyotype

CL. • Humans:  

- Diploid (2n) # = _________

- Haploid (n) # = __________

CM.

 CN. Understand the function of mitosis  

CO. Mitosis

CP.• genetic continuity between generations of cells

CQ. • Parent and daughter cells have __________________ CR. Function of mitosis: replacing skin and hair cells; growing  CS.

CT. Describe steps involved in mitosis 

CU.

CV. Regular Function:  

CW. • ________ – cells performing normal function (ex: blood cells  carrying O2)

CX. • ________ – non-dividing cells withdrawn from the cycle heading  towards duplication; continue carrying out normal function CY.• ________ – where DNA replication takes place

CZ. • ________ – similar to G1; cell is still performing normally but  form of chromosomes has been altered.  

DA.

DB. Mitotic Division (PMAT)

DC. Q: What happens in Mitosis, Prophase?

DD. A: ___________ move to pole and organize microtubules and  spindle fibers, _________________ breaks, and ________________ are visible as pairs of genetically identical sister chromatids  

DE.

DF. Q: What happens in Mitosis, Metaphase?

DG. A: Chromosomes migrate to the central axis and then align along the metaphase plate

- Sister chromatids pulled apart in the ___________ direction  DH.

DI.Q: What happens in Mitosis, Anaphase?

DJ.A: Sister chromatids separate and migrate to opposite poles, separated sisters = ______________. Daughter cells will get ___________ sets of  chromosomes

DK.

DL. Q: What happens in Mitosis, Telophase?

DM. A: Cytoplasm starts to divide (cytokinesis), chromosomes uncoil  back to diffuse chromatin, __________________ reforms, and  _________________ disappear.  

DN. Order of Regular function and Mitosis:  

DO. Interphase

1. _________

2. G0  

3. _________

4. G2

DP. Mitosis

1. Prophase

2. ___________

3. ___________

4. ___________

DQ.

DR. LECTURE 3

DS. Mitosis Recap

1. Starts with _____________ (2n =4)

2. __________ – main events: chromosomes condense, nuclear envelope  degenerates, chromosomes move to opposite sides and lay down  spindle fibers  

3. __________ – 4 chromosomes each consisting of a pair of sister  chromatids line up in middle

4. __________ and __________ – creates 2 new diploid (___) daughter cells  DT.

 DU. Describe Steps involved in Meiosis  

DV.

DW. Function of Meiosis:

DX. • Continuity between generations of cells

DY.• Reduces number of chromosomes by ½: diploid –> haploid  DZ. • Passes genetic material from parent to offspring

EA. • Fusing ½ of each parent DNA allows them to reform to create  the next generation’s makeup

EB.

EC. • Meiotic cell cycle involves ONE ______________________ and TWO  _________________

- PMAT I and PMAT II

ED.

A. Prophase I

B. Metaphase I

C. Anaphase I

D. Telophase I

E. Prophase II

F. Metaphase II

G. Anaphase II

H. Telophase II

EE.

EF.

EG.

1. ___________ – Chromosomes mostly visible, homologous chromosomes  pair up (synapse) together, ________________ from physical contact  between chromatid of one chromosome with the chromatid of another  chromosome allows for crossing over

- synapsed homologs have how many chromatids? ________ 2. ___________ – chromosomes migrate and align in middle; equational  division starts

3. ___________ – tetrads pulled apart (one pair of _____________ to each  pole), poles now have random mix of maternal and paternal sister  chromatids that have been previously modified by crossing over

4. ___________ – starts with 2 separate daughter cells from meiosis I;  chromosomes condense, centrioles move to opposite poles, spindle  fibers laid down; no pairing of homologous chromosomes so no  ____________; chromosomes in each cell have already been duplicated  (are now a pair of sister chromatids)  

5. ___________ – 1st division (___________) ends here; nuclear membrane  forms around chromosome sets at each pole, cytoplasm begins to  pinch; interphase occurs briefly after without DNA replication

6. ___________ – chromosomes migrate and align in middle; reductional  division starts  

7. ___________ – 2nd division (___________) ends here; cytoplasm is divided,  nuclear membrane reforms; ultimately 4 haploid cells (gametes), each  with different genetic makeup, are produced from single meiotic cycle  

8. ___________ – dyads pulled apart (one __________ to each pole); poles  now have random mix of chromatids from each chromosome; each full  (hapoloid) set of the genome  

EH.

 EI.Contrast Mitosis vs. Meiosis

EJ. ** See class handout **

EK.

EL. LECTURE 4

EM. Contrast spermatogenesis v. oogenesis  

EN. Spermatogenesis  

EO. • General procedure of meiosis  

EP.

EQ. Male Gamete Production

ER. • Starts with ______________________________ (spermatogonium) ES. • Becomes an enlarged primary spermatocyte, which undergoes  1st (______________) meiotic division  

ET.• _____________________ undergoes 2nd meiotic division EU. • each make 2 haploid spermatids that develop into motile  sperm = total of 4 _________ sperm  

EV.

EW. Oogenesis  

EX. Female Gamete Production

EY.• Starts from ____________, which enlarges to become the primary  oocyte 

- primary oocyte then splits during ____________

EZ.

FA.• Daughter cells of the primary oocyte receive equal _________________,  but unequal _____________ (cellular volume)

- the cell that gets the least is called either the 1st or 2nd_______________  and does not continue through the process of oogenesis FB. • The cell that gets more cytoplasm is either called the  _____________ (after 1st division) or ______________ (after 2nd division) FC. • Oocytes that get shorted cytoplasm do not become eggs FD. • Just ___________________ (egg) produced per meiotic cycle  FE.

FF. Sperm v. Egg Production

FG. • Initially start off similar

– diploid parent cell undergoes growth and maturation

FH. • End of first division (reductional):

- first polar body in oogenesis

- 2 daughter cells as secondary spermatocytes in spermatogenesis  FI. • A single ovum produced at end of oogenesis vs. 4 haploid sperm at  end of spermatogenesis  

- haploid gametes produced are genetically different from parent cell  FJ.

FK. Spontaneous DNA Mutations

FL.Usually caused by rare erros during DNA replication, at low, on-going  “background rates” like a slow ticking clock

FM. • Most mutations are _________ (such as in non-coding DNA so  there is no phenotypic effect)

FN. • Most mutations in coding regions are detrimental, only some  are beneficial  

- most common outcome: production of an allelic variant that does not  function as well as original DNA  

FO.

FP. Q: What are the different Mutations in Protein-Coding genes?  FQ. A: __________ – change in a codon and in its amino acid –>  changes amino acid/protein and ____________ – change in a codon that  creates a premature STOP codon so protein synthesis is terminated  FR.

FS. Q: What could create new allelic variants that did not previously  exist?

FT.A: Crossing over between parts of homologous chromosomes in  ______________ because it creates “mosaic-like” chromsomes FU.

FV.Q: What is independent (random) assortment?

FW. A: This is the ultimate outcome of meiosis because each haploid  gamete is not a clone of its diploid parent.  

FX.

FY.Q: Why is it important to know about sickle-cell anemia when learning  about genetics?

FZ. A: Sickle cell anemia is caused by a single DNA substitution,  which causes a mutant protein to be produced, which in turn alters the  phenotype of an individual, effecting his or her red blood cells.  GA.

GB. Q: What is the difference between a population and a gene pool  GC. A: ______________ – group of individuals living together that have  capacity to interbreed and can also compete for limited resources;  ______________ – total number of allelic variants of a gene that exist  within a particular group of individuals  

GD.

GE. LECTURE 5

 GF. Understand Mendel’s 4 Postulates 

GG. Q: Who was Gregor Mendel?

GH. A: A monk who experimented with pea plants while working in a  monastery garden – discovered units of inheritance by observing  similarities in ______________ between one generation to the next  without any knowledge of _________________________  

GI.

GJ.Q: What traits did he observe were encoded in the same gene?  GK. A: flower color, position of flower, seed color, seed shape, pod  shape, pod color, and stem length. Pea plants are a great model  organism because they are easy to cross, fast growing, and have  observable traits  

GL.

GM. Q: What does “true breeding” mean?

GN. A: This means that the individuals were homozygous for a gene  GO.

GP. Monohybrid Cross

GQ. • Mating between individuals that differ in ________ trait  GR. • P1 – parentals – true breeding green or true breeding yellow  (RR and rr)

- F1 – 1st generation offspring –> ______ % yellow (all Rr)

- F2 – 1st generation crossed with each other –> ______ % yellow, _______  % green (RR, rr, and Rr)

o _________ ratio

o Mendel’s Conclusion: proposed existence of “___________”  – things that pass unchanged from one generation to the next,  and determine the phenotype that is expressed

GS.

GT. Mendel’s First 3 postulates

GU. 1. ________ of a gene: occur in pairs in an individual; only applies  to diploid organisms (including peas)

GV. 2. ________________ : one phenotype dominates over the other  (not true of all genes)

GW. 3. ________________ : during meiosis, chromosomes on which  alleles reside separate and go to different gametes randomly  GX.

 GY. Use a Punnett Square for Mono- and Di-hybrid crosses GZ.

HA. Q: If a cross between P1 resulted in an F1 generation of all  heterozygotes, what will the F2 generation look like genotypically and  phenotypically?

HB. A: genotypic ratio = 1:2:1 and phenotypic ratio = 3:1 (See  Punnett Squares Below)

HC.

HD.

HE. Punnett Squares

Practice

F1

C. d

HF.

P1

Cro

ss

HG.

D

A.  Cr

HH.

D

os

s

B. D

E. __ _

F. __

_

HI.d

HJ. ____

_

_____D. DHK.

G. d

H. __

_

I. __

_

HL.

d

HM.

_____

HN.

___

HO.

HP. Dihybrid Cross

• Simultaneously follow inheritance of 2 phenotypic characters 

• Generates an expected phenotypic ratio 9:3:3:1 

HQ.

HR. yellow, round X green, wrinkled

HS.

HT.

HU.

HV.

HW. F1

HX. All yellow, round

HY.

HZ.

IA.

IB. F1 X F1

IC. yellow, round X yellow round 

ID.

IE.

IF.

IG.

IH. 9/16 yellow, round 3/16 green, round

II.3/16 yellow, wrinkled 1/16 green, wrinkled 

IJ.

IK.

IL.

• As in monohybrid cross, more genotypes than phenotypes 

• Here, 9 genotypes underlie the 4 phenotypes 

­ GGWW, GGWw, GgWW, GgWw, GGww, Ggww, ggWW, ggWw, ggww • 9:3:3:1 ratio requires that 2 traits are fully independent 

• fig 3­7

IM.

IN.

IO.

 IP. Extend Principles of Independent Asortment into three-trait crosses, suing the forked line method  

IQ. Forked Line Method (Phenotypes)

IR.

IS.

IT. round (3/4) yellow, round IU. 3/4 X 3/4 = 0.57 (9/16) IV. yellow

IW.¾

IX. wrinkled (1/4)  yellow, wrinkled

IY. 3/4 X 1/4 = 0.19 (3/16) IZ.

JA.

JB.

JC. round (3/4)   green, round JD. 1/4 X 3/4 = 0.19 (3/16) JE. green 

JF. 1/4

JG. wrinkled (1/4)     green, wrinkled JH. 1/4 X 1/4 =  0.06 (1/16)

JI.

JJ.

JK. LECTURE 6

JL. Mendel’s 4th Postulate

JM.• ______________: different phenotype traits are inherited independently of one another JN.• The 2 alleles of one gene have an equal chance of segregating with either 2 alleles at  another gene

JO.• Need a min. of 2 different genes, each with 2 alleles 

JP.

JQ.What about a trihybrid cross? 

JR.• identical processes of segregation and independent assortment apply to 3 (or more)  traits 

JS. • Punnet Sqare with 64 boxes (crikey!), or forked line method 

(http://www.youtube.com/watch?v=nBTX2h2GYYw) 

JT. • Based on laws of probability and relationships between expected phenotypic ratios  JU.

JV. Expectations v. Observations

JW. • We can calculate expected phenotypic ratios from Mendel’s principals  (segregation, independent assortment) 

JX. • …and we can make observations of phenotypic ratios from experimental crosses JY.• Next, we need a way to assess the “fit” between observed v. expected ratios (i.e,  equivalent or not?)

JZ.

KA. Recap: Mendel’s Postulates  

KB. 1. Alleles of a gene occur in __________.  

KC. 2. _________/__________ : one phenotype dominates KD. 3. _________________: during meiosis, alleles of a gene separate  and go to different gametes randomly  

KE. 4. ____________________ : different phenotypic traits are inherited  independently of one another

KF.- rule or inference only possible following experiments that follow the  inheritance of 2 different phenotypes separately such as pod color and  pod shape  

KG.

KH. • If we want to assess whether Mendelian rules of inheritance are true, we need a sufficient number of observations to really assess that  - why larger sample sizes converge on true answer better than small  sample sizes  

KI.

 KJ. Perform Chi-Squared Analyses, Determine P-values, and interpret them with respect to a null hypothesis  

KK. Testing Mendel’s Principles

KL. • Assuming dominant/recessive, we can test for  ___________________ and _______________ _____________________.  KM. • A dihybrid cross needs a larger sample size than monohybrid  because there are more combinations of phenotypes  

KN.

KO. Q: What is your null hypothesis when performing chi-square  analyses  

KP.A: The null hypothesis would be whatever ratios you would get under  Mendelian principles (Monohybrid: 3:1 genotype; Dihybrid: 9:3:3:1  genotype)

- you either ____________ your null hypothesis or _______ ____ __________  (never accept a null hypothesis for Mendelian X^2 Analyses)  KQ.

KR. Q: What is the Chi-Square equation?  

KS. A: X^2 = ∑((o – e)^2 / e) where ____ represents observed and  ______ represents expected  

KT.

KU. Q: How do you find your degrees of freedom?  

KV. A: Count the number of classes (ex: round, wrinkled, etc.) and  subtract by 1 (n – 1)  

KW.

KX. Q: How do you use a chi square look up table?  

KY.A: After you determine your chi-square number and your degrees of  freedom number, look at the given table (FIG 3-10) and find which row  represents your degrees of freedom. If your df=1, look on the first row,  etc. Next, find where your X^2 value would fall on that row. If the P  value above that is _____________ 5%, or 0.05, reject the null hypothesis because this means that there is less than 5% chance of obtaining  observed data when the null hypothesis is true (aka we start to think  that there may be other factors such as crossing over and independent assortment messing up the data)  

KZ.

LA. Q: You mean to accept the null hypothesis when your p value is  > 5% (p > 0.05)

LB. A: NO!! This is when you would “Fail to reject” the null hypothesis LC.

LD.Using a X^2 test “look up graph” 

LE.• Chi square of 0.53

• locate area on X axis where 0.53 occurs

LF. • Df = 1

• diagonal lines represent degrees of freedom values

LG.• Find 0.53 on x­axis, draw line up to df 1 diagonal line

LH.• Cut across to y axis; provides estimate of p­value 

LI. FIG 3­10

LJ.

LK.Reasons for Rejecting a Null Hypothesis (P<0.05)

LL. • One or more of the underlying assumptions do not hold: 

LM. 1) dominance/recessiveness, 2) segregation, and 3) independent  assortment  

LN.

LO.

LP. LECTURE 7

LQ. Use Pedigrees to Infer Mode of Inheritance of a Trait LR. Pedigrees

LS. • __________________ – is the phenotype of interest dominant or  recessive? Is it an autosomal or X-linked chromosome?  

- Recorded histories of family trees

- Circles are females, squares are males, diamond for unkown sex - Shade shape if individual is expressing a phenotype (ex: albino) - Parents are unrelated – ________ line

- Related parents – _________ line

- Numbers – represent _______________  

LT.

LU. Pedigree Analysis

LV. • Key point – ususally does not provide same level of certainty as  carefully designed _______________________

LW. • Larger pedigrees allow _________ inferences to be made by  providing greater confidence in __________ of ______________.  LX. • Multiple pedigrees _______________________.  

LY.

LZ. Modification of Mendelian Ratios Ch. 4

MA. After Mendel:  

MB. Research started to focus on traits that do not follow the simple  Mendelian model  

MC. • Although the basic mechanism of inheritance is the same,  other assumptions did not always hold

- Some traits are influenced by more than one gene

- Can also be influenced by the environment  

- Maybe dominant/recessive doesn’t have same effect such as when  occurring in the

MD. heterozygous form  

ME.

MF. • _______________ – the (often dominant) allele carried by the  most common phenotype in a population (common and usually  dominant)

MG. • ___________ is the source of new genetic variants (mutant  alleles) and are more rare

MH.

MI.

MJ.More Genotype Notation

MK. • Capital letter alleles don’t have to represent dominance only;  they could also represent a ________________ allele (e.g. as D+) ML. • Also don’t always have to represent with one alleles (e.g., the  Wr allele) which can also be separated by a slash (e.g. Wr+/Wg MM.

MN.

1. Single-Locus Phenomena

MO. - single locus/gene and one gene is controlling a specific trait  MP. - departures from Mendelian genotypic ratios  

MQ.

MR. Incomplete Dominance  

MS. • Offspring’s phenotype is an ______________________ of the  parents (think calico cats)

MT. • In a monohybrid cross, each of 3 genotypes (homozygous  dominant, heterozygous, homozygous recessive) yields a  ___________________________.  

MU. • Expected phenotypic and genotypic ratios = ____________ MV.

MW. Co-Dominance

MX. • Offspring’s phenotype is different form the 2 parents,  ___________________

- Joint expression of both alleles in ___________  

- Expected genotypic and phenotypic ratios in monohybrid cross = 1:2:1 MY.

MZ. Multiple Alleles  

NA. • Strictly a population-level phenomenon

NB. • Ex: Human ABO blood group – controlled by a single autosomal  locus with 3 alleles (I^A, I^B, and I^O)

- 3 alleles = ________ possible diploid genotypes

NC.

ND. • These 6 genotypes yield 4 alternative phenoytpes  - A&B is _______ dominant to O but _______________ to each other  1 AO or AA – _______ is expressed  

2 AB – both expressed

3 BO or BB – _______ is expressed  

4 OO – neither A nor B expressed

NE.

NF.Recessive Lethal Alleles

NG. • Selective disadvantage (or advantage) of a given allele causes deviation from  Mendelian ratios

NH.      • Some ___________________ can be tolerated by     ______________,     but not  _____________ 

NI. • ____________________: individuals that are homozygous for the recessive allele have a short life expectancy

• sometimes a single copy of the recessive allele is enough to cause problems  NJ.

NK. Agouti has complex inheritance

NL.• This locus affects two traits (hair color and embryonic development);  dominance/recessiveness switches 

• The yellow allele (AY) is a lethal recessive (AY/AY is never seen – embryos die before birth) 1 homozygotes with that genotype die very young

2

• AY is dominant with respect to coat color (AY/A survives but has an unusual yellow coat) • The dual action of this locus on survival and coat color cause __________________ from  Mendelian ratios 

1 ­ we know from a series of crosses that if we cross a pair of agouti mice (AA X AA), all  offspring will show the agouti color, and therefore they all survive. 

2

NM. yellow (AAY) X yellow (AAY) = _____ yellow and _____ agouti, the survivors (the  other 1/3 die due to detrimental homozygote genotype so you do not account for these in ratios) 

NN.

NO. agouti (AA) X yellow (AAY) = ½ agouti and ½ yellow (all survive) NP.

NQ.

NR. Dominant Lethal Alleles

NS. • ____________________: degeneration of nerve cells in the brain, caused by a run of “CAG” repeat, chromosome 4. 

NT.• Usually late onset (around 40 years) in heterozygotes

NU. • Alleles can persist in a population when carrier individuals survive long enough to  reproduce.

NV.

NW. Q: So, Why has Huntington’s diseas not been weeded out of the genepool?  NX. A: because of those who are heterozygous for the trait have a later onset and can  mate before showing signs or symptoms. 

NY.

NZ.Genes on Sex Chromosomes

OA.      • __________: compared to autosomal genes, those on the X­chromosome have  unique inheritance 

OB. • ___________________ only has ONE copy (e.g. XY males, in mammals) OC.

OD. X­linkage and Drosophila White eyes

OE.

OF. Sex­Limited Traits

OG. • Autosomal genes that act differently in sexes

OH. • e.g., feathers of male and female chickens, where a certain phenotype found only  in one of the sexes

• Hormones determine expression 

OI.

OJ.Sex­Influenced Traits

OK. • Autosomal genes that act differently in sexes, but the phenotype is not limited to  only one of the sexes

J.

K.

L. G

e

n

ot

y

p

e

M.

N.

O. F

e

m

a

l

e

P.

Q.

R. M

a

l

e

S.

T.BB

U.

V. B

a

l

d

W.

X. B

a

l

d

Y.

Z. B

b

AA.

AB.

Not

AC.

AD.

Bald

AE.

AF.b

b

AG.

AH.

Not

AI.

AJ. N

o

t

Principles 

PF.

PG. LECTURE 8

PH.

OL.ex: baldness 

OM.

ON.

OO.

OP.

OQ.

OR.

OS.

OT.

OU.

OV.

OW.

OX.

OY.

OZ.

PA.

PB.

PC.

PD. • Mitochondria and Chloroplasts have their  own protein­coding genes, so can affect  phenotype 

PE.• Since they are haploid and uniparentally inherited, ___________ follow basic Mendelian 

2. Locus-by-locus interactions

PI.

PJ. Physical Linkage

- A form of ___________________________

- Two genes that occur closely on the same chromosome will have  alleles hthat exhibit ____________ because they tend to be inherited as a single unit

- Probability of crossing over to separate the 2 is possible but not very  likely  

PK.

PL.Locus-By-Locus Interactions

PM. • Many phenotypic traits are controlled by more than one gene PN. • interactions among genes and their products can be direct:  - In some cases, alleles at one locus __________ the expression of alleles  at another (epistasis)

- __________________ are masked by an epistatic locus  

PO.

PP.Epistasis: The Bombay Phenotype

PQ. • The single locus that determines human ABO blood type can be impacted by a second locus, ___________  

- the recessive allele at ____________ causes individuals that are AA, BB,  or AB to express the O blood phenotype  

- ABO blood group are the _________________ alleles and that locus is the  ___________ locus  

PR. Ch. 5 –> Sex Determination and Sex Chromosomes  PS.

PT. Sexual Differentiation  

• sex chromosomes are the location of many (but not all) sex  determining genes

• mechanisms of sex-determination are _______________ among  eukaryotes (not just ‘XY’ vs. ‘XX’)

• differences between males (♂) and females (♀) (sexual dimorphism)  are separated into 2 classes:

PU. 1. ___________: gonads only

PV. 2. _____________: other traits related to gender

PW.

PX. Homo- vs. Heterogametic  

• the ________________ (XY) has a pair of dissimilar sex chromosomes (vs.  ___________________(XX))

• human males make _____________________ (some with X, others with Y)  that determine sex of offspring

PY.

• ______________________: failure of the X chromosomes to segregate  properly during meiosis

PZ.

QA. Q: What (usually) determines the sex of a human embryo? QB. A: presence v. absence of the Y chromosome

QC.

QD. ____________________________: paring and synapse/  recombination; rest of the Y (male-specific region) has no  counterparts on the X

QE.

QF. Q: What acts as a “switch” for expression of other genes?  QG. A: ___________: (where other 74 genes are) contains genes, incl.  sex-determining region Y (SRY)* present in all (male) mammals

- SRY codes for the testis-determining factor (TDF) protein, a ‘switch’ for  expression of other genes

- SRY has a large effect by either facilitating or preventing transcription  or translation of other genes

QH.

QI.Dosage Compensation

• females have the potential to produce 2x as much protein encoded by X-linked genes than males

• _______________________ mechanisms balance out the expression of X linked genes

- _____________ : inactivated X in diploid cells of females, visible during  mitotic interphase

QJ.

QK. • Mechanism in silencing the other X chromosome is usually by  _______________

- that X chromosome is never going to express any of its genes - this way, each gender has an equal amount of genes  - Early on both chromosomes contribute to gene expression until one is  silenced  

QL.

QM. Lyon Hypothesis  

• one X chromosome per ___________ cell is randomly ___________ during  embryonic development

• ________________ have the same X inactivated as their parent cell, but  not all parent cells are the same

QN.

QO. Temperature Dependent sex Determination  QP. • Not like that across tree of life – sex may not even be  genetically determined  

QQ. - Most common in reptiles that incupation temp sets the path  whehter indivudual will be male or female  

QR.

Page Expired
5off
It looks like your free minutes have expired! Lucky for you we have all the content you need, just sign up here