Genetics ib

Genetics

What are the key components of chromosomes?

A. DNA-heterochromatin-euchromatin

B. ProteinsC. Found in nucleusD. You should

understand the relationship between DNA and proteins (chromatin packing and histones)

Key termsA. Eukaryotic chromosomes-made of DNA

and proteins (histones)B. Gene-heritable factor that controls

specific characteristics-made up of a length of DNA, found on a

specific chromosome location (a locus)C. Allele-one specific form of a gene (all

found at the same locus)-Example: Everyone has the gene for eye color. The possible alleles are blue, brown, green, etc.

More Key Terms

D. Genome-total genetic material of an organism or species (Example: The Human Genome)

E. Gene pool-total of all genes carried by individuals in a population

Mutations

A. Chromosome mutations-involve large sections of chromosomes (or the whole thing)-Ex: Down’s syndrome, Turner’s syndrome

B. Gene mutation-involves changes in single base pairs-Some mutations may not have any effect on the

cell and may involve:1. part of the sense strand of DNA which is not transcribed2. part of the DNA that a cell does not use3. changes in second or third bases of a codon (since the genetic code is

degenerate the same base may still be coded for)

Mutations

MutationsB. Gene mutation-involves changes in single base pairs

Example: Insertion or deletion of single organic bases -changes the DNA sequence that will be

transcribed and translated

original DNA sequence: ATG-TCG-AAG-CCC transcribed: UAC-AGC-UUC-GGG

translated: tyr-ser-phe-gly

addition of base A: ATA-GTC-GAA-GCC-Ctranscribed: UAU-CAG-CUU-CGG translated: thy-glu-leu-arg

A. Hemoglobin-protein that helps RBC carry oxygen

B. Hb is a gene that codes for hemoglobin-made of 146 amino acids

C. In some cases one base is substituted for anothernormal: (HbA) base substitution: (HbS)

CTC CACGAG GUG

-after transcription and translation HbA produces glutamic acid and HbS produces valine

Mutations: Base substitutions and sickle-cell anemia

D. The altered hemoglobin HbS is crystalline at low oxygen levels causing the RBC to become sickled and less efficient at oxygen transport

E. Symptoms of sickle cell anemia-physical weakness-heart or/and kidney damage-death

Mutations: base substitutions and sickle-cell anemia

F. In heterozygous people (one normal allele and one sickle cell allele)-the alleles are codominant, but the normal allele is expressed more strongly

-in codominance both alleles are expressed (one is not dominant to the other)-some sickled cells present, but most are normal-some people show mild anemia (deficiency of the hemoglobin, often accompanied by a reduced number of red blood cells and causing paleness, weakness, and breathlessness)

Mutations: base substitutions and sickle-cell anemia

G. Advantages of being heterozygous-In areas where malaria is infested:

-Plasmodium cannot live in erythrocytes with HbS

-Heterozygous individuals have a reduced chance of contracting the protist that is carried by mosquitoes

Mutations: base substitutions and sickle-cell anemia

KaryotypingA. Karyotyping-process of finding the

chromosomal characteristics of a cell-chromosomes are stained to show banding

and arranged in pairs according to size and structure

-You should be able to look at a karyotype and determine the sex of the individual and if non-disjunction has occurred

Amniocentesis and karyotyping

A. Amniocentesis-performed at around 16 wks-sample of amniotic fluid is taken and cultured-when there are enough cells, a karyotype can be

performed-chromosomes are arranged into pairs to detect

any abnormalities-can be used to detect Down’s syndrome (a.k.a.

trisomy 21) -can be used to recognize sex or non-disjunction

Amniocentesis

Karyotyping and Chorionic villus sampling

Sampling is performed around 11 weeks of pregnancyCells are gathered from chorionic villi (cells from the zygote)Cells are cultured, DNA is extracted and a karyotype is

made

A Side by Side Comparison

Meiosis Key TermsA. Diploid-having two sets of chromosomesB. Homologous chromosomes-matching

pairs of chromosomes-have the same genes-are not identical (one chromosome comes from each parent, thus alleles may be different)-found in diploid cells

C. Haploid-having only one set of chromosomes

More Meiosis Terms

D. Chromatids -two parts of a chromosomeE. Centromere -part of a chromosome that

connects the chromatidsF. Reduction division-in organisms that reproduce

sexually -reduction of the number of chromosomes by half (from a diploid nucleus to a haploid nucleus)-think of eggs and sperm; both are haploid (when they unite the diploid number is restored)

Meiosis

B. Produces gametes (sperm and egg)C. Overview

1. homologous chromosomes pair (diploid)2. two divisions (meiosis I and meiosis II)3. result=4 haploid cells

D. When gametes unite (to produce a diploid cell) a cell with homologous pairs is created

-one set of chromosomes is from the mom and one set of chromosome is from the

dad

Meiosis (details)A. Interphase -DNA replication

B. Prophase I-chromosomes condense-nucleolus becomes visible-spindle formation-synapsis-homologous chromosomes are side

by side (they become a tetrad and are intersected at the chiasmata)

-nuclear membrane begins to disappear

chiasmata

Tetrad

C. Metaphase I: Bivalents move to equator

D. Anaphase I: -Homologous pairs split-One chromosome from each pair moves to opposite pole

E. Telophase I:-Chromosomes arrive at poles-Spindle disappears

Meiosis (details)

F. Prophase II – new spindle is formed at right angles to the previous spindle

G. Metaphase II – Chromosomes move to equator

H. Anaphase II – -Chromosomes separate-Chromatids move to opposite poles

Meiosis (details)

I. Telophase II-Chromosomes arrive at poles-Spindle disappears-Nuclear membrane reappears-Nucleolus becomes visible-Chromosomes become chromatin-Cytokinesis

Meiosis (details)

A. An important source of variation-creates new combinations

B. Happens during prophase I C. Called a synapsisD. Recombination-reassortment of genes into

different combinations from those of the parents

Crossing over of homologous chromosomes

Chiasmata formed during a synapsis

CrossoverABCD

H E

h eABCD

H E

h e

ABCD

H E

h e

ResultsA genotype HEB genotype HeC genotype hED genotype he

StartA genotype HEB genotype HEC genotype heD genotype he

CrossoverB and C become

recombinants

Meiosis and genetic variationA. The number of possible gametes

produced by random orientation of chromosomes is 2n (where n is = to the haploid number of chromosomes)

B. In humans the production of 1 gamete has over 8 million possible combinations (223)

C. Recombination (in prophase I) + Random orientation of chromosomes (in metaphase I) = an infinite number of variations

Meiosis and Non-disjunctionA. Disjunction - when the homologous

chromosomes separate in anaphase IB. Aneuploidy -happens when chromosomes

do not separate (in anaphase I or II)-caused by non-disjunction-result: one cell missing a chromosome and one

cell having an extra chromosome-Total number of chromosomes = 2n ±1

C. Polyploidy- having more than two complete sets of chromosomes (common in plants)

Karyotype of non-disjunction

Normal karyotype (2n=46)Abnormal karyotype (aneuploidy)

2n + 1 = 47

Non-disjunction and Down’s syndrome

A. One of the parent gametes contains two copies of chromosome 21

B. The zygote then ends up with 3 copies-2 from one parent-1 from the other

C. Down’s syndrome = trisomy 21D. Chances of non-disjunction of

chromosomes increases with age in females (in males too, but less of an effect)

Non-disjunction and Down’s syndrome

E. Female age has a greater effect because:-gametes are produced before birth-more exposure to chemicals, radiation, etc.

F. Male age has less effect because they do not produce gametes until puberty

G. Genetics may also increase the likelihood of having a child with Down’s syndrome

Theoretical Genetics Key TermsA. Dominant allele-the allele that always shows in

the heterozygous state (Example: Bb=brown)B. Recessive allele-the allele that only shows in

the homozygous recessive state (Example: bb=blue)

C. Codominant alleles-pairs of alleles where two differing alleles are shown in the phenotype in a heterozygote

D. Homozygous -having two identical alleles of a gene (Example: BB or bb)

E. Heterozygous -having two different alleles of a gene (Example: Bb)

More VocabularyF. Carrier- a person who has a recessive

allele, but does not express it (they are generally heterozygous, Bb)

G. Genotype-alleles that a person has(the letters) Ex: Bb

H. Phenotype- the physical characteristics the a person shows (caused by the genotype) Ex: brown hair or blue eyes

I. Test cross- crossing two or more genotypes to find the possible genetic outcomes

Mendel’s Monohybrid Crosses

A. Punnett square-shows possible outcomes from a test cross

B. Mendel studied characteristics of pea plants

Wrinkled and round peas(round peas are dominant)

TraitDominant

ExpressionRecessiveExpression

1. Form of ripe seed Smooth Wrinkled

2. Color of seed albumen Yellow Green

3. Color of seed coat Grey White

4. Form of ripe pods Inflated Constricted

5. Color of unripe pods Green Yellow

6. Position of flowers Axial Terminal

7. Length of stem Tall Dwarf

Gregor Mendel’s Findings

Mendel’s Monohybrid Crosses

C. Mendel found tall is dominant over shortD. His procedures were:

1. Start with 2 pure breeding homozygous plants (This is the P generation.)

-Plant 1=tall (TT)-Plant 2= short (tt)

2. Cross-breed the plants -Place pollen from the tall plant in the

short plant and vice versa.

Mendel’s Monohybrid Crosses

D. His procedures were:3. The F1 generation is the 1st group of offspring.

-All were tall, and had a heterozygous genotype. (Tt)

-This is an application of the law of segregation. -All offspring had a ‘T’ from one

parent and a ‘t’ from the other parent

Mendel’s Monohybrid CrossesD. His procedures were:

4. The F1 offspring were then crossed. (Tt x Tt)-Possible outcomes can be found using a Punnett squareGENOTYPES-25% TT-50% Tt-25% ttPHENOTYPES-75% Tall -25% Short

T t

T TT Tt

t Tt tt

Multiple allelesA. When genes have more than two allelesB. Example: Blood type has 4 phenotypes based

on three alleles (IA, IB and i)C. IA and IB are codominant and i is recessiveD. This is why parents can have kids with

different blood types

Phenotypes GenotypesA IAIA or IAiB IBIB or IBiAB IAIB

O ii

Multiple allelesD. Perform a test cross for P: mother with O blood

type and father with AB blood type. What are the possible phenotypes?

Phenotypes GenotypesA IAIA or IAiB IBIB or IBiAB IAIB

O ii

Perform a test cross for P: mother with O blood type and father with AB blood type. What are the possible phenotypes for F1?

P = ii x IAIB

i iIA IA i IA i

IB IB i IB i

None of the children can have the same blood type as either of the parents.

Multiple alleles

CodominanceA. When neither allele for a gene is recessiveB. Example: Blood typeC. Alleles A and B are both dominant (both are

expressed)D. i is recessive to alleles A and BE. One letter is chosen and the possible alleles

are written in upper case letters to illustrate codominance Phenotypes Genotypes

A IAIA or IAiB IBIB or IBiAB IAIB

O ii

Sex chromosomes and genderOnly possibility for P generation = XX and XY

-The X chromosome is larger and carries more genes than the Y chromosome -Examples of genes on X, but not Y = color

blindness and hemophilia-Many sex-linked traits are related to the X

chromosome.

X XX XX XX

Y XY XY

The sex of all babies is determined by the chromosomes in the sperm from the man.

Sex linkage

A. Genes carried on sex chromosomes (usually X)

B. Example: Hemophilia-a blood disorder that prevents clotting

-patients do not produce clotting factors that allow coagulation of blood, and thus torn blood vessels are prevented from closing

-most common in boys (they get it from their mother’s X chromosome, as they only get one X, which means only one chance to get the dominant allele)

Sex linkageC. Two parents without hemophilia:

XHXh and XHY

*The XhXh does is very rare*Males cannot be heterozygous carriers because they

only have one X.*Females can be carriers and pass on the trait to the next

generation. They can be heterozyg. or homozyg.* XH -Normal and Xh –Hemophilia

XH Xh

XH XHXH XHXh

Y XHY XhY

Predict the genotypic and phenotypic ratios of monohybrid crosses for each of the following.

1. Sickle cell anemia: HbA=normal and HbS=sickle cellHbAHbS x HbAHbS

2. Colorblindness: XB=normal and Xb=colorblindXBXb x XbY

3. Hemophilia: XH=normal and Xh=hemophiliaXhXh x XHY

4. Blood type: IAi x IBi**You should be able to determine if alleles are codominant

because both alleles will be represented by capital letters. You should also know if the inherited traits are sex-linked.

Mendel’s Law of Segregation

A. States: The separation of the pair of parental factors, so that one factor is present in each gamete. (This is how it is written in the IB book.)

B. The two alleles for each characteristic segregate during gamete production. This means that each gamete will contain only one allele for each gene. This allows the maternal and paternal alleles to be combined in the offspring, ensuring variation. (This is from wikipedia.)

Mendel’s Law of Segregation and Meiosis

A. Mendel looked at genes (on chromosomes)B. Found that each gene appeared twice (in

homologous pairs)C. Figured out that when a synapsis occurs in

prophase I followed by a separation in anaphase I, homologous chromosomes move to opposite poles

D. One chromosome from each pair ends up in a gamete

Mendel’s Law of Independent Assortment

A. States that allele pairs separate independently during the formation of gametes

B. Any one pair of characteristics may combine with any one of another pair of characteristics

C. See p. 163 in Green Book

Independent assortment and meiosis

A. Any combination of chromosomes is possible in metaphase I (there is no ‘master plan’ for the order that they line up on the metaphase plate before separation)

B. Mendel thought all genes were inherited separately and had no relationship

-Ex: Pea plants could be green or yellow and wrinkled or round. Shape and color had nothing to do with each other, because the genes are on separate chromosomes. Any combination could have been produced (wrinkled/green, wrinkled/yellow, round/green, round yellow)

C. This is demonstrated in Dihybrid crosses.D. Today we realize that there are many exceptions

to this law.

Possible chromosome alignments

A Possible random outcomes B

Mendel’s Law of Independent AssortmentA. Any one of a pair of characteristics may combine

with either one of another pairB. Example:

gene=shape of peaalleles=round (R) or wrinkled (r)gene=color of peaalleles = yellow (Y) or green (y)*When crossing two plants that are heterozygous for both traits the offspring will show all combinations. This shows that genes for shape and color are independent (unlinked).

Dihybrid crosses

SY Sy sY sy

SY SSYY SSYy SsYY SsYy

Sy SSYy SSyy SsYy Ssyy

sY SsYY SsYy ssYY ssYy

sy SsYy Ssyy ssYy ssyy

Parent genotypes: SsYy x SsYy-S=smooth s=wrinkled

-Y=Yellow y=green

Possibleallelecombosfrom oneparent

Possible ratios forSsYy x SsYy

SY Sy sY sy

SY SSYY SSYy SsYY SsYy

Sy SSYy SSyy SsYy Ssyy

sY SsYY SsYy ssYY ssYy

sy SsYy Ssyy ssYy ssyy

F1 Genotypic ratios1:SSYY2:SSYy1:SSyy2:SsYY4:SsYy2:Ssyy1:ssYY2:ssYy1:ssyy

F1 Phenotypic ratios9: smooth-yellow3:smooth-green

3: wrinkled-yellow1: wrinkled-green

Perform a test cross for P: SSYY x ssyy

1.What will the genotype and phenotype ratios be for the F1 generation?2. What will the phenotype and genotype ratios be for the F2 generation?3. Determine the recombinants in each generation.

Perform a test cross for P generation: SSYY x ssyy

SY SY SY SYsy SsYy SsYy SsYy SsYy

sy SsYy SsYy SsYy SsYy

Sy SsYy SsYy SsYy SsYy

Sy SsYy SsYy SsYy SsYy

All F1 generation offspring are heterozygous (SsYy). What will the outcome be if you cross two individuals from F2?

Autosomal Gene Linkage

A. Autosome -all chromosomes that are not sex chromosomes

B. Sex chromosomes -determine if an individual is male or female

C. Linkage group -a group of genes whose loci are on the same chromosome

D. Gene linkage is caused by pairs of genes being inherited together. The presence or absence of one can affect the other.

Autosomal Gene Linkage

A. In gene linkage, all of the genes on a chromosome are inherited together

B. Does not apply to Mendel’s law of independent assortment

C. The closer the loci of the two genes on the chromosome, the smaller the chance that crossing over will occur in a chiasmata

D. If no info. is given, assume the genes are not linked

Autosomal Gene Linkage

E. These genes do not follow the law of independent assortment.

F. Example: P=purple, p=red, L=long, l-roundP gen.=PPLL x ppll

PL PL PL PL

pl

pl

pl

pl

F1 generation

*all PpLl

Autosomal Gene LinkageF. Example continued:

P=purple, p=red, L=long, l-roundF1 gen.=PpLl x PpLl

-predicted results would be a ratio of 9:3:3:1

-observed results were very different than predicted because the genes are for color and pollen shape are linked on the same chromosome.

PL Pl pL pL

PL

Pl

pL

pL

F2 generation

See p. 87 in your review guide.

Gene linkage and DrosophilaA. When genes are linked actual outcomes do

not match expected outcome B. This would be evident in a dihybrid cross (a

cross between two genes)C. Linked genes are represented in vertical

pairs with horizontal lines between themExample: b+=tan body b=black body

w+=long wings w=short wingsb+ w+ *The fly is tan and has long wings.

b+ w+ (genotype b+b+w+w+)

Gene linkage in Drosophila

Recombinants and gene linkageA. Recombinants are formed as a result of

crossing over (during prophase I)B. They are found in combinations that did

not exist in the parents

Polygenic InheritanceA. When the inheritance/expression of a

characteristic is controlled by more than one gene

B. Example: Human skin color-involves at least 3 independent genes-A, B and C represent alleles for dark-a, b and c represent alleles for light skin-P: AABBCC x aabbcc-F1: AaBbCc (all are heterozygous for all alleles)-F2:???

Polygenic Inheritance

B. Example: Human skin color-F2:???-to find out you can make a Punnett square-The more dominant alleles there are, the darker the skin-Dominant alleles are codominant (they are all expressed)

Polygenic inheritance (Other traits)

A. HeightB. Eye colorC. Finger prints

Polygenic Inheritance

Example: Flower color of beans-Genes A and B control color expression-AP and BP = purple -AW and BW = white-Each color has two alleles-Purple and white are codominant-Because of the codominance the flowers will be shaded depending on their genotypes

P: APAPBP BP x AWAWBWBW F1: APAWBPBW

AP BP AP BP AP BP AP BP

AWBW

AWBW

AWBW

AWBW

F1: APAWBPBW x APAWBPBW

F2: See Punnett square (fill it in)Determine which ones are white, purple and intermediates.)

AP BP AP BW AWBP AWBW

AP BP

AP BW

AWBP

AWBW

Polygenic inheritance and variation patterns

Two patterns are commonly expressed A. Continuous variation – shows a continuum of

variation among a population in a bell curve format- Example: In height expression people can be short, medium or tall (and everything in between)

B. Discontinuous variation – does not show a continuum (it is one or the other, there is no in-between expression of phenotype)- Example: Blood type can be A, B, AB or O (nothing in between)

Pedigree Charts

A. Used to show inheritance of traits over several generations

B. Affected individuals-shaded blackC. Unaffected individuals- left blankD. Heterozygous individuals (carriers)-

shaded grey or filled in half wayE. Example: Queen Victoria and

hemophilia (recessive/sex-linked)

PCR

A. PCR=polymerase chain reactionB. PCR=DNA photocopierC. Used to make copies of specific DNA

sequences (for study)

How a PCR works

1. DNA is heated (H bonds are broken)2. RNA primers are added to start

replication3. As the DNA cools primers bind to the

single strands (H bonds and complementary base pairing)

4. Nucleotides and DNA polymerase are added

How PCR works

5. Nucleotides bond with ‘exposed’ bases on the DNA strand

6. DNA polymerase joins them7. Complimentary strands are formed8. New strands can be heated, separated

and copied9. Animation

Gel electrophoresis

A. Technique used to separate molecules base on their rates of movement in an electric field-caused by charge and size of molecules

B. Commonly used in DNA profiling

Gel electrophoresis and DNA profiling

A. Scientists cut a mixture of DNA into segments by restriction enzymes

B. Place into a special gel with a current running through it

C. DNA separates into bands according to sizeD. Mixture is compared to a control group E. The more similar the DNA strands are the

more closely they are relatedF. See p. 30 in review guide or p. 66 in IB

textbook for examples

Figure 3.Comparison of known gel results for normal

hemoglobin (AA), sickle cell disease (SS) and sickle cell trait (AS). S represents the

molecular size marker. What results are present in the lanes marked with a question

mark?

Application of DNA profiling

A. Criminal investigations-collect blood or semen from suspect-if enough is not present use PCR-compare gel electrophoresis of suspect and victim

B. Indentify remains of deceased-take blood samples of living-compare with samples from dead

C. Paternal tests

Genetic screening

A. Test individuals in a population for presence of absence of a gene or allele

Genetic Screening

B. Advantages:1. pre-natal diagnostics (seek treatment or

abortion) Ex: PKU or down’s syndrome2. Reduce frequency of alleles that cause

genetic illnesses (opt to not reproduce or use IVF and select embyros without genetic defect)

3. Genetic diseases that show symptoms in later life can be detected earlier (Ex: Huntington’s disease)

Genetic Screening

C. Disadvantages:1. Increased abortion rate2. Harmful psychological effects

-could lead to discrimination (when seeking insurance or medical assistance)

-fear of getting older (depression)-creates a genetic ‘underclass’

Human Genome Project

A. An international cooperative venture established to sequence the complete human genome

B. Hope to determine the location and structure of all genes in the human chromosomes

C. Genome-total genetic material of a cellD. Completed in 2000 (about 10 years

early)

Advantages of the Human Genome Project

A. Understand genetic diseasesB. Figure out if any of them can be

prevented though screening-could also be negative (in terms of

insurance or employers)C. May lead to development of

pharmaceuticalsD. Insight into evolution and migration

patterns

Genetic Modification

A. Deliberate manipulation of genetic material

B. Genetic code is universal -it can be transferred between organisms

because the bases are the sameC. Used to create new combinations of DNAD. Mutation and recombination occur

naturally (often are disadvantageous and do not remain in the population)

Genetic Modification

E. Genetic engineers direct the process of recombination -increase chances of favorable combinations

Technique for gene transferExample-insulin production

A. Must have a vector (bacteriophage), a host (bacteria), restriction enzymes and DNA ligase

B. Plasmids (small circular DNA from bacteria) are cut using restriction enzymes

C. Sticky ends are created by adding C nucleotides

Technique for gene transferExample-insulin production

D. mRNA that codes for insulin is extracted from the pancreas

E. Reverse transcriptase makes DNA from the mRNA

F. Sticky ends are creating by adding G nucleotides

G. Insulin gene and plamsid are mixedH. They join together at the sticky ends (C

pairs with G)

Technique for gene transferExample-insulin production

I. Plasmid with the human insulin gene is called a recombinant

J. Host cell (usually E. coli) receives the geneK. E. coli are cultured w/ new geneL. Insulin produced by modified E. coli is

extracted and eventually given to diabetics

Other examples of genetic modification

A. Insulin production in E. coli (discussed above)

B. Bacteria can be modified to produce growth hormone for cows-cows injected with hormone increase milk production by 10-20%

C. Breeding of plants to increase disease resistance

D. Dog breeding

Benefits of genetic modification

A. Less disease (possibly)-long term effects are unknown

B. More product-Ex: milk

C. Some are beneficial with no know side effects-Ex: Insulin production

Negative aspects of genetic modification

A. Introducing plasmids-hospitals fear them-can carry genes for antibiotic resistance-can be passed from one species to another

B. Don’t know long term effects (Example)-effects of using growth hormone in cows is unknown (some could be in our food)-cows needs extra antibiotics to stay healthy (we get these too)

C. Could create super bacteriaD. Less than perfect becomes unacceptable (if anyone can

be genetically modified before birth)

Gene therapy

A. Treatment of genetic illness by modification of genotype, or base sequence, or allele (dominant for recessive)

B. Best if done with stem cellsC. Many attempts have not been successfulD. Read SCID example on p. 28

Cloning

A. Clone-group of individuals identical in genotype or a group of cells descended from a single parent cell

How Dolly was made1. Udder cells removed from sheep 1

a. cells grown in a culture to turn off their genes2. Embryos removed from sheep 2

a. Nuclei removed from embryos3. Embryos and udder cells fused by electricity to

form zygotes4. Zygotes developed to embryos5. Embryos implanted into Sheep 3 (the surrogate

mother)6. Dolly develops and becomes first born clone

a. Identical to sheep 1

Ethical issues of cloning in humans

A. Possible to only clone a specific organ?B. Does the whole person need to be

cloned for organ donation?-if we take the heart, what do we do with the

rest of the bodyC. Read p. 32 in the review guide for

arguments for and against therapeutic cloning of humans.