biol 1114 unit 2 notes

7/13/2019 BIOL 1114 Unit 2 Notes

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BIOL 1114 Unit 2 Notes

Genetics

Lecture 13

The Central Dogma

Accessing Genetic Information Transcription; Protein Synthesis

6 seconds (60m): ATP reserves depleted

15 sec (150m): Creatine phosphate reserves depleted

Next 45 sec (200m): Fermentation pathways of ATP production

o Lactic acid buildup

o Oxygen debt

o High Altitude Training = more red blood cells/more oxygen

[7.3AB, 7.4AB]

Overview of Protein Synthesis

Amino Acids = monomers of proteins

DNA Codes for proteins

o Transcription:transfer of info from DNAto mRNA

In nucleus

o Translation: transfer of info from mRNAinto a polypeptide

In cytoplasmribosomes

DNA (transcription) -> mRNA (Translation) -> Ribosomes (protein synthesis) ->

RNA Polymerase (-ase = protein)

G1 and G2 in cell cycle

Transcription

ALWAYS read the Template Strand

RNA Polymerase reads the Template Strand

Promoter: [TATA Box], TATA Binding Protein, Transcription Factors, RNA Polymerase

Starts Transcription^

Promoter Region:

o [TATA Box]

o [TATA Binding Protein]: recognizes the TATA box and binds to it

o [Enhancers]

o [Transcription Factors] bind to the enhancers


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o [DNA bends] so the transcription factors connect to the TATA binding

proteins and RNA Polymerase

o [RNA Polymerase]: connects and starts transcription

Always add to the 3 End

DNA template 3 to 5 DNA coding 5 to 3

RNA 5 to 3

[Terminator Region]:

o End of the gene

o Stops the coding of genes

Messenger RNA Processing

Exons (A, B, C)

Introns (1, 2, 3)o Get snipped out

o Non-coding

mRNA cap copied + Poly A tail (AAA)

Splicing

o mRNA contains only Exons (A, B, C)

o mRNA cap --- 5 *Exon A, Exon B, Exon C,+ 3 --- Poly A Tail

o Directed to ribosomes


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Translation

20 Amino Acids; 4 nucleotides (ATCG , AUCG)

o 1:1 code: only 4 amino acids

o Doublet code: only 16 amino acids

o Triplet code: 64 amino acids

mRNA = 3 letter codes for codons

AUG is always the code at the start

Codes for M, or signals the beginning of a protein

UAA, UAG, UGA is always the end

Different from the terminator Signals the end of a protein; release factor connects to

dissemble elements

Peptide bonds are covalent

mRNA = codons (used to read code) AAG = Lysine


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mRNA: language translated to polypeptide language

tRNA is the translator:

o picks up amino acids


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recognizes codons on mRNA (contains anticodons)


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Many ribosomes translate mRNA

Its kinda like a library of books being read by someone

Library = all the cells DNABook = one gene

Reader = RNA Polymerase

Transcription = transposing wiring music from one key to another

Translation = reading music from a page to play music

AUC CUU GCA GUG GAA UGG GAG

I LAV EWE


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Gene Regulation

Mutations

Specialization:cells express different genes in same DNA

(How does a cell know what genes to turn on?)

Ex. Cell type; pancreas eye lens nerveGlycolysis enzyme: on on on

Crystalline gene: off on off

Insulin gene: on off off

Hemoglobin gene: off off off

Regulation of gene expression

Enzyme inductionthe lac operon(in bacteria)

E. coli, exposed to glucose, fat (donut for breakfast)

o Make enzymes that break down glucose and lipids

Repressor protein:binds to DNA and blocks transcription Lactose is not present

E. coli, exposed to lactose, cellulose (salad and milkshake)

o After lactose becomes present, it

o binds to the repressor and pulls it off of the DNA, to continue

transcription;

o enzymes that break down lactose are translated,

o lactose becomes digested, repressor protein binds back to DNA

Gene Regulation in Eukaryotes

Transcription factors are the switches that turn genes on and off

Ovaries: estrogen + progesterone (steroids, lipids made out of cholesterol)

o Lipid-soluble hormones are secreted (non-polar molecules)

Hormone passes through cell membranes and binds to interior

receptor

o Makes mRNA

mRNA Attaches to ribosome

Certain genes activated leading to production of new

proteins

o Hormones change shape by receptor proteins (transcription factors)

o

DNA bends to attach to RNA polymerase and TATA binding proteino Transcribe and translate the genes to make proteins needed to build up

the endometrium


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Mutations

Change in DNA sequence

o Substitution

o Insertion

o

Deletion Base Substitution

o Sickle-cell disease [7.22]

Normal:

DNA: CTC

mRNAGAG

Amino acid: Glutamic Acid

Sickled:

DNA: CAC

mRNA: GUG

Amino acid: Valineo Not Always noticeable (Silent)

Normal:

CTC

GAG

Glutamic Acid

Sickled:

CTT

GAA

Glutamic Acid

Insert/Delete (one or more bases) Ser Gly Leu

o Addition of a letter: Ser Pro Val Ile

o Deletion of a letter: Ser Gly Leu

TAC TAT GTA CGT CAG CTC TCA CGT TCG ACT

AUG AUA CAU GCA GUC GAG AGU GCA AGC UGA

M I HAVE SAS

M I HAVE GAS

Base insertion or deletion

o

What happens to downstream codons?

o Change the whole reading frame

o Near the promotor would change more than by the terminator

o Theyll probably change

o Usually ruins protein shape; fatal if protein is essential

o What is # of bases is a multiple of 3?

o Adds an amino acid to the original coding


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o AT have more chance of mutation than CG

Point: change in a single DNA base

Missense: changes in a codon to code for different Amino Acids

o Sickle-cell

Nonsense: changes a codon to code for stop

Frameshift: add or delete anything exceptmultiples of 3

Figure 7.23 and Table 7.2

Where do these mutations come from?

o You inherit some from your parents

o Some happen at randomin your cells

Physical or chemical mutagensdamage DNA

E.g. radiation, many chemicals, smoking, etc.

Errors in DNA replication

Errors in production of sex cells (meiosis)o Think on this: Your DNA mutates throughout your life

Which mutations can you pass on to your children

Cancer

Two major classes of cancer causing genes

o Oncogens:normally trigger cell division, (such as in would healing), but

are over-exposed in cancer cells

o Tumor suppressor genes: normally prevent a cell from dividing

Arises from mis-timed or misplaced cell division

o Or absence of normal apoptosis

Since proteins control cell cycle, and proteins are coded by DNA, DNA mutations

can lead to inappropriate cell division

10% of human genes code for transcription factors (turn genes on and off)

Meiosis [9.1-9.6]

Figure 9.5

Animal body -> Meiosis (diploid) -> gametes (eggs and sperm) haploid -> Fertilization

(diploid) -> Mitosis (diploid)

Homologous chromosomes: matched pair Both carry genes controlling the same inherited characteristics

o Example: gene for eye color may be located on this chromosome, in this

spot (might be different versions (alleles))

o Not identical; they just carry the same types of genes

o Of our 23 pairs of chromosomes:

22 pairs = autosomes


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1 pair = sex chromosomes (23

rdchromosome)


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Meiosis Stages

Before meiosis is interphase, then:

Meiosis I = reduction division

o Prophase I

Diploid Doubling of DNA

synapsis: homologs line up next to one another to form tetrads

(early)

crossing over occurs(late)

o Metaphase I

Diploid

Spindle fibers

Side by side, alignment; not single file line across equator

Independent Assortment

o

Anaphase I

Diploid

The centromere for the chromosomes are not broken

o Telophase I and cytokinesis

Diploid

Opposite poles

Nuclear envelope

Meiosis II = equational division

o Prophase II

Haploid

2 separate cells

o

Metaphase II Lined up single file

o Anaphase II

Centromeres are broken

o Telophase II and cytokinesis

Haploid

Nuclear envelope

Four nonidentical haploid daughter cells

Create variety

Comparison of meiosis & mitosisMeiosis I

Prophase

o Synapsis; can be long

o Crossing over

Metaphase I

o Homologous pairs align, side by side


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Anaphase

o Centromeres do not divide

Mitosis

Prophase

o No synapsis

o

No crossing over Metaphase

o Individual chromosomes align, single file

Anaphase

o Centromeres divie

Meiosis II is similar to mitosis: sister chromatids separate

Mitosis starts 2n ends 2n (2 identical cells)

Meiosis starts 2n ends n (4 non-identical cells)


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Gamete Formation[9.8a, 34.2b, 34.3b, 10.1]

The Mitosis Waltz

Variation

Diploid # of 4

Haploid # of 2

2 homologous pairs of chromosomes

3 pairs, 8 possible outcomes

number of combinations = 2# of pairs

Meiosis generates genetic variety:

223

= 8,388,606 (form independent assortment)

223

(sperm) x 223

(egg) = 70,368,744,177,664

(this doesnt count crossing over)

If a normal diploid cell has 8 chromosomes, then

There are 4 homologous pairs of chromosomes per pair

There are 16 chromatids per diploid cell between S and the first division of

meiosis

There are four chromosomes per cell after meiosis

Sperm cells made from this diploid cell would have 8 chromosomes

Gamete formation in Males

Chromosomes in the Nucleus

Acrosome = golgi apparatus

o Has enzymes that help burrow through cells to find the eggs

Spiral mitochondria

Tail: to swim


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Sperm are made in the testes

o Seminiferous tubules: 150 meters (place of meiosis)

The walls, stem cells that keep dividing

Spermatogenesiso 2n (diploid): spermatogonium (spermatogonia)

o Mitosis: process another 2n cell (primary

spermatocyte)

begins Meiosis (primary spermatocyte)

o Morph into secondary spermatocytes (2 cells, n =

haploid)

begins Meiosis II

o Morph into early spermatids

o Morph into late spermatids

o Morph into sperm

Production:

50,000/s

70 days

50-500 million/ejaculation

o Epididymis: collect sperm to queue

Cells in the process of metaphase I in the human male are called:

Primary Spermatocytes

If an intestinal cell in a grasshopper contains 12 chromosomes, a grasshopper secondary

oocyte would contain 6 chromosomes


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Gamete formation in females


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Oogenesis

2n = Oogonium (oogonia)

o Primordial follicle

o Primary oocyte

Before birth Mitosis to generate up to 7 million oogonia (potential eggs)

Primary oocyte arrested in prophase I

At birth, about 2 million remain

Childhood

o Primary oocyte arrested

Puberty

o Few get called up to continue

o Prophase I

~400,000

Each month a few follicles begin to grow Growing (nursing) follicle cells

Day 14 = ovulation = n = haploid

Secondary oocyte

Meiosis I complete

First polar body (nucleus growing on

the side


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Secondary oocyte arrested in

metaphase II

If not fertilized, it will stay in

metaphase II

Is fertilized, (w/ sperm cell) meiosis

II is completedo 3

rdand 4

thpolar bodies will

divide (1 fertilized egg and 3

polar bodies)

o Becomes zygote

First cell division

begins ~24 hours


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Single Gene Inheritance [10.1, 10.2A-D, 10.3AB]

Mendelian Genetics

Introduction to Probability

Any event has a probability of occurring between 0 and 1 (inclusive)

o

0 = no chanceo 1= 100% chance

o 0.5 = 50% chance

Flip a coin

Possibility of tossing heads = = 50%

Tossing heads twice = x = chance = 25%

Tossing heads thrice = x x = 1/8

Roll a 6 sided die

Roll a 3 = 1/6 chance

Roll a 3 six times = 1/6

6= 1/46656

Gregor Mendel & Pea Plants

Why garden peas?

o Many distinct varieties

o Lots of offspring, quickly

o Strict control of mating

o Possess either-or traits

o Can cross-fertilizeeasily

Flower

Male = stamens (sperm)

Female = stigma (egg)

Pollen from stamen travels to stigma and begins fertilization

o [10.2

Tracing the Inheritance of One Gene; Principle of Segregation

What Mendel did..

Started with true-breedingparents

Cross-fertilized two true breeding varieties

o Purple flowers; always breed purple flowers (RR)

o

White flowers; always breed white flowers (rr)

P generation = original parents (true-breeding)

o Purple (RR) x White (rr)

F1 Generation hybrids

o All Purple (Rr)

Purple (Rr) x Purple (Rr)

F2 Generation


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o Mostly Purple (RR, Rr) some white (rr)

F2 generation 929 pea plants

Purple 705

White 224

o About purple, white

o

3:1 ratio of purple to white flowers For each cross, Mendel found a 3:1 phenotypic ratio in the F2 generation

Mendel concluded that:

o Alternate forms of genes (alleles) account for variation

o For each character, an individual has two alleles, one from each parent

o If two alleles, differ, the dominantallele is fully expressed, while the

recessivehas no effect

Mendels Principle of Segregation

o Pairs of elementen (alleles) segregate (separate) during gamete formation

Anaphase I

Homologs separate Alleles are on the homologous chromosomes

Replication in interphase

Meiosis: segregates alleles into gametes

Fertilization: gametes combine


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(TT)

o Genotype: homozygous dominant

o Phenotype: tall

(Tt)

o

Genotype: heterozygouso Phenotype: Tall

(tt)

o Genotype: Homozygous recessive

o Phenotype: short

P generation (true-breeding parents)

Phenotype: Purple Flowers (PP) dominant

Phenotype: White Flowers (pp) recessive

P Generation: PP x pp = P (sperm) p(eggs) = Pp

F1 Generation = Pp (heterozygous) Purple Flowers

Pp x Pp

Pp (sperm) x Pp (eggs) [meiosis]

F2 Generation

o PP (purple), Pp (purple), Pp (purple), pp (white)

o 3:1 phenotype ratio

o 1:2:1 genotype ratio

Punnett Squares

F1: Pp x Pp

PP, Pp, Pp, pp

3:1 phenotype ratio (F2 generation)

1:2:1 genotype ratio (F2 generation)

P generation: PP (tall) x pp (short)

F1 generation: Pp, Pp

F2 generation: PP, Pp, Pp, pp; 3 tall : 1 short

Test CrossUsing a Punnett Square to determine unknown parental genotype

Chance is a male, liver-spotted Dalmatian

Spot color is genetically determined

Chances genotype is bb

Probability has black spots; we dont know for certain what her genotype is


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She could be heterozygous (Bb) or homozygous dominant (BB)

99 puppies

48 black, 51 brown = heterozygous (Bb); about 50% chance

Homozygous dominate

bb x BBo Bb 100% black

Heterozygous

bb x Bb

o Bb 50% black

o bb 50% brown

Could she have a litter of 8 black puppies even if she were heterozygous (50% black)?

x x x x x x x = 1/28= 1/256; slim chance, but yes.

The probability of one of her puppies having black spots is, 1(bb) x (Bb)= (50%black)

P generation: RR (homozygous dominant) x rr (homozygous recessive)

F1 generation: Rr (heterozygous) offspring

Principle of Independent Assortment

Each pair of alleles segregates independently of each other

o Metaphase 1


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Meiosis I

Metaphase I

4 possible gamete types

Tracing two genes

Solid color (HH) is dominant to points (hh)Long hair (LL) is dominant to short hair (ll)

HHLL x hhll

How to determine the gametes:

for one set of genes:

o Genotype: Bb

o Gametes: Bb

For two genes, remember independent assortment

o

Genotype: AaBbo Gametes: AB, Ab, aB, ab

4 sperm, 4 eggs = 16 punnett quare

P Generation:

HHLL x hhll

o HL x hl

F1 Generation:


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HhLl (heterozygous for both traits)

Solid color, long hair

F2 Generation (16 squares)

HL, Hl, hL, hl x HL, Hl, hL, hl

o HHLL HHLl HhLL HhLl

o

HHLl HHll HhLl Hhllo HhLL HhLl hhLL hhLl

o HhLl Hhll hhLl hhll

3:1 ratio still exists

12:4 ratio

Solid Colored, Long Haired (H,L): 9 (dominant, dominant)

Solid Colored, Short Haired (H,ll): 3 (dominant, recessive)

Points, Long Hair (hh,L): 3 (recessive, dominant)

Points, Short Hair (hh,ll): 1 (recessive, recessive)

o 9:3:3:1 (phenotypic ratio)

o

Di-hybrid cross

Tracing Inheritance of Multiple Genes; Product Rule [10.4AB]

Tracing Inheritance of Two Genes

The Product Rule:

3 genes: 8 different gametes; 64 boxes

o AaBbCc

o ABC Abc ABc AbC aBc abC aBC acb

4 genes; 16 different gametes; 256 genes


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The chance that 2 independent events will both occursuch as 2 alleles both

being inheritedequals the product of the individual chances that each event

will occur

o Note: you can use the Product Rule if you see statements like this:

Assume the genes assort independently

The genes are on different homologous pairs of chromosomes The genes are not linked

Cross 2 trihybrid plants that are tall, and have round, yellow, seeds. What is the chance

that the offspring is also a trihybrid. [Assume the genes assort independently]

TtRrYy (heterozygous) x TtRrYy (heterozygous)

TT Tt 2/4 = (Tt)

Tt tt

RR Rr 2/4 = (Rr)

Rr rr

YY Yy 2/4 = (Yy)

Yy yy

(Tt) x (Rr) x (Yy) = 1/8 TtRrYy


that the offspring will be short with wrinkled, green seeds?

(tt) x (rr) x (yy) = 1/32 ttrryy


that the offspring will be short, with round seeds and heterozygous for seed color?

(tt) x (RR or Rr) x (Yy) = 3/32 ttRrYy or ttRRYy


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Cross a fish that is heterozygous for big body (Bb), homozygous for long tail (LL), and has

green scales (rr) with a fish that is heterozygous for big body (Bb), has a short tail (ll),

and is heterozygous for red scales (Rr). Big body size, long tail, and red scales are

dominant alleles for these traits. What proportion of the offspring do you expect to be

heterozygous for all three traits?

BbLLrr x BbllRr

BB Bb 1/2

Bb bb

Ll Ll 1

Ll Ll

Rr rr 1/2

Rr rr

(Bb) x 1 (Ll) x (Rr) = Chance(25%) BbLlRr

----

BbLLrr x BbllRr

bbLLRr

(bb) x 1 (Ll or LL) x (Rr or RR) = 1/8 chance

If 80 fish produced, 10will be bbLLRr


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----

Refer to the table and cross a pea plant that is heterozygous for seed form (Rr),

homozygous recessive for seed color (yy), heterozygous for pod form (Vv), and has

yellow pods (gg) with (x) a pea plant that has wrinkled seeds (rr) and is heterozygous for

seed color (Yy), pod form (Vv), and pod color (Gg). If 100 offspring are produced,

approximately how many do you expect to show the recessive phenotype for all fourtraits? [Assume these genes assert independently]

RryyVvgg x rrYyVvGg

Rr rr

Rr rr

Yy yy

Yy yy

VV Vv

Vv vv

Gg gg

Gg gg

x x = 1/32

1/32 x 100 offspring = 3 offspring rryyvvgg

----

Complications to Mendelian Ratios [10.6AB, 10.9AB]

All of these things complicated matters:

Multiple alleles:more than 2 alternative forms of a gene;

o e.g. human blood type, rabbit coat color

Co-dominance:both alleles are equally expressed; more than one allele may be

expressed

Human blood type

o

3 alleles: IA

, IB

, i I

A: red blood cells coated with carbohydrate A

IB: red blood cells coated with carbohydrate B

I: no carbohydrate

Immune system reacts to carbohydrates not found in your own body [Fig. 29.5]

o 6 genotypes

o 4 phenotypes

o Figure 10.15


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Anti-B

Antibodies

Anti-A

Antibodies

Co-dominant

No antibodies

Universal

Recipient

Both antibodies

(Anti-A + Anti-B)

Universal Donor

Recipients blood in test tube: Type A

Anti-B Antibodies

o

Donor has Type Bo The Anti-Bs attack the Bs from the donors blood

o Clumping= bad

If type a gives to type b, would there be clumping? Yes. Bad

A to AB. No. Good

B to AB. No. Good

O to AB. No Good

B to O. Yes. Bad

A to O. Yes Bad

AB to O. Yes. Bad

AB to A Yes. Bad AB to B. Yes. Bad

O to A. No. Good

O to B. No. Good

Couple 1: O (IAIB) x A (I

AIAor I

Ai) = Type A (I

AIB)

Couple 2: B (IBIBor I

Bi) x AB (I

AIB) = Type AB (I

AI

Aor I

Ai)

Couple 3: B (I

BIBor I

Bi) x B (I

BIBor I

Bi)= Type O (ii)


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Epistasis:one gene masks anothers phenotype

o Gene 1 codes for protein 1

o Gene 2 codes for protein 2

If protein 1 is not present, gene 1 and gene 2 cannot be present

H Gene

Incomplete Dominance:

o Snapdragon flowers: red flowers, white flowers, pink flowers

P generation: red and white r1r1 x r2r2

F1: pink r1r2 x r1r2

F2: 1 red (r1r1), 2 pink (r1r2), 1 white (r2r2)

Lethal alleles:causes development to stop before birth

o Chihuahuas:

HH lethal (stop development)

Hh hairless Hh hairy

Hh x Hh

o HH, Hh, Hh, hh

o Phenotype ratio: 2 hairless : 1hairy

o Genotypic ratio: 2 Hh: 1 hh


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Polygenic Inheritance:skin pigmentation/coloration

o More Dominant genes = darker color

Environmental Effects: Siamese Cats

o Dark points on Siamese cats occur on the extremities of the body

Feet, ears, tail

o Effected by the temperature of the cells that dictate the pigment of the

skin at the time of creation

Chromosomes & Gene Linkage [10.1, 10.5AB]

How to solve genetics problems (p 226-227) + emailed problems

Chromosomal Basis of Inheritance

Linked Genes

Genetic Maps Based on Crossover Data

Thomas Hunt Morgan(early 1900s) Drosophila melanogaster (fruit fly)

Morgan associated a specific gene with a specific chromosome

Fruitfly has 8 chromosomes, 4 pairs; male has hooked chromosome

Prophase 1, Metaphase 1 [independent assortment], (diploid 2n=4)

If genes are on the same homologous chromosome


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Linked Genes:

Located close together on the same chromosome

Tend to be inherited together

o If genes are on the same chromosome, only 2 allele combinations are

expected in the gametes, and the number of phenotypes is reduced

b+b vg

+vg (Female) x bb vgvg (Male)

BbVv (Female) x bbvv (Male)

b: refers to black body (mutant body type)b

+: dominant allele (gray body

b: recessive allele (black body)

vg: refers to vestigial wings (mutant type)

vg+: dominant allele (normal wings)

vg: recessive allele (vestigial wings)

BbVv bbVv Bbvv bbvv

965 185 206 944


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So these genes are linkedon the same chromosome

If the genes are lined, why do all the other combinations occur at all?

Crossing over, Prophase 1

S phase: replication of DNA in chromosomes

Prophase 1: BV BVbv bv

End of Meiosis 1: B

End of Meiosis 2: BV -- BvbV -- bv

Genetic Maps Based on Crossover Data

Recombinants / Total = Recombination Frequency

Total: 965 + 185 + 206 + 944 = 2300

Recombinants: 185 + 206 = 391

391 / 2300 = .17 = 17% = Recombination Frequency


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A.H. Sturtevant (Morgans student)

The probability of crossing over between two genes is directly proportional to

the distance between them

Genes are farther apart, genes are more likely to crossover

o Distance between linked genes determines how often they will be split

up Map unit = 1% recombination frequency

Loci/Locus (address) Recombination Frequency Approximate Map Units

b vg 17% 17 (18.5)

cn b 9.0% 9.0

cn vg 9.5% 9.5

32 DE

18 +22

40/100 = 40% (DE)

23 AD

2+3

5/50 = 10% (AD)

36 AE

15+15

30/100 = 30% (AE)

E A D

Or

D A E


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Sex-linked Traits [10.7A-C, 10.8]

Assume genes A and B are linked and are 40 map units apart. An individual

heterozygous (AaBb) at both loci is crossed with an individual who is homozygous

recessive and (aabb) at both loci. If 100 offspring are produced, how many would be

(aabb)?

A40B

40% AB = 20 + 20 for recombinants

100-40=60 /2 = 30 for parentals

AaBb aaBb Aabb aabb

30 20 20 30

Chromosomes and Sex Determination

Mammals:o Male is heterogametic (XY)

o Female is homogametic (XX)

Birds and many other animals: female is heterogametic

Extremely diverse

SRY specifies maleness in humansproduces protein that directs development

of male structures

In humans, female by default until this SRY gene is activated


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X-linked:genes transmitted on X chromosomeo Red-green colorblindness; males more likely to have it

Y-linked:genes transmitted on Y chromosomeextremely rare in humans


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X Inactivation

Inactivation of one X chromosome (randomly) per nucleus in female humans Barr Body

o All subsequent cells will have the same X chromosomes inactivated

In which cats did X inactivation happen earliestin development?

Large patchesor small patches?

A genetic counselor is asked for a probability that the male offspring of a husband and

wife will be color-blind. The woman has normal vision, but her father is color blind, the

man also has normal vision, but his father is color-blind as well. (note: color-blindness is

sex-linked and recessive)

Half the male offspring will be color-blind, and half will have normal color vision

XX (carrier) * XY

XX XY

XX XY

--


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Review

Transcription: in the nucleus

Translation: in the cytoplasm

Introns are spliced from sections of mRNA before exiting the nucleus

Which of the following forms hydrogen bonds with the anticodon?

MRNA

In translation. mRNA contains the codon

How many of the following terms apply to events that take place during the processes

of transcription and translation in eukaryotes before the mRNA transcript leaves the

nucleus

Release factor attaches to stop codon TAT binding protein binds to promoter

5 cap added

Peptide bond formed

Small rRNA subunits joins mRNA

tRNA binds opposite codon

Introns removed

Transcription factors bind to enhancer

Which of the following early-in-the-strand DNA template strand substitutions would

have the worst results?

CTT -> CTC

GAA (E) -> GAG (E)

GGA -> GGC

CCU (P) -> CCG (P)

ATG -> ATT

UAC (Y) -> UAA (stop codon)

Given that the haploid number (n) for this species is 2 and the diploid number (2n) is 4

Aligned single file as whole (diploid) chromosomes

Metaphase of mitosis

If an intestinal cell in a grasshopper contains 24 chromosomes, a grasshopper secondary

spermatocyte would contain 12 chromosomes (meiosis II; haploid chromosomes: 24 /2

= 12 chromosomes)


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Meiosis occurs in the walls of the seminiferous tubules of male mammals

What is the chance of flipping 5 coins at the same time and getting the result of heads

on each of them?

1/32 = (1/25)

If a man and a woman have four children (none of them twins), the chance that all four

will be boys is 1 in 16.

1/24

= 1/16 True

Assume free ear lobes are dominant to attached ear lobes. A woman with attached ear

lobes mates with a heterozygous male. Their first two offspring have free ear lobes.

What is the probability that their third offspring will have attached ear lboes

P1: Rr x rr

F1: Rr =50%

rr =50%

A heterozygous individual is more likely to pass on a dominant allele than a recessive

allele

False = equal chance Rr

A female fruitfly that is heterozygous for wing type is mated with a male fruit fly that is

homozygous recessive for wing type as shown by the following drawing

Which of the following sets of numbers would be most likely for the numner of

genotypes represented by their offpring?

Vv x vv

432 Vv + 441 vv

50/50 chance

Assume the allele for long pinkie finger is dominant to short pinkie fingers. If a man who

is heterozygous (Ll) for this allele has a child with a woman who is homozygous recessive

(ll) for this same allele, what are the possible genotypes of this offspring

Ll x ll

Ll ll Ll ll

To perform a test crossfor independently assorting traits, you mate an organism with

an unknown genotype with an organism that is homozygous recessive


39/40

What is the expected ratio of the gametes from the following genotype: AABb (assume

that these genes assort independently)

AABb

1 AB : 1 Ab

In a species known as squeaks white ears are dominant to red ears and short hair is

dominant to long hair. You want to breed a red-eared, long-haired female with a white-

eared, shorthaired male. The female and male are homozygous for both traits. These

genes assort independently. If the mating produces 16 offspring, how many will you

expect to have red ears and long hair?

0

--

How many different kinds of gametes would you expect from an individual with the

following genotype? AABbCcDdee

1 (AA) x (Bb) x (Cc) x (Dd) x 1 (ee) = 8

--

Cross a true breeding female that is homozygous dominant for both characters with a

true breeding male that is homozygous recessive for both traits. Which of the following

represent the genotypes and phenotypes of their offspring (F1)

P1: BBWW (black eyes, wide gape) x bbww (orange eyes, narrow gape)

F1: BbWw black eyes and wide gape

Now cross two of the F1 individuals is 1000 individuals are produced in the F2

generation, approximately how many do you expect to have black eyes and a narrow

gape?9:3:3:1 ratio

black eyes, narrow gape = 3/16 x 1000 = 187.5

3/16 x 1000 = 188

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A couple has three children, all of whome have brown eyes and blond hair. Both parents

are homozygous for brown eyes (BB), but one is a blond (rr) and the other is a redhead

(Rr). What is the probability that their next child will be a brown-eyed redhead

BBrr x BBRr

BBrr (50) BBRr (50)

--

Cross 2 trihybrid plants that are tall and have round yellow seeds (TtRrYy). What is the

chance that the offspring will be tall, with wrinkled, yellow seeds

(TT,Tt) x (rr) x (YY,Yy) = 9/64

--


40/40

ABO blood types in humans are determined by a single gene having three alleles, IA, IB,

and i. A mother with blood type A has a child with blood type O. What is the mothers

genotype? IAi

Mom: IAi

Dad: i?Baby: ii

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Baby Making couples

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Recombination frequency mapping

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Genes D and F are on the same chromosome. The Dross DdFf x ddff yields 1000

offspring. 150 of which are genotype ddFf. About how many map units apart of D and F

150 x 2 = 300

300/1000 = 30%

biol 1114 unit 2 notes

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