today: imprinting and linkage exam #3 t 12/2 in class final sat. 12/6 from 9am – 12noon in bio 301

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Today: Imprinting and Linkage Today: Imprinting and Linkage Exam #3 T 12/2 in class Exam #3 T 12/2 in class Final Sat. 12/6 from Final Sat. 12/6 from 9am – 9am – 12noon in BIO 301 12noon in BIO 301

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Page 1: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Today: Imprinting and LinkageToday: Imprinting and LinkageExam #3 T 12/2 in classExam #3 T 12/2 in classFinal Sat. 12/6 from Final Sat. 12/6 from 9am – 12noon in BIO 9am – 12noon in BIO 301301

Page 2: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

PhenotypeGenotype

Genes code for proteins (or RNA). These gene products give rise to traits…

It is rarely this simple.

Page 3: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Epigenetics: http://www.pbs.org/wgbh/nova/sciencenow/3411/02.html

Lamarck was right? Sort of…

Image from: http://www.sparknotes.com/biology/evolution/lamarck/section2.rhtml

Page 4: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Genomic Imprinting

• Genomic imprinting is a phenomenon in which expression of a gene depends on whether it is inherited from the male or the female parent

• Imprinted genes follow a non-Mendelian pattern of inheritance

– Depending on how the genes are “marked”, the offspring expresses either the maternally-inherited or the paternally-inherited allele **Not both

Page 5: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

A hypothetical example of imprinting

A=curly hair

a=straight hair

B=beady eyes

b=normal

*=methylation

A* in males

B* in females

A*abB*

A*abB*

A*abB

AabB*

A*b, A*B,ab, aB

Ab, AB*,ab, aB*

aB*

aB* A*

bA*b

similar to Fig 7.10

Page 6: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

• Genomic imprinting must involve a marking process

• At the molecular level, the imprinting is known to involve differentially methylated regions–They are methylated either in the oocyte or

sperm• Not both

Imprinting and DNA Methylation

Page 7: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

• For most genes, methylation results in inhibition of gene expression

–However, this is not always the case

Page 8: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Haploid female gametes transmit an unmethylated gene Haploid male gametes transmit

a methylated gene

Fig 7.11Changes in methylation during gamete development alter the imprint

Page 9: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Thus genomic imprinting is permanent in the somatic cells of an animal

– However, the marking of alleles can be altered from generation to

generation

Page 10: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

To date, imprinting has been identified in dozens of mammalian genes Tbl 7.2

Page 11: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Tbl 7.2

Page 12: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Imprinting plays a role in the inheritance of some human diseases: Prader-Willi syndrome (PWS) and Angelman syndrome (AS)

–PWS is characterized by: reduced motor function, obesity, mental deficiencies

–AS is characterized by: hyperactivity, unusual seizures, repetitive muscle movements, mental deficiencies

Usually, PWS and AS involve a small deletion in chromosome 15

–If it is inherited from the mother, it leads to AS–If it is inherited from the father, it leads to PWS

Page 13: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

• AS results from the lack of expression of UBE3A (encodes a protein called EA-6P that transfers small ubiquitin molecules to certain proteins to target their degradation)

– The gene is paternally imprinted (silenced)

• PWS results (most likely) from the lack of expression of SNRNP (encodes a small nuclear ribonucleoprotein that controls gene splicing necessary for the synthesis of critical proteins in the brain)

– The gene is maternally imprinted (silenced)

Page 14: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Fig 7.12The deletion is the same in males and females, but the expression is different depending on who you received the normal version from.

Page 15: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

The relationship between genes and traits is often complex

Complexities include:

• Complex relationships between alleles

Page 16: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

The relationship between genes and traits is often complex

Complexities include:

• Multiple genes controlling one trait

Page 17: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Two genes control coat color in mice

Fig 4.21

Page 18: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Variation in Peas

Fig 3.2

Page 19: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Fig 2.8

Inheritance of 2 independent genes

Page 20: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Y y

r R

Gene for seed color

Gene for seed shape

Approximate position of seed color and shape genes in peas

Chrom. 1/7 Chrom. 7/7

Page 21: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

There must be a better way…Fig 2.9

Page 22: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Inheritance can be predicted by probability

Section 2.2, pg 30-32

Page 23: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Sum rule

• The probability that one of two or more mutually exclusive events will occur is the sum of their respective probabilities

• Consider the following example in mice

• Gene affecting the ears– De = Normal allele– de = Droopy ears

• Gene affecting the tail– Ct = Normal allele– ct = Crinkly tail

Page 24: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

• If two heterozygous (Dede Ctct) mice are crossed• Then the predicted ratio of offspring is

– 9 with normal ears and normal tails– 3 with normal ears and crinkly tails– 3 with droopy ears and normal tails– 1 with droopy ears and crinkly tail

• These four phenotypes are mutually exclusive– A mouse with droopy ears and a normal tail cannot have

normal ears and a crinkly tail

• Question– What is the probability that an offspring of the above

cross will have normal ears and a normal tail or have droopy ears and a crinkly tail?

Page 25: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

• Applying the sum rule– Step 1: Calculate the individual probabilities

9 (9 + 3 + 3 + 1) = 9/16 P(normal ears and a normal tail) =

1 (9 + 3 + 3 + 1) = 1/16 P(droopy ears and crinkly tail) =

– Step 2: Add the individual probabilities

9/16 + 1/16 = 10/16

• 10/16 can be converted to 0.625– Therefore 62.5% of the offspring are predicted to have

normal ears and a normal tail or droopy ears and a crinkly tail

Page 26: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Product rule

• The probability that two or more independent events will occur is equal to the product of their respective probabilities

• Note– Independent events are those in which the

occurrence of one does not affect the probability of another

Page 27: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

• Consider the disease congenital analgesia – Recessive trait in humans– Affected individuals can distinguish between sensations

• However, extreme sensations are not perceived as painful

– Two alleles• P = Normal allele

• p = Congenital analgesia

• Question– Two heterozygous individuals plan to start a family– What is the probability that the couple’s first three children will all have

congenital analgesia?

Page 28: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

• Applying the product rule– Step 1: Calculate the individual probabilities

• This can be obtained via a Punnett square

1/4 P(congenital analgesia) =

– Step 2: Multiply the individual probabilities

1/4 X 1/4 X 1/4 = 1/64

• 1/64 can be converted to 0.016– Therefore 1.6% of the time, the first three offspring of a

heterozygous couple, will all have congenital analgesia

Page 29: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Crossing-Crossing-overover

Meiosis I

Meiosis II

4 Haploid cells, each unique

(Ind. Assort.)(Ind. Assort.)

Different genes are not always independent

Page 30: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

The haploid cells contain the same combination of

alleles as the original chromosomes

The arrangement of linked alleles has not been altered

Fig 5.1

Page 31: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

These haploid cells contain a combination of alleles NOT

found in the original chromosomes

These are termed parental or non-recombinant cells

This new combination of alleles is a result of

genetic recombination

These are termed recombinant cells

Fig 5.1

Page 32: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Linked alleles tend to be inherited together

Page 33: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Recombinants are produced by crossing over

Page 34: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Crossing over produces new allelic combinations

Page 35: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Only 2 chromosomes cross-over, and so the maximum number of recombinants that can be

produced is 50%.

Page 36: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

For linked genes, recombinant frequencies are less than 50 percent

Page 37: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Homologouspair of chromosomes

Page 38: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Does this show recombination?

D/dM1/M2

d/dM1/M2

D/dM1/M2

d/dM2/M2

D/dM2/M2

d/dM2/M2

Page 39: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Does this show recombination?

D/dM1/M2

d/dM1/M2

D/dM1/M2

d/dM2/M2

D/dM2/M2

d/dM2/M2

Page 40: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Longer regions have more crossovers and thus higher recombinant frequencies

Page 41: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Some crosses do not give the expected results

Page 42: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

=25%

8%9%41%42%

Page 43: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

These two genes are on the same chromosome

Page 44: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301
Page 45: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

By comparing recombination frequencies, a linkage map can be constructed

Page 46: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

By comparing recombination frequencies, a linkage map can be constructed

= 17 m.u.

Page 47: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

Linkage map of Drosophila chromosome 2:This type of map, with mapping units more than 50, can only be put together by making comparisons of linked genes.

Page 48: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

The probability of crossing over can be used to determine the spatial relationship of different genes

Page 49: Today: Imprinting and Linkage Exam #3 T 12/2 in class Final Sat. 12/6 from 9am – 12noon in BIO 301

similar to Fig 5.3,also see Fig 5.9,and pg 115-117

What is the relationship between these 3 genes? What order and how far apart?