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Genome 351, 15 April 2013, Lecture 5. Today…. Meiosis: how the genetic material is partitioned during the formation of gametes (sperm and eggs) Probability: -the product rule -the sum rule Independent assortment of nonhomologous chromosomes during meiosis. - PowerPoint PPT Presentation
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Today… Genome 351, 15 April 2013, Lecture 5 •Meiosis: how the genetic material is partitioned during the formation of gametes (sperm and eggs) •Probability: -the product rule -the sum rule •Independent assortment of nonhomologous chromosomes during meiosis
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Page 1: Today…

Today…Genome 351, 15 April 2013, Lecture 5

•Meiosis: how the genetic material is partitioned during the formation of gametes (sperm and eggs)• Probability:

-the product rule-the sum rule

• Independent assortment of nonhomologous chromosomes during meiosis

Page 2: Today…

Meiosis: the formation of gametes

DNAReplication

DNARecombination

Meiotic Division 1• Copied chromosomes (sister chromatids) stay joined together at the centromere.• Homologous chromosomes pair up and physically join at sites of recombination• Proteins pull the two homologs to opposite poles

Meiotic Division 2• Proteins pull the two sister chromatids to opposite poles• Each gamete gets a copy of only one homolog (usually a maternal-paternal hybrid).

(crossing over)

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Mitosis vs. Meiosis1m 1p

2 x 1m 2 x 1p

1m

1p1m1p

1m

1p1m

1p

2 x 1m 2 x 1p

exact copies

DNAReplication

DNARecombination

1m 1p

2 x 1m 2 x 1p

2 x 1m/p 2 x 1p/m

1m 1p1m/p1p/m

2 x 1m/p 2 x 1p/m

2 x 1m/p 2 x 1p/m

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Meiotic Division ICrossovers hold the homologues together until all of the chromosomes are attached to the spindle

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The homologues then separate from one another, exchanging corresponding portions as they do so

Meiotic Division ICrossovers hold the homologues together until all of the chromosomes are attached to the spindle

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The two daughter cells from meiotic division I go directly into meiotic division II

Sister chromatids separate during meiotic division II

Meiotic Division II

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• One round of DNA synthesis with one cell division

• Two genetically identical daughters

• Sister chromatids segregate

• Homologs do not line up or separate

• Homologs do not exchange corresponding segments (no crossing over)

• Final products are diploid (2n)

Mitosis vs. Meiosis

Page 8: Today…

Probability in genetics

Page 9: Today…

Probability is important in genetics

- testing hypotheses

- mapping disease genes

- genetic counseling

Needed for…

Page 10: Today…

Pedigree of a family segregating phenylketonuria (PKU)

What can we infer from the pedigree?

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A couple has a first child who tests positive for PKU. What can you infer about their genotypes?

What is the probability that their next child will have PKU?

What is the chance the next child, if he or she is not affected, will be a carrier?

Use of the Product and Sum rules

Page 12: Today…

DNAReplication

DNARecombination

2 copies PAH+ 2 copies PAH-

2 copies PAH+ 2 copies PAH-

PAH+ PAH-

PAH-

PAH-PAH+

PAH+

PAH-

PAH-PAH+

PAH+

PAH-

PAH-PAH+

PAH+

PAH+ PAH+ PAH- PAH-

Following the fate of the PAH gene in a PAH+/- heterozygote during meiosis

Page 13: Today…

DNAReplication

DNARecombination

PAH+

PAH-PAH+

PAH-

2 copies PAH+ 2 copies PAH-

2 copies PAH+ 2 copies PAH-

PAH+ PAH-

PAH+

PAH-

PAH+

PAH-PAH+

PAH-

PAH+

PAH-

PAH+ PAH- PAH+ PAH-

Following the fate of the PAH gene in a PAH+/- heterozygote during meiosis

Page 14: Today…

Genetic accounting

What are the possible genotypes and phenotypes of the children of parents who are both phenylketonuria carriers?

eggs

sper

m

Page 15: Today…

Product rulePAH+/- PAH+/-

PAH-/-

What is the probability that their next child will have PKU?

PAH+

PAH+

eggs

sper

m ½ PAH+

½ PAH-

½ PAH+ ½ PAH-

PAH+

PAH-

PAH+

PAH-PAH-

PAH-Probability of PAH-/-?

Product Rule: The probability of 2 or more independent events occurring simultaneously

Page 16: Today…

Sum rulePAH+/- PAH+/-

PAH-/-

What is the chance the next child, if he or she is not affected, will be a carrier?

PAH+

PAH+

eggs

sper

m ½ PAH+

½ PAH-

½ PAH+ ½ PAH-

PAH+

PAH-

PAH+

PAH-PAH-

PAH-Probability of PAH+/-?

= sum of the separate probabilities

Sum Rule: The probability of an event that can occur in 2 or more ways

Page 17: Today…

Punnett Square

Determine types of gametes from each parentCombine each type of female gamete with each male gamete

Advantages of Punnett SquareOrganized and systematicGives all possible combinations of genotypes automatically

Slow and labor intensive, especially for complex genotypes (e.g., AaBbCc X AabbCc)

Disadvantages of Punnett Square

Execution

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Two events necessary:II-3 must be Aa and they must have aa child

Example: Albinism…a = no pigmentWhat is the probability that III-1 will be albino?

Using the product and sum rules

Page 19: Today…

Independent assortment of nonhomologous chromosomes

during meiosisWhat happens to non-homologous chromosomes during meiosis?

Page 20: Today…

1m 1p

2m 2p

1m

2m1p

2p

1m2m

1m2m

1p

2p1p

2p

1p 1m

2m 2p

1p

2m

1p

2m1p

2m

1m2p

1m2p

1m2p

Two equally probable arrangements:

Independent assortment of nonhomologous chromosomes

during meiosis

Page 21: Today…

1 pair of homologous chromosomes gives 2 types of gametes (21 = 2)2 pairs of homologous chromosomes gives 4 types of gametes (22 = 4)n pairs of homologous chromosomes gives 2n types of gametes23 pairs of homologous chromosomes gives 223 (8 million) types of gametes

Examples:

Meiosis and independent assortment of nonhomologous chromosomes can create many different types of gametes

Page 22: Today…

An example of independent assortment

• Following the fate of genes on different (nonhomologous) chromosomes– Cystic fibrosis on chromosome 7– A gene that influences ABO blood

types on chromosome 9

Page 23: Today…

Some background on ABO blood groups

= A antigen= B antigen

A red cells

B red cells

AB red cells

O red cells

Page 24: Today…

The ABO (I) gene

There are 3 different versions (alleles) of the I gene:

IB

i

IA A adds A sugar to red cell surface

B adds B sugar to red cell surface

adds no sugar to red cell surface

I alleles

Page 25: Today…

The ABO gene - dominance relationships

IA is dominant to i

The I gene lies on chromosome 9q34

IA/IA or IA/i - A blood type

IB is dominant to i IB/IB or IB/i - B blood type

IA is co-dominant with IB IA/IB - AB blood type

i is recessive i/i - O blood type

Page 26: Today…

Independent assortment of genes on nonhomologous chromosomes

CFTR+ CFTR-

IA iCFTR+CFTR-

IA i

CFTR+

IACFTR-

iCFTR-

IACFTR+

i

CFTR+

IACFTR+

IACFTR-

iCFTR-

iCFTR-

IACFTR-

IACFTR+

iCFTR+

i

Gametes formed from a CFTR+/- IA/i double heterozygote:

Page 27: Today…

CFTR+ IA

CFTR- i

CFTR- IA

CFTR+ i

CFTR+ IA

CFTR- i

Gametes that arise from a CFTR+/- IA/i double

heterozygote:

CFTR+ i

CFTR- IA

1/4

1/4

1/4

1/4

Page 28: Today…

CFTR+

IA

CFTR+

iCFTR-

IA

CFTR-

i

CFTR+

IACFTR-

iCFTR+

iCFTR-

IAC+/+

IA/IAC+/-

IA/iC+/+

IA/iC+/-

IA/IA

C+/-

IA/iC+/+

IA/-C+/-

IA/IA

C-/-

i/iC+/-

i/iC-/-

IA/i

C+/-

i/iC-/-

IA/iC+/+

i/iC+/-

IA/iC+/-

IA/iC-/-

IA/IA

Eggs

Sper

m

What genotypes give CF and A blood type?What genotypes give nonCF and A blood type?What genotypes give nonCF and O blood type?What genotype gives CF and O blood type?

Possible genotypes and phenotypes from a mating of CFTR+/- IA/i double

heterozygotes

Page 29: Today…

CFTR+

IA

CFTR+

i

CFTR-

IA

CFTR-

i

CFTR+

IACFTR-

iCFTR+

iCFTR-

IA

C+/+

IA/IAC+/-

IA/iC+/+

IA/iC+/-

IA/IA

C+/-

IA/i

C+/+

IA/-

C+/-

IA/IA

C-/-

i/i

C+/-

i/i

C-/-

IA/i

C+/-

i/iC-/-

IA/i

C+/+

i/iC+/-

IA/i

C+/-

IA/iC-/-

IA/IA

9 N, A 3 N, O 3 CF, A1 CF, O

1/4

1/4

1/4

1/4

1/4 1/4 1/4 1/4


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