Midterm 1

• Mean = 74.6%

> 100: | |

> 90: | | | | | | | | | |

> 80: | | | | | | | | | | | | |

> 70: | | | | | | | | | | | | |

> 60: | | | | | | | | | | | |

> 50: | | | | | | | | | | | > 40: |

Extra Credit, To Date: = 18 pts

Please see me if you have questions about arithmetic, or obvious misgrades.

If you want to argue a point, I reserve the right to re-grade

the entire test.

Midterm 2

• Lectures,

• Assignments,

• Through Chapter 5, Chromosomal Mutations.

Linkage

• Genes located on the same chromosome do not recombine,

– unless crossing over occurs,

• The recombination frequency gives an estimate of the distance between the genes.

Recombination Frequencies

• Genes that are adjacent have a recombination frequency near 0%,

• Genes that are very far apart on a chromosome have a recombination frequency of 50%,

• The relative distance between linked genes influences the amount of recombination observed.

Linkage Ratio(How do you determine it?)

recombinant

total progeny

GW Gw gW gw

? ? ? ?

= Linkage Ratio

P GGWW x ggww

Testcross F1: GgWw x ggww

Experimentally

Linkage Ratio(If Sorting Independently)

50 (Gw) + 50 (gW)

200 (all classes)

GW Gw gW gw

50 50 50 50

= .5

P GGWW x ggww

Testcross F1: GgWw x ggww

Linkage Ratio Units

% = mu (map units)

- or -

% = cm (centimorgan)

Fly Crosses(white eyes, minature, yellow body)

• In a white eyes x miniature cross, 900 of the 2,441 progeny were recombinant, yielding a map distance of 36.9 mu,

• In a separate white eyes x yellow body cross, 11 of 2,205 progeny were recombinant, yielding a map distance of 0.5 mu,

• When a miniature x yellow body cross was performed, 650 of 1706 flies were recombinant, yielding a map distance of 38 mu.

Simple Mapping

• white eyes x miniature = 36.9 mu,

• white eyes x yellow body = 0.5 mu,

• miniature x yellow body = 38 mu,

my

38 mu

36.9 mu

w

0.5 mu

Three Point Testcross

Triple Heterozygous

(AaBbCc )

x

Triple Homozygous Recessive

(aabbcc)

Three Point Mapping Requirements

• The genotype of the organism producing the gametes must be heterozygous at all three loci,

• You have to be able to deduce the genotype of the gamete by looking at the phenotype of the offspring,

• You must look at enough offspring so that all crossover classes are represented.

Representing linked genes...

W G D w g d

x

w g dw g d

w g d

P

Testcross

= WwGgDd

= wwggdd

Representing linked genes...

+ + + w g d

x

w g dw g d

w g d

P

Testcross

= WwGgDd

= wwggdd

Phenotypic Classes

W-

ww

G-

gg

G-

gg

D-

dd

D-

dd

D-

dd

D-

dd

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

wwggdd

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

Crossovers

0

W G D

w g d1

1

2

#

179

52

46

4

22

22

2

wwggdd 173 0

1

1

2

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

#

179

52

46

4

22

22

2

wwggdd 173Parentals

Recombinants,double crossover

Recombinants 1 crossover, Region I

Recombinants 1 crossover, Region II

W G D

w g d

III

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

#

179

52

46

4

22

22

2

wwggdd 173Parentals

Recombinants,double crossover

Recombinants 1 crossover, Region I

Recombinants 1 crossover, Region II

W G D

w g d

I

Total = 500

Region I:

46 + 52 + 2 + 4

500x 100

= 20.8 mu

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

#

179

52

46

4

22

22

2

wwggdd 173Parentals

Recombinants,double crossover

Recombinants 1 crossover, Region I

Recombinants 1 crossover, Region II

W G D

w g d

II

Total = 500

Region II:

22 + 22 + 2 + 4

500x 100

= 10.0 mu

20.8 mu

W G D

w g d

W-gg-D

wwG-dd 4

2Recombinants,double crossover

Total = 500

10.0 mu 20.8 mu

0.1 x 0.208 = 0.0208

6/500 = 0.012

NO GOOD!

Interference

…the affect a crossing over event has on a second crossing over event in an adjacent region of the chromatid,

– (positive) interference: decreases the probability of a second crossing over,

• most common in eukaryotes,

– negative interference: increases the probability of a second crossing over.

Gene Order in Three Point Crosses

• Find either double cross-over phenotype, based on the recombination frequencies,

• Two parental alleles, and one cross over allele will be present,

• The cross over allele fits in the middle...

#

2001

52

46

589

990

887

600

1786

Which one is the odd one?

A C B

a c b

II I

A-B-C-

A-B-cc

A-bb-C-

A-bb cc

aaB-C-

aaB-cc

aabbC-

aabbcc

A-B-C-

A-B-cc

A-bb-C-

A-bb cc

aaB-C-

aaB-cc

aabbC-

#

2001

52

46

589

990

887

600

aabbcc 1786

Region I

A C B

a c b

I

990 + 887 + 52 + 46

6951x 100

= 28.4 mu

A-B-C-

A-B-cc

A-bb-C-

A-bb cc

aaB-C-

aaB-cc

aabbC-

#

2001

52

46

589

990

887

600

aabbcc 1786

Region II

A C B

a c b

28.4 mu

600 + 589 + 52 + 46

6951x 100

= 18.5 mu

II18.5 mu

Master

• Problems 1, 2

• Questions 4.1 - 4.4, 4.6 - 4.16, 4.19 - 4.20

Quantitative Traits

Optional; Klug and Cummings (at the library),

– Chapter 5, pp. 115 - 120, Insights and Solutions #1 (pp. 130), Questions 1-6 (pp. 131-132)

Reading Assignments

Available as a PDF, online.

Will answer question on Monday.

Read 15.5

Quantitative Traits

…traits that show a continuous variation in phenotype over a range,

…often result from multiple genes and are further termed polygenic traits.

Ribera, 1642

The Club Footed Boy

•Heart disease

•Spina bifida

•Neural tube disorders

•Diabetes

•Etc.

Environmental Factors

Optional: Chapter 15

Discontinuous Traits

• Discontinuous Traits,– exhibit only a few distinct phenotypes and can

be described in a qualitative manner,

• Mendel and others worked with “inbred” lines,Father: DD BB cc aa EE FF GG

Mother: dd BB cc aa EE FF GG

Continuous Traits

• “Outbred” transmission,

Father: dd BB CC AA EE FF GG Mother: DD bb CC aa EE ff gg

• Continuous Traits,

– Display a spectrum of phenotypes, and must be described in quantitative terms,

F3 #1: Cross Two Individuals With 10 Cm Ears.

F3 #2: Cross Two Individuals With 17 Cm Ears.

Polygenic Traits and Mendel

• It is possible to provide a Mendelian explanation for continuous variation by considering numbers of genes contributing to a phenotype,

– the more genes, the more phenotypic categories,

– the more categories, the more the variation seems continuous.

Melanin Pigmentation

Calculating the Number of Genes

• If you know the frequency of either extreme trait, then the number of genes (n) can be calculated using the formulae…

1

4n

= ratio of F2 individuals expressing

either extreme phenotype.

1/4 show the extreme traits,

1 4n = 1

41

n = 1

1 4n =

n = 2

1/16 show either extreme traits,

1 42

1 16

=

Continuous Variation

• Two or more genes that influence the same phenotype, often in an additive way,

– additive allele: contributes a set amount to the phenotype,– non-additive allele: does not contribute to the phenotype.

• The affect of each allele on the phenotype is relatively small, and roughly equivalent,

• Substantial variation is observed when multiple genes control a single trait,

• Must be studied in large populations.

• Inbred strain #1, mean height = 24 cm,• Inbred strain #2, mean height = 24 cm,• F1 mean height = 24 cm,• F2, 12 - 36 cm range, mean = 24 cm,

– 4/1000 are 12 cm, smallest class

– 4/1000 are 36 cm, largest class

a. What mode of inheritance?

b. How many gene pairs?Quantitative.

1/4n = 1/250, n = 4

4/1000 = 1/250 are the extreme class,

solve 1/4n = ratio of extreme class

42 = 16, 43 = 64, 44 = 256, etc.

• Inbred strain #1, mean height = 24 cm,• Inbred strain #2, mean height = 24 cm,• F1 mean height = 24 cm,• F2, 12 - 36 cm range, mean = 24 cm,

– 4/1000 are 12 cm, smallest class

– 4/1000 are 36 cm, largest class

c. How much does each allele contribute?

range: largest (36 cm) - smallest (12 cm) = 24 cm

alleles: 4 genes, 8 potential additive alleles

24 cm / 8 additive alleles = 3 cm / additive allelle

• Inbred strain #1, mean height = 24 cm,• Inbred strain #2, mean height = 24 cm,• F1 mean height = 24 cm,• F2, 12 - 36 cm range, mean = 24 cm,

– 4/1000 are 12 cm, smallest class

– 4/1000 are 36 cm, largest class

d. Indicate one possible set of P1, F1.

P1 AABBccdd x aabbCCDD

F1 AaBbCcDd x AaBbCcDd

Base Height is 12 cm, so each P1, F1 has 4 additive alleles.