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Announcements. 1. You should be finishing “Monk in the garden” and starting to answer questions for writing assignment. We will discuss this next week. 2. Dr. Brad Swanson is guest lecturing on conservation genetics on Monday. This material will be on Exam 2 .

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slide1

Announcements

  • 1. You should be finishing “Monk in the garden” and starting to answer questions for writing assignment. We will discuss this next week.

2. Dr. Brad Swanson is guest lecturing on conservation genetics on Monday. This material will be on Exam 2.

3. “Problem set 2” answers due today at start of class.

  • If you’re in Group A for presentations, I encourage you to discuss your main source/reference with me. Plan ahead! Group A presents in 11 or 12 days.
  • Pay special attention to “Insights and Solutions” - Ch .4, p. 104-106.

6. Study this weekend! Your exam next week is Thursday and Friday. Avoid lines, go early in the day on Thursday. Give yourself enough time to finish the exam.

slide4

Review of last lecture

1. Incomplete dominance vs. Codominance

  • 2. Multiple alleles of a gene -studied in a population
  • ex. Blood group alleles, A, B, AB, O
  • Lethal alleles - recessive or dominant

-can lead to modified Mendelian ratios such as 2:1 genotypic and phenotypic ratios in the F1

4. Modification of the 9:3:3:1 ratio: examining inheritance of two gene pairs that have different modes of inheritance

slide5

Modified Dihybrid Cross: First Consider Each Trait on its Own

Ex. 2 humans heterozygous for albinism and are blood type AB

slide7

Outline of Lecture 8

I. Epistasis - genetic interaction affecting expression of one trait

II. Novel phenotypes due to gene interaction

III. Complementation analysis

  • X-linkage and color blindness

V. Is phenotype always completely due to genotype?

slide8

I. Epistasis

  • Two or more separate genes (not separate alleles at one genetic locus) interact to control a phenotypic character.
  • If one gene locus prevents the expression of a second gene, the first locus is epistatic to the second, and the second is hypostatic to the first.
slide9

Example 1: H/h and AB pathway

A antigen

A allele

HH or Hh

Precursor H substance

H Substance

hh

B allele

B antigen

slide10

Epistasis example 2: Coat color in mice

Wt coat color is agouti - A (dominant to black); Nonagouti (black) coat color - a

Pigmentation expression - B (dominant to albino); No pigmentation (albino) - b

If individual is bb, then is albino regardless of allele at a locus - due to gene interaction.

slide11

Agouti Markings

  • A- hair made with bands of black pigment and yellow pigment.
  • Aa hair all black.
slide12

Coat color example of epistasis, cont.

AABB (agouti) x aabb (albino)

AaBb (all agouti)

Genotype Phenotype F2 ratio Final phenotypic ratio

AaBb x AaBb

A-B- agouti 9/16 9/16

A-bb albino 3/16 4/16

aaB- black 3/16 3/16

aabb albino 1/16

Due to gene interaction, we see a 9:3:4 F2 ratio. The b locus is epistatic to the a locus.

slide13

Example of dominant epistasis, a 12:3:1 ratio

Inheritance of fruit color in summer squash: two loci together control color and a dominant allele at one locus can mask the expression of the alleles at the second locus.

A--- white

aaB- yellow

aabb green

A is dominant to a, and the a locus is epistatic to the b locus.

Therefore, if AaBb is crossed to AaBb, the F2 is as follows:

A-B- white 9/16

A-bb white 3/16

aaB- yellow 3/16

aabb green 1/16

12/16 white

3/16 yellow

1/16 green

slide14

Novel phenotypes and F2 ratios

due to gene interaction

Summer squash fruit shape inheritance:

AABB - disc shape x aabb - long shape

F1 = AaBb - disc shape

F2 = A-B- disc 9/16 9/16 disc

A-bb sphere 3/16 6/16 sphere

aaB- sphere 3/16

aabb long 1/16 1/16 long

slide15

Summary of modified F2 ratios - was Mendel just wrong?

No, none of these cases has violated the principles of segregation and independent assortment - just added complexity.

slide16

Practice Problem

In a plant, a tall variety was crossed with a dwarf variety. All F1 plants were tall. When two F1 plants were interbred, 9/16 of the F2 were tall and 7/16 were dwarf.

Explain the inheritance of height by a) indicating the number of gene pairs involved and b) by designating which genotypes yield tall and which yield dwarf.

When studying a single character, a ratio expressed in 16 parts suggests that two gene pairs are “interacting” during the expression of the phenotype.

A 9:7 ratio implies a dihybrid condition with epistasis.

slide17

3. From any dihybrid cross of double heterozygotes, we see the following genotypic ratio:

A-B- 9/16

A-bb 3/16

aaB- 3/16

aabb 1/16

tall

dwarf

dwarf

dwarf

4. In our problem, we see a 9:7 phenotypic ratio. If the 3:3:1 groups above were lumped together, a 9:7 ratio emerges.

Assign tall to any plant with both A-B- and dwarf to any plant that is homozygous recessive for either or both the recessive alleles.

slide18

III. Complementation Analysis Reveals Whether 2 Mutations are at the Same Locus

The mutations do complement each other, so they are in different complementation groups.

Mutations do not complement, so

in same complememtation group.

slide19

IV. X Linkage

  • Affects males and females differently
  • Example: red-green color blindness
  • Other examples in humans:
    • Hemophilia
    • Duchenne’s muscular distrophy

“3” or “8”??

slide20

X-linkage means genes on X chromosome

  • Sex chromosomes are “unlike” in many species; all others are autosomes.
  • In humans and Drosophila, males contain XY and females contain XX.
  • Y chromosome contains region of pairing homology with X: pseudoautosomal region.
  • Y contains few genes; X contains many.
slide21

Color Blindness Pedigree

Males are hemizygous

for this gene locus; only

females can be carriers.

slide22

Hemophilia in the European Royal Families

  • Disorder is lethal or debilitating prior to reproductive maturation.
  • X-linked recessive: heterozygous females are carriers with 50% chance of passing trait to sons only.
  • Arose from spontaneous mutation in Queen Victoria?
  • Hemizygous males were affected in Russian, German, Spanish royal families, trait not passed on in British royal family.
slide23

Is phenotype always due to genotype?

Penetrance vs. Expressivity

If 70% of the individuals show the mutant phenotype, then the penetrance is 70%.

Expressivity reflects the range of expression of the mutant genotype.

Variable penetrance or expressivity?

slide24

Temperature effects on phenotype

Temperature can affect coat coloration.

Siamese cat fur in the extremities is darker due to cooler temperatures. The enzyme making darker pigment doesn’t work well at the higher temperatures in the rest of the body.

Temperature also affects primrose flower color and fur of Himalayan rabbits.

slide25

Influence of chemicals on phenotype

PKU: phenotype is mental retardation due to metabolic disorder; severity is affected by diet and whether phenylalanine is restricted in the diet.

Phenocopies: non-hereditary phenotypic modification that mimics a phenotype caused by a known gene mutation. example: deafness can be caused in a developing baby if a mother is infected with rubella during the first 12 weeks of pregnancy; also caused by homozygous, recessive mutation

slide26

Summary of section V

Variation in most of the genetic traits that we have considered so far is determined predominantly by differences in genotype.

For many traits, however, the traits are influenced by both genes and the environment.

Nature vs. Nurture debates for human behaviors:

Ex. Considerations for the weekend/ alcoholism:

genes influence how susceptible you are to alcohol abuse but they do not force you to drink alcohol!