Lecture 13

Question 1

Gene Interaction: Important Points

Answer 1

• Mutations have various functional effects: dominant/recessive and loss of function/gain of function.
• Gene interaction can modify the di-hybrid mendelian ratios (9:3:3:1)
• Different Loci can influence the same phenotype: complementation.

Question 2

Functional consequences of mutations

Answer 2

a. Wild type: The expression of the products of wild-type alleles produces wild-type phenotype.
b. Loss of function (leaky/hypomorphic mutations): Leaky mutant alleles produce a small amount of wild-type gene product. Homozygous organisms have a mutant (hypomorphic) phenotype.

Question 3

Functional Consequences of Mutations pt2

Answer 3

a. Wild type: The expression of the products of wild-type alleles produces wild-type phenotypes.
b. Loss of function (Null/amorphic mutation): Null alleles produce no functional product. Homozygous null organisms have mutant (amorphic) phenotype due to absence of the gene product..
d. Loss of function: dominant negative mutation: The formation of multimeric proteins is altered by dominant negative mutants whose products interact abnormally with the protein products of other genes, leading to malformed multimeric proteins.

Question 4

Wild type vs Gain of Function

Answer 4

e. Gain of Function (hypermorphic mutation): Excessive expression of the gene product leads to excessive gene action. The mutant phenotype may be more severe or lethal in the homozygous genotype than in the heterozygous genotype.
f. Gain of function: Neomorphic mutation: The mutant allele has novel function that produces a mutant phenotype in homozygous and heterozygous organisms, and may be more severe in homozygous organisms.

Question 5

Loss vs. Gain of function

Answer 5

Refers to the molecular activity as compared to the wild type gene product

Question 6

Dominant vs Recessive

Answer 6

• Refers to the phenotype:
• dominant mutation will lead to a phenotype in the heterozygous situation.
• a recessive mutation wont: the heterozygous will be the same as the homozygous wild-type.
• Those are two different concepts, but they are often related.

Question 7

Loss of function is often…

Answer 7

RECESSIVE, but not always.

Question 8

Haploinsufficiency

Answer 8

• The Heterozygous null +/- shows a phenotype

- This is due to incomplete gene activity from a single wild-type allele.

Question 9

Dominant Negative Alleles

Answer 9

• These are often seen with multimeric proteins.
• ALL subunits must be wild-type for full activity
• A single mutant subunits leads to reduced or lost activity
• Think of it as a “spoiler effect” or “poisonous” subunit.

Question 10

Gain of Function if often DOMINANT

Answer 10

• It can result from increased transcription or increased intrinsic gene product activity.
• Just like loss of functional alleles, gain of function alleles have various degrees of functional consequence:
• A dominant allele can be STRONGLY dominant, or WEAKLY dominant.
• Alleles can have various degrees of PENETRANCE.

Question 11

Co-dominance:

Answer 11

this term is for the wild-type alleles when both alleles are detectable into the heterozygous situation.
- Classic example: BLOOD GROUPS (ABO LOCUS).

Question 12

Co-dominance: ABO locus

Answer 12

• In model genetic systems such as yeast, Drosophila, mouse, etc:
• The lab lines have been INBRED and there is usually one allele that is the wild-type allele.
• In this case: the wild type allele is the one present in the reference organism. A mutant allele then is anything different from the wild-type (reference) allele.

Question 13

Allelic Series

Answer 13

• The lethality is due to loss of Raly, itself an essential gene.
• Chromosomes carrying wild-type A alleles produce Raly protein required for mouse embryonic development, and a moderate amount of yellow pigment.
• Chromosomes carrying the mutant Ay allele produce no Raly protein and a very high level of yellow pigment due to the hypermorphic mutation.
• The notion of essential gene is important: A gene absoulutely required for viability, at the cellular level (e.g. Yeast) or at the level of the organism (e.g. during mouse development).
• NOTE: AA^Y = Raly +
• A^Y A^Y = Raly -

Question 14

Pathway Analysis

One gene- one enzyme hypothesis (Beadle and Tatum)

Answer 14

• Genetics, through mutant analysis has helped discovery and study of PATHWAYS:
• Biosynthetic: anabolic
• Degradation: catabolic
• Signal transduction
• Developmental

Question 15

Order of Intermediates in the Biosynthetic Pathway

Answer 15

• Homoserine -> Cysteine -> Cystathionine -> Homocysteine -> Methionine. (Met 4 -> Met 3 -> Met 2 -> Met 1)
• Additions of specific intermediates in the biosynthetic pathway allows the genetic dissection of Methionine biosynthesis in Neurospora (Horowitz)

Question 16

Epistasis

Answer 16

• Interaction between genes influencing the same phenotype
• Without epistasis: two genes influence different phenotype and Dihybrid inheritance is characterized by the 9:3:3:1 ratio.
• Ex: green/yellow versus smooth/wrinkled peas.
• With epistasis, two different genes influencing the smae phenotype will lead to specific DISTORSION of the 9:3:3:1 ratio. This is known as epistasis.
• Interaction between genes influencing the same phenotype depending on the nature of the gene interaction involved between the two genes, the ratios obtained differ.

Question 17

Epistasis Example #1

Answer 17

• Two genes act in a pathway and must both be active to produce a product: the anthocyanin pigment in the sweet pea.
• If either one is defective, there is no pigment and the flowers are white.
• The obtained ratio is NOT 9:3:3:1.
• Why? Because three genotypic classes combine into ONE phenotypic class: 9: (3+3+1) = 9:7 phenotypic ratio.

Question 18

Epistasis Example #2

Answer 18

• Two genes act in the SAME step to produce the pigment- think of it as two REDUNDANT GENES: duplicate gene action.
• In this case, the activity of either gene is sufficient to produce the wild-type phenotype.
• Only the double mutant has the mutant phenotype (white flowers)
• So, phenotypically, we get (9+3+3):1 = 15:1 ratio.

Question 19

Epistasis Example #3

Answer 19

• Two genes act in the SAME step but a defect in either gene produces a phenotype different from the double mutant: Wt/wt =disk; wt/a or wt/b =sphere; a/b= long.
• In this case, we add the two genotypic classes A_bb and aaB_:
  9: (3+3):1 = 9:6:1.

Question 20

Epistasis Example #4

Answer 20

• One gene is required for the activity of the other.
• In this case, ee: the dogs are yellow regardless of B: recessive epistasis.
• In this case, we add the two genotypic classes B_ee and bbee.
• 9:3:(3+1) = 9:3:4
• Note: with E_, the alleles in B matter. B_ is different from bb (black vs chocolate)

Question 21

Epistasis Example #5

Answer 21

• One gene masks the activity of the other.
  -In this case, W_: the squash are white regardless of Y.
  -This is dominant epistasis.
• In this case, we add the two genotypic classes W_Y_ and W_yy:
  (9+3):3:1 = 12:3:1.
• Note: with ww, the alleles in Y matter. Y_ is different from yy. (green vs yellow).

Question 22

Epistasis Example #6

Answer 22

• One gene masks the activity of the other.
• In this case, D_: this is a special case of dominant epistasis called Dominant Supression.
• The difference with the previous case is that wih dd (no supression is occuring).
• The alleles at L MATTER: L_(blue) is different form II (white)
• The ratio becomes 13:3.

Question 23

Epistasis (cont)

Answer 23

• In all cases, the genomic classes are unchanged, but epistatic interactions change the phenotypic ratios from 9:3:3:1.
• The observed ratios are informative on the NATURE of gene interactions in determining the phenotypes.

Question 24

The complementation test: How many genes?

Answer 24

• Several independent mutations have the same phenotype.
• For example, mutations affecting eye color in Drosophila.
• White, apricot, buff, carnation, coral, vermillion, cherry, brown All affect the eye color but…
• Are they in the same gene, or are they in different genes? And if in different genes, how many?

Question 25

The complementation test: How many genes? (pt2)

Answer 25

• To address this question, you cross two homozygous mutant flies and ask whether the progeny is wild-type or mutant.
• If two independent alleles are in the SAME gene, there will be NO complementation.
• If the progeny is wild-type, there is complementation and the two loci represent DIFFERENT GENES.
• Note #1: the mutations must be recessive
• Note #2: there are exceptions.

Question 26

The complementation test:

How many genes? The results of these crosses can be represented in a complementation table:

Answer 26

For example: a cross between apricot -/- and cherry -/- gives a progeny: apricot ‘-/+’ cherry ‘-/+’

You see in the table that there is a - : NO complementation: there is a defective gene product in the double heterozygous. Therefore, apricot and cherry are independent mutations in the SAME gene.