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Each chromosome has one DNA molecule Each chromosome has many genes
Lecture 10: What is a gene? Each chromosome has one DNA molecule Each chromosome has many genes A gene produces a protein that give rise to a phenotype A gene has many forms- alleles Different alleles are caused by different changes in the same gene Mutations in different genes CAN give you the same phenotype chromosome yellow Blanco eye Shaven body Forked bristle White eye Many genes Genes on DNA w1 w2 w3 Mutations in white b1 b2 Mutations in blanco
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Complementation Glutamic acid-
The complementation test is a rapid method of determining whether two independently isolated mutants with the same phenotype (in the same pathway) are in one or two (or more) genes. Ornithine Citruline Arginine Enzyme1 Enzyme2 Glutamic acid- Both mutant1 and mutant2 cannot make arginine. If you did not know the pathway you would wonder if these two mutants were mutations in the same gene or mutations in two different genes If you are working with Neurospora, you can feed the intermediate (Citruline) to the mutants and see if they can now make arginine. You are “complementing” the mutants with chemical intermediates Mutant1+ citruline=cell makes arginine Mutant2+citruline=cell cant make arginine- mutant
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B----> E----> A----> N
This process might also identify multiple mutants for the steps in the pathway! B----> E----> A----> N Mut3 Mut1 Mut4 Mut2 mut1 and mut4 might be two different mutations in the same gene. OR Mutations in two different genes that work in the same pathway B----> E----> S--- A----> N Mut3 Mut1 Mut2 Mut4 Feeding intermediates to complement genetic mutations is not possible all the time.
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Two individuals with the same phenotype
The wildtype eye color in flies is red. You start with red eyed flies. You mutagenize flies. You isolate TWO mutant flies that have white eyes. You cant feed flies eye pigment precursor to figure things out! You want to know if these two flies have mutations in the same gene or in two different genes
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One or two genes? Are two independently isolated mutations THAT HAVE THE SAME WHITE EYE PHENOTYPE disrupting the same or different genes? Precursor (white) Product (red pigment) Enzyme1 Gene1 OR Precursor (white) Intermediate (white) Product (red pigment) Enzyme1 Enzyme2 Gene1 Gene2
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If White1 and White2 are mutations in the same gene ….
Precursor (white) Product (red pigment) Enzyme1 w1w1 x w2w2 F1 w1w2 In the F1 will the flies be red eyed or white eyed? All white 1 2 1 2
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F1 to F2 w1 w1w1 w1w2 w1w2 w1w2 Precursor -------> product
White enzyme1 red white x white w1w1 w2w2 F1 w1w2 Phenotype = White What happens if you do a self cross with the F1 w1 w1w1 (white) w2 w1w2 (white) F2 w1w2 (white) w1w2 (white) w2 This is multiple alleles of the same gene
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If White1 and White2 are mutations in two different genes……
Precursor (white) Intermediate (blanco) Product (red pigment) Enzyme1 Enzyme2 Gene1 (W) Gene2 (B) wwBB x WWbb (white eye) (white eye) F1 Ww Bb phenotype= 1 2 1 2
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F1 to F2 = 9:7 Precursor----> intermediate----> product white white red EnzW EnzB White x white wwBB WWbb F1 WwBb (phenotype= Red) What happens if you do a self cross with the F1 WB Wb wB wb WWBB WWBb WwBB WwBb WWbB WwbB wWBB wWbB wWBb 9 W-B- red 3W-bb white 3wwB- white 1wwbb white WWbb Wwbb wwBB wwBb wWbb wwbB wwbb F2 This is Recessive epistasis
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Complementation in F1 Precursor -------> product White enzyme1 red
white x white wwwb wwwb F1 wwwb Phenotype = White Self cross with the F1 ww wwww (white) F2 wb wwwb wbwb Multiple alleles of the same gene Precursor----intermediate----product white white red EnzW EnzB White x white wwBB WWbb F1 WwBb Phenotype= Red Self cross with the F1 wB F2 WB Wb wb WB WWBB WWBb WwBB WwBb WWbB WWbb WwbB Wwbb Wb wWBB wWBb wwBb wwBB wB wb wwbB wWbB wWbb wwbb This is Recessive epistasis
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Map Genes Mapping genes takes lots of crosses and is time consuming
You could map each mutation. If white1 = white2 then the two mutations WILL map to the same spot on the chromosome. That would indicate that they are the SAME GENE -two different alleles! yellow singed Shaven body forked white1 white2 If on the other hand the two mutations map to different regions of the chromosome (or different chromosomes) then that would indicate that they are two different genes. yellow singed Shaven body forked white1 white2 Mapping genes takes lots of crosses and is time consuming There is an easier way!
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How many genes? What is the eye-color of the F1 w/b fly? Depends on whether the w and b mutations disrupt same gene or two different genes. If the w1 and w2 mutations disrupt the same gene? w1 w2 w1 w2 F1 white=blanco If w1 and w2 mutations disrupt two genes? w1 W2+ W1+ w2 W1+ w2 w W2+ F1 If the F1 flies are red-eyed, then white1 and white2 mutations disrupt two genes. You say that these two mutations complement one another. They complement because normal function is restored
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Xeroderma Pigmentosum
aaBB AAbb AaBb Xeroderma pigmentosum is caused by mutations in genes that are involved in repairing damaged DNA. Extreme sensitivity to UV light It is a recessive disease More than half of all cases in the United States result from mutations in the XPC, ERCC2 or POLH genes. Mutations in the other genes generally account for a smaller percentage of cases. Complementation gives rise to unaffected children Human deafness for example can be caused by genes that affect the structure of the cochlea and those that effect the formation of the auditory nerve. In this case you would have one a/a homozygote crossed with a b/b homozygote and the progeny would all be a/+; b/+
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Molecular basis of mutations
Now what is the molecular basis for two mutations within the same gene? Lets say that w1 and w2 both disrupt geneW What is a gene? It’s a piece of DNA with a specific sequence DNA consists of a linear array of the four nucleotides Adenine- Cytosine- Guanine- Thymine ACGT Specific DNA sequence = gene = protein ATGCCCCCCCCCCCCCCCCCCCTAA = GeneW = proteinW The nucleotides in the sequence of the gene is critical for its proper function.
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Mutant genes Nucleotide sequence of the normal W gene:
---ATGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCTAA--- ---TACGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGATT--- The sequence of the w1 mutation of gene W ---ATGCCCCCCCCCCCCCCCCCCCCCCCCCCTCCCTAA--- ---TACGGGGGGGGGGGGGGGGGGGGGGGGGGAGGGATT--- The sequence of the w2 mutation of gene W ---ATGCCCCACCCCCCCCCCCCCCCCCCCCCCCCCTAA--- ---TACGGGGTGGGGGGGGGGGGGGGGGGGGGGGGGATT---
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Mutant genes A w1/w1 homozygous will have the DNA sequence:
---ATGCCCCCCCCCCCCCCCCCCCCCCCCCCTCCCTAA--- ---TACGGGGGGGGGGGGGGGGGGGGGGGGGGAGGGATT--- A w2/w2 homozygous will have the DNA sequence: ---ATGCCCCACCCCCCCCCCCCCCCCCCCCCCCCCTAA--- ---TACGGGGTGGGGGGGGGGGGGGGGGGGGGGGGGATT--- the w1/w2 heterozygote would be: ---ATGCCCCCCCCCCCCCCCCCCCCCCCCCCTCCCTAA--- ---TACGGGGGGGGGGGGGGGGGGGGGGGGGGAGGGATT--- ---ATGCCCCACCCCCCCCCCCCCCCCCCCCCCCCCTAA--- ---TACGGGGTGGGGGGGGGGGGGGGGGGGGGGGGGATT---
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Two White Genes Precursor white Intermediate Product red Enzyme1
Lets make things more complicated. w1 and w2 disrupt one gene (geneW). w3 disrupt a second gene (geneB Precursor white Intermediate Product red Enzyme1 Enzyme2 Gene W Ww1 Ww2 Gene B Bw3 Disruptions in geneA (Ww1 and Ww2) and geneB (Bw3) give rise to white eyes. HOW DO YOU FIGURE OUT THAT w1 and w2 disrupt gene W and w3 disrupts geneB
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Genes Precursor white Intermediate Product red Enzyme1 Enzyme2 Gene W
Gene B Disruptions in geneW and geneB both give rise to white eyes. Cross a Ww1 fly with a Ww2 fly and see if you get red eyes. If w1 and w2 disrupt geneW, they will not complement . What about Bw3?
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Precursor white Intermediate Product red w1/w1 B/B w2/w2 B/B w1/w2 B/B
enzyme Blanco white blanco w1/w1 B/B w2/w2 B/B w1/w2 B/B F1=
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Precursor white Intermediate Product red w1/w1; B/B W/W; b(w3)/b(w3)
enzyme Blanco white blanco w1/w1; B/B W/W; b(w3)/b(w3) W/w1; b(w3)/B F1=
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Precursor white Intermediate Product red w2/w2; B/B W/W; b(w3)/b(w3)
enzyme Blanco white blanco w2/w2; B/B W/W; b(w3)/b(w3) W/w2; b(w3)/B F1=
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Genetic Complementation analysis
Genotype eye color complementation Ww1/Ww2 white - N Ww1/Bw3 red + Y Ww2/Bw3 red + Y w1, w2 = geneW= complementation groupW (multiple alleles) w3,= geneB= complementation groupB
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Suppose we isolate 5 delta wing mutations
We want to know how many genes are disrupted in these mutations and which mutations are in the same complementation group
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Complementation crosses
We systematically perform crosses First we perform the cross d1/d1 x d2/d2 F1 d1/d2 are produced Phenotype of F1 wings = flat or delta? If they are flat, they disrupt 2 genes Then we perform d1/d1 x d3/d3 F1 d1/d3 Phenotype of F1 wings=flat or delta? d1/d1 x d4/d4 F1 d1/d You construct a complementation table
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Genetic Complementation Table
You construct a complementation table + is flat wing - is delta wing Mutation complement mutation don’t complement Different genes same gene (diff complementation grp) (same complementation grp) Phenotype of F1 progeny: d1 d2 d3 d4 d5 d d d d d Complementation Grp1= Gene1= alleles (d1, d2, d5) Complementation Grp2= Gene2= alleles (d3, d4)
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The pathway Precursor delta Intermediate Product flat Enzyme1 Enzyme2 Gene1 (allele d1, d2, d5) GeneD2 (allele d3, d4)
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