Download doc Summer 2010 Lecture 4

BIOL 202
5.6 Phenotypes and gene interactions
In humans, look at families that are similar; look for traits as they are distributed in the
family—how the phenotype moves
LOD scores
- Log Of Difference
- Look for polymorphisms (e.g. 2 different alleles) in a family
- Let M be a molecular marker
o M has 2 forms, M1 and M2
 Differ by 1 bp
 M1 is GAATTC (restriction enzyme sequence for EcoR1
 M2 doesn’t have this sequence
- Let D represent the dominant allele responsible for the disease
o Dad, e.g. is heterozygous
o Test if disease locus linked to molecular marker
o Mated with “test cross”—homozygous recessive at both alleles
 Look at each indiv born and see probability of getting that
indiv
 Look at probability at different RF
o See the odds ratio; then find LOD
o Not significant, but do the same for more
families with the trait, and see their LODs; then
add the LODs together
o Get the probability of getting the end locus and
the disease
o As get closer to the gene, see higher probability
 Look at different markers; the ones with
highest probability of linkage show the
highest LOD scores
 LOD of 3 is statistically significant to
p=0.1
Phenotypes and gene interactions
- different kinds of dominance
o complete dominance (recessive mutations)
o incomplete dominance
o codominance
- Haplosufficiency
o 1 (recessive) allele that codes a gene LOF, the other normal gene allele
will suffice for completing the function
 if homozygous for recessive allele, disease/otherwise different
phenotype will appear; if heterozygous, will have normal
phenotype
- Haploinsufficiency
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o Dominant mutation
 Even if 1 normal allele, 1 bad allele  (usually just as) bad
phenotype
 If you have t/t, e.g., will have function, but not if have T/t or
T/T
 LOF for all but homozygous recessive
o Sometimes need proteins to work as dimers
 Homo/heterodimers
 If have protein that functions as dimer, and mutation alters one
of the proteins’ shape, could screw up the function
 Being heterozygous makes proteins completely nonfunctional though have both normal and bad (?)
proteins—is dominant
Incomplete dominance
o Gene encoding an enzyme or protein that, e.g. makes colour
 If have 0 WT, will have white; 1 WT, will get a lighter colour;
if have 2 WT, will get dark colour
o E.g. blood type
 Based on the I gene
 Encodes different forms of a surface protein that have 3
major alleles
o A, B, nothing (i)
o People with B blood would be IB/IB for B allele
o i makes no protein
 can see co-dominance: can recognize protein B and A using
different antibodies
 AB has both A and B proteins
How characterize phenotype can influence type of dominance seen
o Consequence of pleiotropy
o E.g. sickle cell allele
 Encodes less water-soluble protein that keeps blood cells from
puffing up
 Red blood cells have most efficient surface areas for oxygen
transfer
 Heterozygous has both normal and sickle-cell blood cells
 Wrt anemia, HbS (sickle cells) is recessive to normal
HbA
 But wrt blood cell shape, see incomplete dominance
o At molecular level, HbA and HbS are
codominant
Recessive lethals
o E.g. yellow mice
 Normal mice have dark pigmentation; yellow mice have lighter
coats
 Never get pure yellow mice
 Couldn’t get purebred
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Got 2:1 ratio
o One of the genotypes is lethal in utero
 Mating yellow with normal mice gives 1:1 ratio
o Also occurs in other organisms
 E.g. cats with no tails: end up with 2:1 ratio again; can’t get
truebreeders
o Some mutations present at different conditions
 E.g. lethal at higher temperatures; normal at lower temp (e.g.
because protein is on verge of instability)
 Useful for determining mutations
 E.g. auxotrophs
 Can screen for these by putting them in minimal media
o Just has basic salts and sugars
o Supplement media with vitamins and amino
acids
 E.g. find auxotroph of arginine (needs arginine to
survive)
o Systematically put in medias with and without
arginine
o If can produce arg from precursor, have 3
proteins in the pathway
 Look for single genes encoding the
enzymes in the pathway (1 gene, 1
enzyme theory)
 Bombard WT with X-rays to mutagenize
o Process with WT neurospora
o Put ascospores into individual tubes with media
(complete media—permissive conditions for
growth)
 Transferred into minimal media
 Look for a lack of growth; go back to the
original stock of that which didn’t grow
in minimal media
 See if grows if amino acids or
vitamins added to minimal media
o Grew in presence of
amino acids
o  what’s missing is
ability to make one of the
amino acids
o took same strain out and
put in different conditions
 added single
amino acids
 found that grew in presence of arginine
(was deficient in arg)
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then looked to see if 1 gene affected all the reactions, or
if needed different genes
o (found several mutants that didn’t grow; picked
out the arg mutants)
o found 3 types of arg mutants
 mated them together
 arg-1 defect was in enzyme X that
formed ornithine: grew if added
ornithine; citrulline, arg
 arg-2 defect was in enzyme y, so grew if
added citrulline and arg
 arg-3 grew only if added arg mutation
was in enzyme z
Interactions between genes in pathways
- Several varieties of, e.g., true-breeding white-flowered mutants can be found;
are they mutations of the same or different alleles?
o E.g. find in Japan a white version of nightshade, as well as in Nova
Scotia find another white nightshade—probably different mutations
 How prove mutations are/aren’t in the same gene?
 Using complementation
o Only works for recessive mutations
o Always start with true-breeding strain
o 2 genes control colour of flower, e.g.
 have different areas with white flowers;
know that they’re all recessive. Are they
mutations in the same genes?
 if they’re mutations in the same gene:
 interbreed and see if end up with
white flowers or blue flowers
 if end up with blue flowers, the 2
complement each other
(mutations are in different genes)
o get 1 good copy from the
other gene—presuming
complete dominance
 if end up with white flowers,
mutations don’t complement
each other (same gene mutation)
o get 2 bad copies of the
gene
o see that the 2 genes interact genetically to
produce blue phenotype (by producing different
enzymes that process precursors)
- Independence of actions of each gene
o If took truebreeding black snake and mated with truebreeding orange
snake, end up with camouflage snake
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 End up with a 9:3:3:1 ratio in the next cross
o Back to flowers from before: crossing, e.g. british with Japanese
strains, have a 9:3:3:1 genotypic ratio, but not a 9:3:3:1 phenotypic
ratio
 Get blueness if have 1 dominant allele encoding proper enzyme
2 and 1 dominant allele encoding proper enzyme 1
 But if homozygous for british mutation (is on W1
gene), end up with homozygous WT for W2 gene, but
enzyme 1 blocked (LOF)  white flower
 Same idea if W2 gene is blocked/LOF: end up with
white flower again
 Get 9:7 ratio—still adds up to 16
Interaction b/w regulator genes and target genes
o Mutations in regulatory gene, recessive homozygous mutation: gene
for protein won’t make protein
o But if mutation in protein gene making non-functional protein, same
phenotype
o Mating dihybrid ends up with 9:3:3:1 genotype ratio but not phenotype
ratio because a few genotypes cause the same phenotype
Recessive epistasis
o If, e.g. enzyme 1 makes pink flower, and enzyme 2 makes blue;
enzyme 2 block makes pink; if enzyme 1 blocked, remains white
(colourless)
o 1 gene affects ability of other gene to make phenotype = epistasis
o e.g. Laborador retrievers
 black, chocolate, golden
 B = dominant, black ; b = recessive, chocolate
 Truebreeding chocolate mated to truebreeding golden 
classic heterozygous; then end up with 9:3:3:1 in selfed cross
 Dihybrid cross  9 black, 3 chocolate, 4 golden phenotypic
ratio
 B/--; E/--  black
 b/b; E/--  chocolate
 B/--; e/e  golden
 b/b; e/e  golden
o recessive phenotype of one gene affects what you see from the other
gene
Dominant epistasis
o E.g. flowers: D gene affects intensity of redness of pigment
 D = red; d = pink
 W gene affects whether pigment can be made
 w/w = pigment production; W/-- = no pigment
production
 W/-- ; D/--  white
 Bred to a truebreeding red (w/w; D/--)  white
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If have W allele, doesn’t matter whether have D or d, or
1 w  will be white because of dominant epistasis