Transmission Genetics Download

Transcript
Transmission (Classical, Mendelian)
Genetics Ch 11
• Gregor Mendel
– Experiments in Plant Hybridization, 1865
• Simple, controlled, data collection,
mathematical analysis
Pisum sativum, the garden pea
•
•
•
•
•
What makes this a good model organism?
easy to grow
hundreds of offspring per cross
short generation time
can self fertilize or cross
– Paint pollen (sperm) from one plant onto the
female parts of another (emasculated)
Mendel’s conclusions
1. Unit factors in pairs
- Genes are physical units
– 2 alleles for each gene
– 1 allele inherited from each parent
Genes and alleles of Pisum sativum
Gene
• Pea color
• Flower color
• Pod shape
• Pea surface
• Stem height
Alleles
?
white, purple
constricted, inflated
?
tall, dwarf
2. Principle of Dominance
- One allele is dominant the other is recessive
- The dominant allele is expressed in the
phenotype
Gene for flower color
P allele = purple
p allele = white
GENOTYPES
Homozygous dominant =
Heterozygous =
Homozygous recessive =
PHENOTYPE
3.
Random segregation of alleles into
gametes
– gamete receives ONE allele per gene
– random segregation of alleles 50/50
P generation
PP
pp
What is the phenotype of all offspring in
F1 generation?
Note that the P generation is true breeding
P
p
Genotype
Phenotype
p
P
How did Mendel do it?
The Monohybrid cross
YY yy
Which allele is dominant?
What is the genotype of the f1 generation?
Cross 2 f1 plants (or let one self-fertilize)
What is the ratio of phenotypes?
Results of Mendel’s monohybrid
crosses
Parental Strains
Tall X dwarf
Round seeds X wrinkled
Yellow seeds X green
Violet flowers X white
Inflated pods X constricted
Green pods X yellow
Axial flowers X terminal
gene = ?
alleles = ?
F2 progeny
Ratio
787 tall, 277 dwarf
5474 round, 1850 wrinkled
6022 yellow, 2001 green
705 violet, 224 white
882 inflated, 299 constricted
428 green, 152 yellow
651 axial, 207 terminal
Test cross (one gene)
• A mouse has black fur, what are its 2
possible genotypes?
Test cross mouse to homozygous recessive
mouse
If black mouse is BB 
If black mouse is Bb 
Mouse was test crossed and 7 offspring were
black 2 were white. What is mouse’s
genotype?
Autosomal recessive inheritance (bb)
• unaffected parents can have
affected offspring
• May “skip” a generation
• Two affected parents cannot
have an unaffected child
• Not sex related
Examples of autosomal recessive
traits
•
•
•
•
Sickle cell disease
Albinism
Cystic fibrosis
O blood type
Phenylketonuria (Ch. 4 pg 73)
• PKU (1/12,000) Mutation in gene encoding
phenylalanine hydroxylase enzyme needed for
phe metabolism
missing
phenylalanine
hydroxylase
enzyme
If plasma phe level is too high, phe is converted
into a phenylpyruvate toxic to brain tissue
Why are these babies normal when born?
Pleiotropic effects
no tyrosine (little melanin)
slow growth
retardation
blue eyes
low adrenaline
No nutrasweet
low phe diet
($5K/yr)
Page 68
1902 Archibald Garrod:
One gene: one enzyme
“Inborn errors of metabolism”
PKU
Albinism
Alkaptonuria
Tyrosinemia
Black urine
arthritis
One gene/one enzyme
• Garrod’s work on alkaptonuria
“Inborn Errors of Metabolism” 1902
Autosomal recessive metabolic disease
Fill in genotypes. If II,1 and II, 4 mate, what is the
chance of offspring having PKU?
How do we know
this is autosomal
recessive?
II, 1
X
II, 4
•p(aa AND a girl)?
p(aa)
If III-3 and II-1 mate p (normal child)
p (affected boy)?
All people have harmful recessive alleles, small
chance
That 2 people with same rare alleles will mate
Consanguinous marriage increases the chance
Bedoin intermarriage
Autosomal dominant disorders
Aa and AA =affected
aa =unaffected
•Tend to show up in every generation
•2 affected parents can have unaffected child
•2 unaffected parents cannot have an affected
child
Dominant pedigree
Achondroplasia -1/20,000 births
• Mutation in one allele of FGFR3
gene Chromosome 4
• Affects cartilage growth needed for
bone lengthening
• Most affected individuals are Aa
why?
• Most cases are spontaneous
(Parents are aa X aa)
P(III, 3 and III, 5 have a child of normal
height)
P ( II, 3 and III, 7 have a boy with
achondroplasia)
Fruit fly nomenclature
pg 317 box 12.1
Red eyes is wildtype phenotype, brown is mutant
bw+ = wildtype allele
bw = brown allele
genotype phenotype
red
brown
• Try it:
• Wingless is recessive mutant (wg allele)
• Genotype of wildtype, heterozygote,
mutant?
Autosomal
Sex-linked
Genetics Home Reference page
(National Library of Medicine)
• Collagen
• Blood clotting factor
• Red blood cell
enzyme
• Dystrophin muscle
protein
• Color vision gene
Sex-linked genes
Ch 12 pg 314 – 317, 326 - 328
• Human Female = XX
– two alleles for each X-linked gene
– normal application of recessive and
dominance
X HX H
X HX h
X hXh
• Human Male XY
XHY
XhY
X-linked genes
• Hemophilia (recessive) 1/5000 males
– Mutation in gene for clotting
factor
Xq28
•Mate III 13 with III 1 Probability of a hemophiliac son?
•Mate IV 2 with homozygous normal female p (hemophilia)?
*Criss cross inheritance of X linked traits
Dihybrid cross – 2 genes
Mendel’s Law of Independent assortment each allele for a trait is inherited independently
of other alleles
Seeds:
G = yellow allele
g = green allele
W = round allele
w = wrinkled allele
gene?
gene?
Parents =
GGWW X
phenotype?
gametes?
F1 genotype
?
F1 phenotype
?
F1 Gametes?
ggww
Note that each gene gives the 3:1 ratio of a
monohybrid cross
Yellow/green ratio =
Round/ wrinkled =
Forked line method for phenotypes
GgWw X GgWw
Test cross
A pea is round and yellow. What is its genotype?
G-W-
X
ggww
Note the cross of the “unknown” to a homozygous recessive
If all yellow and round:
If all yellow and some wrinkled:
If all round and some green:
If 1:1:1:1:
Probability
Product rulethe probability that two outcomes occur simultaneously is
product of their individual probabilities
assumes independent assortment of genes
GgWw X GgWw
• What is the probability of a yellow AND wrinkled?
p(G-ww)
Trihybrid cross
AaBbCc X AaBbCc
p(A-B-cc)
AabbCcDD X AaBbCcDd
p(triply recessive)
Modified Mendelian Ratios
1. INCOMPLETE DOMINANCE
R = red flower (snapdragon)
R’ = white flower
* allele symbols do not connote dominance
* phenotypic ratio = genotypic ratio = ?
P
CrCr
X
Cw Cw
F1
F2
Incomplete dominance
2. Codominance
Each allele encodes separate gene
product distinct in phenotype of heterozygote
L gene for human blood cell surface protein
LM = M antigen
LN = N antigen
• A man with the M bloodtype has a child
with a woman of the MN bloodtype
• Expected ratio of offspring?
3. Multiple alleles (more than 2 alleles for
gene in population)
• Example: Blood Groups
Karl Landsteiner 1900s
Chromosome 9 I gene
ABO blood system = polymorphic I gene
Blood type
A
B
AB
O
genotype
IAIA or IAi
IBIB or IBi
?
ii
What is the mechanism of inheritance of A, B, AB, O?
Autosomal or sex chromosome?
ABO dominance and codominance
4. Dominance series – C series/ rabbits
c+ = full color
cch = chinchilla (hypomorphic)
ch = himalayan (hypomorphic)
c = albino (apomorphic allele = nonfunctional)
Chinchill
a
Himalayan
Albino
Genotype
cch cch
cch ch
ch c
c+ cch
c+ = full color
cch = chinchilla (hypomorphic)
ch = himalayan (hypomorphic)
c = albino (apomorphic allele = nonfunctional)
phenotype?
5. Lethal alleles
MM = normal spine
MM’ = manx cat (no tail)
M’M’ = lethal
Cross two manx, what is ratio of phenotypes in
offspring?
How do breeders obtain manx cats?
6. Epistasis- gene product interactions. Table 13.4
page 355 (look at 4 phenotypic classes and fewer than 4)
• A product of one gene influences, or masks,
the expression of another gene(s)
• Modification of dihybrid cross ratio
AaBb
X
AaBb
9:3:3:1
Epistasis in Cats
• W = white
w = not white
• B = black
b = brown
Mate 2 heterozygous cats
What is the expected ratio?
Epistasis in labrador retrievers
• B and E color genes (labs)
B black
E color
b brown
e no color (yellow)
ee is epistatic
Cross two double heterozygotes
Phenotypes of parents?
Phenotypes of offspring? ratio?
Polydactyly, dominant
7. Penetrance
• % individuals that exhibit
phenotype corresponding to
genotype
Pp
5,5
pp
6, 5
6, 6
8. Expressivity (ex. Piebald spotting) –
the extent to which a trait is exhibited
osteogenesis imperfecta pg. 359
Penetrance AND expressivity
•
•
•
•
NF-1 = Neurofibromatosis1
Autosomal dominant trait N50 – 80% penetrance
Expressivity
– Pigmented skin to tumors on nerve CT coverings
(neurofibromas) on skin, eyes, organs, face
– Speech, blood pressure, spine curvature, headaches
9. Quantitative (multifactorial) traits
• Vary continuously
– Weight, height, IQ
Gene expression also affected by:
•
•
•
•
Sex (baldness)
Temperature (melanin in Siamese cats)
Chemicals (PKU)
Diet (height, cancer)
+ many other factors!
Similar