Transcript Nerve activates contraction

CHAPTER 14
MENDEL AND THE
GENE IDEA
Gregor Mendel 1822-1884
“My scientific labors have brought
me a great deal of satisfaction,
and I am convinced that before
long the entire world will praise
the result of these labors."
A Cloning Poem
Mary had a little lamb, its fleece was slightly gray,
It didn't have a father, just some borrowed DNA.
It sort of had a mother, though the ovum was on loan,
It was not so much a lambkin, as a little lamby clone.
And soon it had a fellow clone, and soon it had some more,
They followed her to school one day, all cramming through the door.
It made the children laugh and sing, the teachers found it droll,
There were too many lamby clones, for Mary to control.
No other could control the sheep, since their programs didn't vary,
So the scientists resolved it all, by simply cloning Mary.
But now they feel quite sheepish, those scientists unwary,
One problem solved, but what to do, with Mary, Mary, Mary!
History of genetics: Leeuwenhoek (1632-1723)
Homunculus – contained inside a sperm (spermist)!!!!!!!
History of genetics:
Charles Darwin
Body cells shed 'gemmules'
which collect in the
reproductive organs prior to
fertilization
History of genetics:
“Blending” hypothesis
–Genetic material
contributed by each parent
mixes in a manner
analogous to the way blue
and yellow paints blend to
make green.
Mendel brought an experimental and
quantitative approach to genetics (1800’s)
• Pea plants have several
advantages for genetics.
Many phenotypes
(physical traits)
Can control fertilization
Short lifecycle
Mendel’s Pea Plant Experiments
Phenotype/trait: feature of
an organism that can be seen.
Ex. flower color (purple/white)
Vocab
Chromosome 1 from Mom
Chromosome 1 derived from Mom and Dad
Chromosomes
A gene locus
A pair of alleles
3 gene pairs
A and a
IMPORTANT: There are at least 2
genes for every trait (called alleles)
Genotype: Genetic make-up of
individual for a given phenotype.
Ex. AA
or
aa
Alleles: Alternative forms of the
same gene (A and a). Gene is - Flower Color.
Homozygous—having 2 of the same allele
A
A
Homozygous
Dominant
(AA)
a
a
Homozygous
Recessive
(aa)
Heterozygous—having 2 different alleles
A
a
Heterozygous
(Aa)
Dominant Allele: Allele that is fully expressed
in a heterozygote (A)
Aa
Recessive Allele: Allele that is masked in a
heterozygote (a)
Aa
True-breeding—always producing offspring
with the same traits as the parents when the
parent plants are self-fertilized
Which plant/s are true breeding?
AA plant, Aa plant, aa plant
True Breeding
AA X AA will always give AA
Aa X Aa - Hmmm
aa X aa - will always give aa
True Breeding
How to cross
pea plants
Punnett
Square:
used to predict
probabilities of
offspring
genotype and
phenotype
Parent Cross:
Parents are true
breeding: this
means they are AA
(pink) or aa (white)
Mendel isolated true
breeding plants by
growing them are selffertilizing to check for
the offspring that
resembled parents
All offspring in the F1
will be:
HETEROZYGOUS
Parent cross produces F1
A
A
a Aa
Aa
a Aa
Aa
Genotype ratio: 0AA : 1Aa : 0aa
Phenotype ratio: 1 pink: 0 white
Probability: 1 pink : 0 white
Frequency : 100% pink : 0% white
Trait = flower color
Pink = dominant
White = recessive
Female parent
Male Parent
A
A
Both parents are heterozygous for the trait
a
A
a
a
a
Aa
A
F1 = A a
A
a
A
A
Monohybrid Cross
A
a
A
a
Aa
aa
A
aa
A
a
AA
Aa
a
Monohybrid cross Offspring (F2):
Phenotype ratio: 3 pink: 1 white
Genotype ratio: 1 AA: 2 Aa: 1 aa
a
Aa
aa
a
Aa
aa
Probability:3/4 pink: 1/4 white
Test or Back Cross:
Given: Mom is not
Deaf (only 2 alleles)
50% pups are deaf
What is the genotype of
Mom ?
Test Cross - Unknown
parent is crossed with
1) Assume unknown is DD
D
D
d Dd
Dd
d Dd
Dd
HOMOZYGOUS
RECESSIVE
Genotype ratio: 0DD : 1Dd : 0dd
Phenotype ratio: 1 norm : 0deaf
Probability: 1 norm : 0 deaf
Frequency : 100% norm : 0% deaf
Test or Back Cross:
Given: Mom is not Deaf
(only 2 alleles)
50% pups are deaf
2) Assume unknown is Dd
D
d
d Dd
dd
d Dd
dd
Unknown parent is
Crossed with
HOMOZYGOUS
RECESSIVE
Now the unknown
Parent (mom) is Dd
Genotype ratio: 0DD : 1Dd : 1 dd
Phenotype ratio: 1 norm : 1 deaf
Probability: 1/2 norm : 1/2 deaf
Frequency : 50% norm : 50% deaf
Mendel’s Law of Segregation
Alleles (A and a) separate in
meiosis (gamete formation)
since homologous
chromosomes are separated
AA X aa
No segregation
All AA (purple)
Or aa (white)
AA X
aa
Meiosis segregation
gametes
The paired condition is
restored by
the random fusion of
gametes at
fertilization.
F1
A
A
a
All Aa (purple)
Aa Aa
Aa Aa
a
AABB
purpleflowered
tall
AB
X
ab
F1 OUTCOME: All F1 plants purple-flowered, tall
(AaBb heterozygotes)
AaBb
aabb
whiteflowered
dwarf
Dihybrid Cross
AaBb X AaBb (Hybrids for both traits)
AaBb
AaBb
1/4 AB 1/4 Ab 1/4 aB
1/4 ab
1/4
AB
1/16
AABB
1/16
AABb
1/16
AaBB
1/4
Ab
1/16
AABb
1/16
AAbb
1/16
AaBb
1/16
Aabb
1/16
AaBB
1/16
AaBb
1/16
aaBB
1/16
aaBb
1/16
AaBb
1/16
Aabb
1/16
aaBb
1/16
aabb
1/4
aB
1/4
ab
1/16
AaBb
purple-flowered, tall
purple-flowered, dwarf
white-flowered, tall
white-flowered, dwarf
Phenotype ratio:9:3:3:1
Mendel’s Law of independent assortment :when
there are 2 or more allelle pairs, each pair of
alleles segregates into gametes independently
Mendelian inheritance reflects rule of
probability
Rule of multiplication - chance that two or more
independent events will occur together
– Compute probability of each independent event.
– Multiply the individual probabilities
– The probability that two coins tossed at the same
time will land heads up is 1/2 x 1/2 = 1/4
– The probability that a heterogyzous pea plant (Aa)
will produce a white-flowered offspring (aa) is
1/2 x 1/2 = 1/4
• The rule of multiplication also applies to
dihybrid crosses.
– For a heterozygous parent (YyRr) the
probability of producing a YR gamete is
1/2 x 1/2 = 1/4
– The probability that an F2 plant will have a
YYRR genotype from a heterozygous parent
is 1/4 YR x 1/4 YR = 1/16
– The probability that an F2 plant will have a
YyRr genotype from heterozygous parents is
(1/4 YR x 1/4 yr) +
(1/4 yR x 1/4 Yr) +
(1/4 yr x 1/4 YR) +
(1/4 Yr x 1/4 yR) = 4/16 = 1/4
• The rule of addition - the probability of an
event that can occur two or more different
ways is the sum of the separate probabilities
of those ways.
• The probability of a heterozygote in a
Monohybrid cross is
1/4 + 1/4 = 1/2
• Probability of finding two recessive phenotypes for at
least two of three traits resulting from a trihybrid
cross between PpYyRr and Ppyyrr.
– There are five possible genotypes that fulfill this
condition: ppyyRr, ppYyrr, Ppyyrr, PPyyrr, and
ppyyrr.
•The probability of producing a ppyyRr offspring:
–The probability of producing pp = 1/2 x 1/2 = 1/4.
–The probability of producing yy = 1/2 x 1 = 1/2.
–The probability of producing Rr = 1/2 x 1 = 1/2.
–Therefore, the probability of all three being present
(ppyyRr) in one offspring is 1/4 x 1/2 x 1/2 = 1/16.
•For ppYyrr: 1/4 x 1/2 x 1/2 = 1/16.
•For Ppyyrr: 1/2 x 1/2 x 1/2 = 2/16
•for PPyyrr: 1/4 x 1/2 x 1/2 = 1/16
•for ppyyrr: 1/4 x 1/2 x 1/2 = 1/16
•Therefore, the chance of at least two recessive traits is 6/16.
Exceptions to Mendelian Rules:
1) Incomplete dominance - heterozygotes
show a distinct intermediate phenotype.
Genotype ratio: 1RR : 2Rr : 1 rr
Phenotype ratio: 1 Red: 2 Pink: 1 white
R
r
R
RR Rr
r
Rr
rr
Exceptions to Mendelian Rules:
2) Codominance - two alleles affect the
phenotype in separate, distinguishable ways..
Genotype ratio: 1RR : 2Rr : 1 rr
Phenotype ratio: 1 Red: 2 Pink: 1 white
R
r
R
RR Rr
r
Rr
rr
How does a genotype become a
phenotype?
• Gene = DNA segment
• DNA codes for a mRNA => protein
• Protein is directly or indirectly
responsible for phenotype
Blue pigment gene makes enzyme that
makes Blue Pigment
White areas - gene for Blue pigment
enzyme is inactive = no Blue pigment
• Dominant gene is not overpowering the
recessive gene
• Cystic fibrosis - chloride transport protein is
missing in cells that have the recessive
gene
• Dominant gene is not always more
common than recessive
• Polydactyly, Huntington’s Disease, Achondroplasia
• Blood group Inheritance:
Plasma - clear liquid
portion of blood (has Antibodies)
RBC (Red Blood Cell) carries Antigens on its surface
• Blood group Inheritance:
Antigen A
Antigen B
Antigen A and B
• Blood group Inheritance:
• Blood group Inheritance:
Blood group AB = Universal
Recipient; O – Universal donor
During blood transfusion only RBCs
are transferred from donor to recepient
Agglutination
with O group
because the
NO
plasma has A
agglutination
and B antibody
with AB group
– so no other
because the
groups can
plasma has no
donate to ‘O’
antibody
• Blood group Inheritance:
Gene
Antigen
Blood
Group
IA IA or IA i A
A
Allele
Relation
IA > i
IB IB or IB i B
B
IB > i
(Dom - rec)
(Dom - rec)
IA IB
A and B
AB
ii
None
O
IA = IB
(Codominance)
Fig. 14.10
Exception to Mendelian Rules
3) Pleiotropic genes- affecting more than one
phenotypic character.
– Sickle-cell disease - one gene mutation,
many symptoms .
•Anemia
•Jaundice
•Blindness
•Infections
•Delayed growth
Exception to Mendelian Rules
4) Epistasis - one gene alters another genes
phenotypic expression
Cc or Cc = coat color
cc = no coat color
BB, Bb = Black
bb = brown
What phenotype will
ccBB bb?
Not 9:3:3:1 in a Dihybrid Cross
but 9 black: 3 brown: 4 white
Fig. 14.11
Exception to Mendelian Rules
6)) Quantitative characters - Trait depends on
HOW MANY genes are expressed
Skin color
3 allelles
AABBCC- dark
AaBBCC
NORMAL DISTRIBUTION
AABbCC
AABBCc
AaBbCc
aaBBCC
AAbbCC
AABBcc
What kind of Dominance?
aaabbcc - light
Fig. 14.12
7) Phenotype depends on environment and
genes (nature vs nurture).
Pedigree Analysis
• Map out family traits and predict genotypes
• Parents are WwFf, what are the chances of
a wwff child?
• 1/4 ww x 1/4 ff = 1/16
Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings
Predict the probability that a child with WwFf
parents will have a widow’s peak and
attached earlobes.
• The chance of having a widow’s peak is 3/4 (1/2 [Ww]
+ 1/4 [WW]).
• The chance of having attached earlobes is 1/4 [ff].
• This combination has a probability of 3/4 + 1/4 = 3/16.
• Cystic Fibrosis is a recessive disorder. If a
carrier marries a person with Cystic Fibrosis ,
what is the probability that their child will be
of Normal phenotype?  Ff X ff
 50% chance of Normal phenotype
Ff
?
Ff
Ff
? Ff
?
ff
?
• Amniocentesis- beginning at the 14th to
16th week of pregnancy (assess the
presence of a gnetic disease).
– Fetal cells extracted from amniotic fluid are
cultured and karyotyped
• Chorionic villus sampling (CVS) can
faster karyotyping, 8th - 10th week of
pregnancy.
– This technique extracts a sample of fetal tissue
from the chrionic villi of the placenta.
Fig. 14.17b
• Other techniques, ultrasound and
fetoscopy, allow fetal health to be assessed
visually in utero.
• Both fetoscopy and amniocentesis cause
complications in about 1% of cases.
– These include maternal bleeding or fetal death.
• Some genetic tests can be detected at birth
by simple tests that are now routinely
performed in hospitals.
• One test can detect the presence of a
recessively inherited disorder,
phenyketonuria (PKU).
– This disorder occurs in one in 10,000 to 15,000
births.
– Individuals with this disorder accumulate the
amino acid phenylalanine and its derivative
phenypyruvate in the blood to toxic levels.
– This leads to mental retardation.
– If the disorder is detected, a special diet low in
phenyalalanine usually promotes normal
development.