Download AP Inheritance

Inheritance
&
Mendelian Genetics
Gregor Mendel
 Modern genetics began in the
mid-1800s in an abbey garden,
where a monk named Gregor
Mendel documented inheritance
in peas


used experimental method
used quantitative analysis
 collected data & counted them
 Most traits in most species do
not follow the simple Mendelian
pattern, but it was a starting
point
Mendel’s work
 Bred pea plants

Pollen transferred from white
flower to stigma of purple flower
P
cross-pollinate
true breeding parents (P)
 P = parental

raised seed & then
observed traits (F1)
 F = filial

allowed offspring
to self-pollinate
& observed next
generation (F2)
anthers
removed
all purple flowers result
F1
self-pollinate
F2
What did Mendel’s findings mean?
 Traits come in alternative versions


purple vs. white flower color
alleles
 different alleles vary in the sequence of nucleotides
at the specific locus (locus = location on a
chromosome) of a gene
 some difference in sequence of A, T, C, G
purple-flower allele &
white-flower allele are two DNA
variations at flower-color locus
different versions of gene at
same location on homologous
chromosomes
Traits are inherited as discrete units
 For each characteristic, an organism
inherits 2 alleles, 1 from each parent

diploid organism
 inherits 2 sets of chromosomes,
1 from each parent
 homologous chromosomes - same genetic
loci (i.e. same genes), different alleles at
those loci
What did Mendel’s findings mean?
 Some traits “mask” others

purple & white flower colors are
separate traits that do not blend
 purple x white ≠ light purple
 purple masked white

dominant allele
 functional protein
 masks other alleles

wild type
allele producing
functional protein
mutant
allele producing
malfunctioning
protein
recessive allele
 allele typically makes a
malfunctioning protein
homologous
chromosomes
Genotype vs. phenotype
 Difference between how an organism “looks”
& its genetics

phenotype
 description of an organism’s trait
 the “physical,” the result of gene expression

genotype
 description of an organism’s genetic makeup
 Its combo of alleles, like “Pp”
X
P
purple
white
F1
all purple
Dominant ≠ most common allele
 Because an allele is dominant
does not mean…

it is more common, healthier, stronger,
better, more likely, etc.
Polydactyly
dominant
allele, yet rare!
Making crosses
 Can represent alleles as letters



flower color alleles  P or p
true-breeding purple-flower peas  genotype
PP
true-breeding white-flower peas  genotype
pp
 In research, alleles are usually
letter/number/symbol combinations (like
ser83psE)
X
x
P
PP
purple
pp
white
Pp
F1
all purple
Discussion
 Which of these are phenotypes and which








are genotypes?
1. Curly hair
2. Jj
3. PE1PE2
4. Arthritic knees
5. Type B blood
6. Spotted fur and a pink nose
7. HHGg
8. Purple leaves and spiny stem
Punnett Square reminders
 The side and top boxes = parents’
potential gametes, each equally likely.
 Inner boxes = potential zygotes.
 Punnett Squares predict the odds of
each offspring being born with a given
genotype/phenotype.

Does not ensure that, say, 50% of the
children will definitely be freckled.
Genotypes
 Homozygous = same alleles = PP, pp

“True-breeding” = homozygous
 Heterozygous = different alleles = Pp

“Carrier”
homozygous
dominant
heterozygous
homozygous
recessive
Test cross
 Method to determine genotype in case of
dominant phenotype
 Breed the dominant phenotype with a
homozygous recessive (pp) to determine the
identity of the unknown allele
How does
that work?
x
is it
PP or Pp?
pp
Discussion
 Suppose that the Y allele codes for orange fins and the y
allele codes for yellow fins.






The heterozygous genotype: __
The homozygous dominant genotype: __
The homozygous recessive genotype: __
A fish with yellow fins must have a _____________ genotype.
A fish with orange fins could be either _____________ or
___________________.
If a fish has orange fins, test-crossing it with a ______-finned
fish will produce either 100% _____ or 50% orange/50% yellow.
If the former result, the orange fish was _________. If the
latter result, the orange fish was _________.
Mendel’s 1st law of heredity
 Law of segregation

P
PP
during meiosis, alleles segregate
P
 homologous chromosomes separate

each allele for a trait is packaged into
a separate gamete
 You only give 1 allele per gene to your
p
pp
child
p
P
Pp
p
Law of Segregation
 Suppose there’s an eye color locus, with
bb
bb
B
BB
Normal cell
in G1
Meiosis II
b
b
Meiosis I
S Phase
the alleles B for brown eyes or b for blue
eyes.
 A man has the genotype Bb, which gives
him the phenotype brown eyes.
 Meiosis produces his gametes…
b
B
BB
B
Four Gametes
He can make
gametes that are
EITHER B or b.
Half of his gametes
will be one, half will
be the other.
That’s segregation!
Discussion
Monohybrid cross practice! Show Punnett Squares
to support your answer.
 If two black bees (bees A and B) have 676 babies,
all black; two red bees (bees C and D) have 983
babies, all red; and a different two black bees
(bees E and F) have 524 babies, 220 red and 304
black, what was each bee’s genotype? Use any
letter for the alleles that you want.
 What generation were bees A,B,C,D,E, and F a
part of? What generation were their children a
part of?
Dihybrid cross
 Other of Mendel’s
experiments followed
the inheritance of 2
different characters
seed color and
seed shape
 dihybrid crosses

Mendel
was working out
many of the
genetic rules!
Dihybrid cross
P
true-breeding
yellow, round peas
Y = yellow
R = round
true-breeding
green, wrinkled peas
x
YYRR
yyrr
y = green
r = wrinkled
yellow, round peas
F1
100%
generation
(hybrids)
YyRr
self-pollinate
F2
generation
9:3:3:1
9/16
yellow
round
peas
3/16
green
round
peas
3/16
yellow
wrinkled
peas
1/16
green
wrinkled
peas
YyRr
Dihybrid cross
YR
YyRr x YyRr
YR
Yr
yR
yr
yr
YR YYRR YYRr YyRR YyRr
Yr
YyRr
Yyrr
yR YyRR YyRr yyRR
yyRr
yr
YYRr
YyRr
YYrr
Yyrr
yyRr
yyrr
or
YyRr
YR Yr
yR
yr
9/16
yellow
round
3/16
green
round
3/16
yellow
wrinkled
1/16
green
wrinkled
Mendel’s 2nd law of heredity
 Law of independent assortment

yellow
different loci (genes) separate into gametes
independently
 non-homologous chromosomes align independently
green
 classes of gametes produced in equal amounts
 YR = Yr = yR = yr
round
wrinkled
YyRr
Yr
Yr
1
yR
:
yR
1
YR
:
YR
1
yr
:
yr
1
Discussion
 Complete a Punnett Square for this
dihybrid cross problem!
 If A = tall and a = short, while B = fuzzy
and b = smooth…
 What are the odds that a parent
heterozygous for both traits and a short
smooth parent will have a tall and fuzzy
offspring?
Law of Independent Assortment
EXCEPTION
 If genes are on same
chromosome & close together
 will usually be inherited
together
 rarely crossover separately
 “linked”
Rules of Probability
 Probability scale ranges from 0 to 1
 Rule of Multiplication: determine the

chance that two or more independent
events will occur together in some
specific combination.
 Compute the probability of each
independent event.
 Then, multiply the individual
probabilities to obtain the overall
probability of these events occurring
together.
Rule of Addition: probability of an
event that can occur two or more
different ways is the sum of the
separate probabilities of those ways.
Rules of Probability
 For instance, if I roll a six-sided die,
what are the odds I’ll get a number that
is equal to or less than 2? Which law
did you use?
 If I roll two dice, what are the odds I’ll
get a 1 both times? Which law did you
use?
Discussion
 You have been using both rules all
along!
 How does the rule of multiplication

come into play in a monohybrid cross?
The rule of addition?
Rules of Probability
 What are the odds that a homozygous
red-haired, heterozygous green-eyed,
white-chinned cat (AAEeww) and a
dark-haired, heterozygous green-eyed,
white-chinned cat (aaEeww) would have
a kitten with the genotype AaEeww?

We can solve each gene as a separate
monohybrid problem, then multiply!
Discussion
 AAEeww x aaEeww = ?% AaEeww
Discussion
 Determine the probability of finding two recessive
phenotypes for at least two of three traits resulting
from a trihybrid cross between pea plants that are
PpYyRr and Ppyyrr.
 There are five possible genotypes that fulfill this
condition: ppyyRr, ppYyrr, Ppyyrr, PPyyrr, and
ppyyrr.

Hint: Use the rule of multiplication to calculate the
probability for each of these genotypes, and then use
the rule of addition to pool the five probabilities.
Answer:
 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 = 3/8.
Beyond Mendel’s Laws
of Inheritance
Mendel chose peas luckily
 Relatively simple genetically

most characters are controlled by a single
gene with each gene having only 2 alleles
 one completely dominant over
the other
 All the genes he chose happened to be on
different chromosomes - whew!
Extending Mendelian genetics
 The inheritance of traits can rarely be
explained by simple Mendelian genetics
Various patterns of inheritance: incomplete
dominance, codominance, pleiotropy,
lethality, epistasis, polygenetic traits,
multiallelic genes, sex-linked traits…
 Not all traits just determined by nuclear
DNA: environmental effects, gene
regulation, mitochondrial DNA…

Incomplete dominance
 Heterozygote shows a NOVEL,
intermediate, blended phenotype

example:
 RR = red flowers RR
 rr = white flowers WW
 Rr = pink flowers RW
 make 50% less color
RR
RW
WW
Incomplete dominance
P
X
true-breeding
red flowers
true-breeding
white flowers
100% pink flowers
F1
100%
generation
(hybrids)
self-pollinate
25%
red
F2
generation
50%
pink
25%
white
1:2:1
Co-dominance
 2 alleles affect the phenotype equally
& separately



Phenotype is not blended, it’s both of
the true-breeding phenotypes
simultaneously
Speckled chickens, Roan cows,
human ABO blood groups
3 alleles
 IA, IB, i
 IA & IB alleles are co-dominant
 glycoprotein antigens on RBC
 IAIB = both antigens are produced
 i allele recessive to both
Genetics of Blood type
phenogenotype
type
A
B
AB
O
antigen
on RBC
antibodies
in blood
donation
status
IA IA or IA i
type A antigens
on surface
of RBC
anti-B antibodies
__
IB IB or IB i
type B antigens
on surface
of RBC
anti-A antibodies
__
IA IB
both type A &
type B antigens
on surface
of RBC
no antibodies
universal
recipient
ii
no antigens
on surface
of RBC
anti-A & anti-B
antibodies
universal
donor
Pleiotropy
 Most genes are pleiotropic

one gene affects more than one trait
 dwarfism (achondroplasia)
Lethal pleiotropy
Aa
x aa
Aa
x Aa
dominant
inheritance
A
a
a
a
Aa
Aa
achondroplastic
achondroplastic
aa
aa
typical
typical
50% affected:50% typical or 1:1
A
A
a
AA
Aa
lethal
a
Aa
achondroplastic
achondroplastic
aa
typical
67% affected:33% typical or 2:1
Discussion
 What if an allele is lethal recessive?
 Suppose that in a plant, the recessive allele
for yellow seeds is lethal, the dominant allele
for green seeds is not.
 What phenotypic ratios would you get from a
cross of…



GG x Gg?
Gg x Gg?
Gg x gg? (gg produced by genetically
engineering gametes while leaving the somatic
cells intact)
Epistasis
 One gene completely masks another gene

coat color in mice = 2 separate genes
 C,c:
B_C_
bbC_
_ _cc
pigment (C) or
no pigment (c)
 B,b:
more pigment (black=B)
or less (brown=b)
 cc = albino,
no matter B allele
 9:3:3:1 becomes 9:3:4
How would you know that
difference wasn’t random chance?
Chi-square test!
Epistasis in Labrador retrievers
 2 genes: (E,e) & (B,b)


pigment (E) or no pigment (e)
pigment concentration: black (B) to brown (b)
eebb
eeB–
E–bb
E–B–
Polygenic inheritance
 Some traits determined by additive
effects of 2 or more genes
phenotypes on a continuum
 human traits

 skin color
 height
 weight
 intelligence
 behaviors
Skin color: Albinism
 However, albinism can be
inherited as a single gene trait

tyrosine
aa = albino
enzyme
melanin
albinism
Sex linked traits
 Genes are on sex chromosomes



as opposed to autosomal chromosomes
first discovered by T.H. Morgan’s “Fly Lab” at
Columbia U.
Drosophila breeding
 good genetic subject
 prolific
 2 week generations
 4 pairs of chromosomes
 XX=female, XY=male
Classes of chromosomes
autosomal
chromosomes
sex
chromosomes
Discovery of sex linkage
P
F1
true-breeding
red-eye female
X
true-breeding
white-eye male
100%
red eye offspring
Huh!
Sex matters?!
generation
(hybrids)
F2
generation
100%
red-eye female
50% red-eye male
50% white eye male
Let’s reconsider Morgan’s flies…
P
x
XR XR
Xr
XR
F1
XR
XR Xr
XR Xr
x
F1
XrY
XR Xr
Y
XRY
XRY
100% red eyes
XR
BINGO!
Xr
XRY
XR
Y
XR XR
XRY
XR Xr
X rY
F2
100% red females
50% red males; 50% white males
Genes on sex chromosomes
 Y chromosome

few genes other than SRY
 sex-determining region
 master regulator for maleness
 turns on genes for production of male hormones
 many effects = pleiotropy!
 X chromosome

other genes/traits beyond sex
determination
 mutations:
 hemophilia
 Duchenne muscular dystrophy
 color-blindness
Human X chromosome
 Sex-linked
Duchenne muscular dystrophy
Becker muscular dystrophy
usually
means
“X-linked”
 more than
60 diseases
traced to
genes on X
chromosome

Chronic granulomatous disease
Retinitis pigmentosa-3
Norrie disease
Retinitis pigmentosa-2
Ichthyosis, X-linked
Placental steroid sulfatase deficiency
Kallmann syndrome
Chondrodysplasia punctata,
X-linked recessive
Hypophosphatemia
Aicardi syndrome
Hypomagnesemia, X-linked
Ocular albinism
Retinoschisis
Adrenal hypoplasia
Glycerol kinase deficiency
Ornithine transcarbamylase
deficiency
Incontinentia pigmenti
Wiskott-Aldrich syndrome
Menkes syndrome
Androgen insensitivity
Sideroblastic anemia
Aarskog-Scott syndrome
PGK deficiency hemolytic anemia
Anhidrotic ectodermal dysplasia
Agammaglobulinemia
Kennedy disease
Pelizaeus-Merzbacher disease
Alport syndrome
Fabry disease
Immunodeficiency, X-linked,
with hyper IgM
Lymphoproliferative syndrome
Albinism-deafness syndrome
Fragile-X syndrome
Charcot-Marie-Tooth neuropathy
Choroideremia
Cleft palate, X-linked
Spastic paraplegia, X-linked,
uncomplicated
Deafness with stapes fixation
PRPS-related gout
Lowe syndrome
Lesch-Nyhan syndrome
HPRT-related gout
Hunter syndrome
Hemophilia B
Hemophilia A
G6PD deficiency: favism
Drug-sensitive anemia
Chronic hemolytic anemia
Manic-depressive illness, X-linked
Colorblindness, (several forms)
Dyskeratosis congenita
TKCR syndrome
Adrenoleukodystrophy
Adrenomyeloneuropathy
Emery-Dreifuss muscular dystrophy
Diabetes insipidus, renal
Myotubular myopathy, X-linked
Discussion
 Hemophilia is X-linked recessive. If a
carrier and her healthy (unaffected)
husband have a child, what are the
odds that their child will be:
Healthy?
 Hemophiliac?
 A carrier?

X-inactivation
 Female mammals inherit 2 X chromosomes

one X becomes inactivated during
embryonic development
 condenses into compact object = Barr body
 which X becomes Barr body is random
 patchwork trait = “mosaic”
patches of black
XH 
XH Xh
tricolor cats
can only be
female
Xh
patches of orange
Male pattern baldness
 Sex influenced trait

autosomal trait influenced by sex hormones
 age effect as well = onset after 30 years old

dominant in males & recessive in females
 B_ = bald in males; bb = bald in females
Environmental effects
 Phenotype is controlled by
both environment & genes
Human skin color is influenced by
both genetics & environmental
conditions
Color of Hydrangea flowers
is influenced by soil pH
Coat color in arctic
fox influenced by
heat sensitive alleles
Pedigrees
1
3
4
2
5
6
Pedigree analysis
 Pedigree analysis reveals Mendelian
patterns in human inheritance

= male
data mapped on a family tree
= female
= male w/ trait
= female w/ trait
Studying Human Genetics
 Circle – female
 Square – male
 Shaded – afflicted with
trait
 Half shaded or Dot –
carrier
 Horizontal line –


mating “marriage line”
“sibling line”
 Vertical line – children
 Dotted vertical line -
adopted children
 (Diagonal lines – twins)
Discussion
 Draw a pedigree of your immediate
family (if adopted, draw your choice of
relatives)
Pedigrees
 Pedigree analysis can reveal the
inheritance pattern of the trait under
consideration…
Autosomal Dominant
 Autosomal dominant – allele is dominant
and on an autosomal chromosome
 Every person with the trait, also had a
parent with it.

Not necessarily a child with it, though!
Why?
Autosomal Recessive
 Autosomal Recessive - allele is recessive
and on an autosomal chromosome
 Trait only appears when two alleles are
present, so there can be carriers.

Trait often “skips” several generations or
shows up seemingly out of nowhere. Why?
X-linked Recessive
 X-linked Recessive – allele is recessive and is
located on the X chromosome
 Males are more likely to show trait – Why?
 Skips generations
X-linked Dominant
 X-linked Dominant - allele is dominant and is
located on the X chromosome
 An afflicted father’s daughters will all be afflicted
too – Why?
 No male to male transmission
 No skipped generations
Y-Linked
 Y-Linked (recessive vs dominant
doesn’t matter): Locus is on the Y
chromosome
 Only males have it, and all sons of an
affected male are also affected – Why?
What’s the
likely inheritance
pattern? Label
genotypes
using A/a
Discussion
11
33
44
22
55
66
Genetic counseling
 Pedigree can help us understand the past

& predict the future
Thousands of genetic disorders are
inherited as simple recessive traits

from benign conditions to deadly diseases
 albinism
 cystic fibrosis
 Tay sachs
 sickle cell anemia
 PKU
Genetic testing
sequence
individual genes
Tay-Sachs (recessive)
 Great example of how pedigrees and genetic

counseling have made a difference!
Primarily Jews of eastern European (Ashkenazi)
descent & Cajuns (Louisiana)
 strikes 1 in 3600 births
 100 times greater than incidence among
non-Jews

non-functional enzyme fails to breakdown lipids
in brain cells
 fats collect in cells destroying their
function
 symptoms begin few months
after birth
 seizures, blindness &
degeneration of muscle &
mental performance
 child usually dies before age 5
Tay-Sachs
 Israel became the 1st country to offer
free genetic testing to couples, in
large part to eliminate TSD

Haredi communities in the US often
required couples to be tested before
marriage
 Incidence of TSD declined by 90%!


Before 1970, 50-70 Ashkenazi infants
born with TSD per year in US
By 2000s, only 1 or 2 per decade
Non-Nuclear Inheritance
 Not all eukaryotic genes are
in the nucleus!

Found in mitochondria, plastids
 In animals, all cytoplasmic (non-nuclear) genes
come from which parent, maternal or paternal?
 Randomly assorted to gametes and daughter cells

Therefore, traits determined by plastid DNA and mtDNA do
NOT display Mendelian inheritance
Non-Nuclear Inheritance
 In humans, mitochondrial DNA (mtDNA) encodes
mostly mitochondrial proteins (such as ETC
proteins, mt-ribosomes)

Mutations cause mitochondrial disorders, including lactic
acidosis, some myopathies (muscle disorders)