Transcript BDOL Interactive Chalkboard

CHAPTER 11/14
I.
Mendel
A. Gregor Mendel
1. Austrian monk-carried out impt
studies of heredity (passing on
chars. from parents to offspring)
a. traits are inherited
2. Father of Genetics
a. branch of bio that studies
heredity
3. Used garden pea due to several good
traits/easy to grow
a. cross pollination
b. true breeding plants (P1-parents)
* F1- 1st gen
* F2- 2nd gen
4. Laws of Heredity
a. Law of Segregation
b. Law of Independent Assortment
II. Crosses
A. Genes exist in alternative forms =
alleles
1. Dominant allele
a. written 1st
b. capital letter (J)
c. usually use letter of the dom
trait
* black hair = B
2. Recessive allele
a. lower case (j) letter of dom trait.
B. Phenotypes & Genotypes
1. Phenotype = way an org looks
(descriptions)
a. blue eyes
2. Genotype = allele/gene combi an org
has (letters)
a. BB
b. Heterozygous genotype
* Tt
c. Homozygous genotype
* TT (homoz dom)
* tt (homoz rec)
C. Punnett Squares
1. Diagram that shows the outcome of
genetic combis (checkerboard)
a. genes of parents written on the TOP
& LEFT side of Punnett
2. Determine geno & pheno of offspring
a. Ratios/ percentages/fractions
3. Monohybrid Crosses
a. cross between inds involving one pair
of contrasting traits (HH x hh)
* 4 boxes
4 Dihybrid Crosses
a. cross between inds involving 2 pairs of
contrasting traits (HhFf x HhFf)
* 16 boxes
* pheno ratio-9:3:3:1
* geno ratio-4:1:1:1:1:2:2:2:2
III. Probability
A. In reality you do not get the exact ratio
of results shown in the square
(chance plays a role)
1. like flipping a coin
2. probability of getting a T vs. a t
is 50:50
IV. Incomplete dominance
A. alleles neither dominant nor recessive (use
2 different letters-CAPITAL)
1. 4 boxes
B. traits are inherited in an incomplete
dominance pattern, the phenotype of
heterozygous individuals is intermediate
between those of the 2 homozygotes
1. blending
C. Example:
Homozygous Red-flowered snapdragon plant
(RR) crossed w/ homozygous White-flowered
snapdragon plant (WW):
ALL of the F1 offspring will be PINK (RW)
Incomplete
dominance
Red
White
All
pink
Pink
RW
Red
(RR)
White
WW
Pink
(RW)
All pink flowers
1/4 red: 2/4 pink: 1/4white
V. Codominance
A. Codominant alleles cause the phenotypes
of both homozygotes to be produced in
heterozygous individuals.
1. both alleles are expressed equally
a. cattle-red hair codominant with
white hair = ROAN color
2. Sickle Cell Disorder=defective hemoglobin
forms crystal-like structures that change shape
of the RBC’s
a. Normal (NN) red blood cells are discshaped, but abnormal red blood cells
are shaped like a sickle (half-moon).
b. homozygote for trait (SS) =sickle cell
c. heterozygotes (SN) have both normal &
sickle blood cells, but more normal
Sickle-cell disease
 change in shape occurs in the body’s narrow
capillaries after the hemoglobin delivers oxygen
to the cells
Normal red
blood cell
Sickle cell
VI. Multiple Alleles
A. common for more than 2 alleles to control
a trait in a population
1. traits controlled by more than 2 alleles
have Multiple Alleles
B. ABO blood group is a classic example of a
single gene that has multiple alleles in
humans.
Multiple Alleles Govern
Blood Type
Human Blood Types
Genotypes Surface Molecules Phenotypes
A
A
lA lA or lAi
B
B
lB lB or lBi
lA lB
A and B
AB
None
ii
O
1. Determining blood type is necessary before a
person can receive a blood transfusion
because the RBC’s of incompatible blood types
could clump together, causing death (immune
system will attack).
2. Gene for blood type, gene l, codes for a
molecule that attaches to a membrane protein
found on the surface of RBC’s.
a. lA & lB alleles each code for a different
molecule
b. Phenotype A
* lA allele is dominant to
i, so inheriting either
the lAi alleles or lA lA
alleles from both
parents will give you
type A blood.
Surface molecule A
c. Phenotype B
 lB allele is also
dominant to i
 B blood: must inherit the
lB allele from one parent
& either another lB allele
or the i allele from the
other
Surface molecule B
d. Phenotype AB
Surface molecule B
 lA and lB alleles are
codominant
 Universal receiver
Surface molecule A
e. Phenotype O
 Homozygous ii
 Universal donor
VII. Sex Determination
A. Humans = 23 pairs
1. 22 pairs of homolog. chrms
(autosomes)
2. 23rd pair differs in males & females
(sex chrms)
a. XX = girl
b. XY = boy
3. X chroms larger
a. Barr body
XY
Male
Sex
determination
X
XX
Female
X
Y
XX
Female
XY
Male
XX
Female
XY
Male
X
B. Sex-linked inheritance
1. Traits controlled by genes located on
sex chromosomes = sex-linked traits
a. alleles for sex-linked traits are
written as superscripts of the X or
Y chromosomes
b. many found on the “X”
* males just one “X” thus all “X-linked”
alleles are expressed even if
recessive (from ma)
Sex-linked inheritance
White-eyed
male (XrY)
F2
Females:
all red eyed
Redeyed
female
(XRXR)
Males:
1/2 red eyed
1/2 white eyed
F1 All red eyed
c. Females, who are XX, pass one of their X
chromosomes to each child.
Male
Female
Female
Sperm
Eggs
Eggs
Female Female
Male
Male
Female
Male
Sperm
Male
Female
Male
d. sex-linked traits in humans
 2 traits that are governed by X-linked
recessive inheritance in humans = red-green
color blindness & hemophilia
Red-Green Color Blindness
People who have redgreen color blindness
can’t differentiate these
two colors. Color
blindness is caused by
the inheritance of a
recessive allele at either
of two gene sites on the
X chromosome.
b. Determining gamete letters for parents
* HhJj (heterozygous for both traits)
HJ
Hj
hJ
hj
XI. Karyotype
A. Metaphase chromosomes are
photographed; the chromosome pictures
are enlarged & arranged in pairs by a
computer according to length & location of
centromere.
1.
Chart of chromosome pairs =karyotype (it’s
valuable in identifying unusual chromosome
numbers in cells)
a. longest pair #1
b. pairs match (banding, length, traits)
c. 23rd pair = X, Y(smaller chroms)
*1-22 pairs = autosomes
d. Normal=23 pairs or 46 chrms
Sex determination
 If you are female,
your 23rd pair of
chromosomes are
homologous, XX.
X
X
Female
X
Y
Male
 If you are male, your
23rd pair of
chromosomes XY,
look different.
B. Nondisjunction
1. failure of homologous chroms to
separate properly during meiosisAnaphase
a. monosomy 2N - 1
b. trisomy 2N + 1
C. Sex Chromo. Anomalies
1. Klinefelter’ Syndrome-XXY (47)
2. Turner’s Syndrome-X?(45)
3. XYY(47)
4. Downs-three 21(47)
Trisomy (3’s)
Down syndrome
is caused by
autosomal
trisomy (#21)
X. Pedigree
A. Family tree that traces family thru successive
generations
B. Pedigree is a graphic representation of
genetic inheritance.
1. made up of a set of symbols that
identify males/females,
individuals affected by the trait
being studied, & family
relationships
Male
Parents
Female
Siblings
Affected
male
Affected
female
Mating
Known
heterozygotes
for recessive
allele
Death
Pedigrees
illustrate
inheritance
Female
Male
I
1
2
II
2
1
3
4
5
III
?
1
2
4
3
IV
1
2
3
4
5
2. Circle
represents a
female; a
square
represents a
male
I
1
2
II
3
2
1
4
5
III
?
1
2
4
3
IV
1
2
3
4
5
3. Highlighted
circles/squares
represent
individuals
showing the trait
being studied.
I
1
2
II
2
1
3
4
5
III
?
1
2
4
3
IV
1
2
3
4
5
4. Circles & squares
that are not
highlighted
designate
individuals that do
not show the trait.
5. A half-shaded
circle or square
represents a
carrier, a
heterozygous
individual.
I
1
2
II
2
1
III
?
IV
1
2
1
3
4
4
3
2
5
3
4
5
6. Horizontal line
connecting a circle &
a square indicates
that the individuals
are
parents(marriage), &
vertical line connects
parents with their
offspring.
I
1
2
II
1
III
1
?
IV
2
1
3
2
4
4
3
2
5
3
4
5
7. Ea. horizontal
row of circles &
squares in a
pedigree
designates a
generation, with
the most recent
generation shown
at the bottom.
I
1
2
II
1
3
2
4
5
III
?
1
2
4
3
8. The generations
are identified in
sequence by
Roman
numerals, & ea.
individual is
given an Arabic
number.
a. oldest kid: left
IV
1
2
3
4
5
XI. Disorders
A. Cystic fibrosis
1. Due to a defective protein in the
plasma membrane, cystic
fibrosis results in the formation
/accumulation of thick mucus in
lungs/digestive tract.
B. Tay-Sachs disease
1. Tay-Sachs disease is a recessive disorder
of the CNS.
a. recessive allele results in the absence
of an enzyme that normally breaks
down a lipid produced & stored in
issues of the CNS
b. lipid fails to break down properly, it
accumulates in the cells
C. Phenylketonuria
1. Phenylketonuria (PKU), is a recessive
disorder that results from the absence of an
enzyme that converts 1 aa, phenylalanine, to
a different aa-tyrosine.
a. phenylalanine cannot be broken down,
it & its by-products accumulate in the
body & result in severe damage to the
CNS
b. Infants affected by PKU are given a diet that is
low in phenylalanine until their brains are fully
developed.
Phenylketonuria
Phenylketonurics: Contains Phenylalanine
D. Huntington’s disease
1.Huntington’s disease is a lethal genetic
disorder caused by a rare dominant allele.
a. It results in a breakdown of certain
areas of the brain.
The end