Transcript Topic_4_ - rlsmart.net

Topic 4 Genetics
4.1 Chromosomes, genes, alleles,
mutations
 Eukaryotic chromosomes are
structures made up of DNA and
proteins.
 Animation of chromosome structure
 Second Animation
Some definitions
 Gene: a heritable factor that controls a specific
characteristic. Coded by a portion of DNA.
 Allele: Alternate form of a gene. Codes for the same
characteristic and is at the same location (locus) on the
chromosome but the DNA sequence is slightly different.
 Genome: The entire set of genetic information of an
organism
 Gene Mutation: A change in the base sequence of a
gene. This can be a small as one base change in an
entire gene.
Sickle Cell Anemia
 An example of just one change in a base that causes a
significant change in the organism and a life
threatening disease.
 The triplet code on the mRNA GAG is changed to GTG.
 This means that the amino acid Valine is translated
instead of Glutamic acid
 These 2 amino acids are very different chemically and
so affect the final protein, in this case hemoglobin.
 The Hbs hemoglobin does not carry oxygen as well as
normal, Hba hemoglobin
Sickle Cell Anemia
 This causes the red blood cell to have a sickled shape.
 These RBCs do not pass through the small capillaries
well and get stuck, occluding blood flow
 This causes problems in many body
and causes severe pain
 Animation 1
 Animation 2
systems
4.2 Meiosis
 Meiosis is a reduction of the number of chromosomes
by half and division of the nucleus.
 A diploid nucleus (one with a pair of each chromosome)
is divided into a haploid nucleus (one that has only one
of each chromosomes.
 So for humans meiosis reduces a gamete to 23
chromosomes from the original 46.
Homologous chromosomes
 In a diploid cell (all cells but gametes) the pairs of
chromosomes are called homologous because they
have the same genes on each chromosome.
 Although the gene is the same the genetic information
may be different.
 Each gene has two possible types (alleles).
 Homologous chromosomes have the same size and
shape and have the same genes at the same place.
Process of meiosis
 Meiosis happens in 2 steps: meiosis I and II
 In the first step the chromosome number is halved
 In the second step the chromosomes are replicated,
this is just like mitosis
 The process is sometimes called reduction-division.
 The parent cell is 2n and the result is 4 daughter cells
with n # of chromosomes.
 How is this different than mitosis?
Meiosis I overview
Prophase and Metphase I
Prophase and Metphase I
•In Prophase homologous
chromosomes are paired
together, called tetrads
•Crossing over is the exchange of
DNA between non sister
chromatids
•Points where crossing over
occurs is called chiasmata
•In metaphase the homologous
pairs line up at the equator
•The way a pair of chromosomes
is oriented is by chance, this is
called Independent Assortment.
There is a 50/50 chance which
allele will end up in which cell.
(remember one allele is maternal,
one is paternal)
Anaphase, Telophase,
Cytokinesis
•Homologous chromosomes are
pulled apart
• Sister chromatids are still
attached
Anaphase I, Telophase I,
Cytokinesis
•In telophase each ½ of the
cell has a haploid set of
replicated chromosomes,
each chromosome I made
up of 2 sister chromatids.
•One or both chromatids
have regions of nonsister
DNA
•Cytokinesis happens
during telophase
•There is no additional
replication between MI and
MII
Prophase II, Metaphase II
Prophase II, Metaphase II
•Spindle fibers reform in
prophase
•Single chromosome line up at
the equator
•The sister chromatids are NOT
identical
•The assortment of non identical
sister chromatids further adds to
genetic variation
•At the end of meiosis there
are 4 daughter cells
•Each cell is genetically
different from the others and
from the parent cell
Meiosis II
 Animation/quiz
 Animation 2/ questions
Compare mitosis and meiosis
Non-disjunction
 Non-disjunction means that in either meiosis I or II, the
chromosomes do not split apart.
 What would happen in MI if the homologous pairs did
not split?
 What would happen if they didn’t split in MII?
 In both cases this leads to disease or more often the
fertilized egg does not even develop.
 Some cells will be missing that chromosome and some
will have an extra copy.
 Cells without the chromosome do not develop at all.
Trisomy 21
 If it has an extra copy of a chromosome it will also not
usually develop.
 In the case of chromosome #21 the fetus does develop
but has significant problems.
 Other trisomies: 13, XY
Karyotyping
 A karyotype is a display of condensed chromosomes
arranged in pairs.
 Cells are grown in a culture and then stopped when
they are in metaphase.
 They are then stained and photographed
 Chromosomes are paired based on size and shape
 A karyotype can be used to screen for abnormal
number of chromosomes or for defective chromosomes
 Sex can also be determined
Human karyotype
4.3 Theoretical Genetics
 Genotype: the alleles of an organism
 Phenotype: the characteristic of an organism
 Homozygous: 2 identical alleles
 Heterozygous: 2 different alleles
 Dominant allele: an allele that has the same effect on the phenotype
whether it is present in homozygous or heterozygous state
 Recessive allele: an allele that only has an effect when it is in
homozygous state
 Codominant allele: both alleles are expressed
Apply Definitions
 Using our Rebop creatures, give an example of each of
the definitions:
 Genotype:
 Phenotype:
 Homozygous:
 Heterozygous:
 Dominant allele:
 Recessive allele:
 Codominant allele
 Genotype: Tt
 Phenotype: curly
 Homozygous: TT or tt
 Heterozygous: Tt
 Dominant allele: T- curly
 Recessive allele: t- straight
 Codominant: QQ= red, Qq= orange, qq= yellow
Punnet Square
 A way to determine genotypes and phenotypes of
genetic crosses.
 We will follow some genetic crosses
 Pure breeding individuals are homozygous
 P generation is parent
Practice with Punnet squares
 Cross a true breeding pea plant with yellow seeds with a
true breeding plant with green seeds. Yellow is dominant.
 Cross one of the offspring from above with a green seeded
plant.
 You can deduce some rules:
1. Cross homozygous dominant with recessive and get
2. Cross 2 heterozygous and get
3. Cross a heterozygous with a recessive and get
Test Cross
 What is you have an individual with a dominant
phenotype, how do you know if it is homozygous or
heterozygous?
 Cross it with a recessive phenotype.
 Determine the genotypic and phenotypic ratios for the 2
possibilities
Practice
 problems
 Locus: position of gene on chromosome
 Test cross: a way to test if a dominant phenotype is
heterozygous or homozygous dominant. Cross the
individual with a recessive phenotype if any offspring
have recessive phenotype then the individual was
heterozygous
Some genes have more than
2 alleles
 This is called multiple alleles
 Any one individual can only have 2 alleles, but there
can be more that 2 in the population
 An example is blood type in humans:
 There are 3 alleles: A, B and O
 A and B are codominant and O is recessive to both A
and B
 So blood type in humans is an example of
codominance AD multiple alleles
 What possible genotypes could a person have if they
had type B blood? Type O blood?
 Type B: BB or Bi, O only ii
 A baby girl is born with blood type O. Her brother is
blood type AB and her mother is blood type A. What is
the father’s blood type? What are the genotypes of
everyone.
 Baby girl: ii, brother: IAIB, father: IBi, mother:Iai.
Inheritance in Corn
 We will look at phenotypes of corn seeds and analyze
the results to determine the genotypes and inheritance
patterns.
 Because we will not be counting
enough
seeds to approach exact
theoretical numbers
we need a
way of knowing if our actual
results are close enough to the
theory.
 We will use a test called the chi-square test.
Chi square test
 Test compares observed and expected outcomes.
 In biology we need to be 95% sure that any differences
we see are due to chance and not some alternate
mechanism.
 We will look at some of Mendel”s actual results:
 Flower position in F1 generation. Axial:651, terminal:
207.
 What do you think the dominant trait is, explain the
inheritance pattern, what is the theoretical ratio you
would expect? What is the real ratio?
 Axial is dominant, the pattern is based on 2
heterozygous parents, produces a theoretical ratio of
3:1 and an actual ratio of 3.14 to 1.
 Are the differences due to chance or some other
inheritance pattern?
 Use chi square to find out.
 First determine the expected numbers: total # is 858,
so we would expect 214.5 terminal and 643.5 axial.
 x2= sum of (0-E)2/E
 (207-214.5)2/214.5= .262
 (651-643.5)2/643.5= .087
 Sum of these= .347
 Go to table of p values, look for .05 across the top then
go down to the row with correct degrees of freedom (n1).
 If the value is less than your x2 then the results you see
are not significantly different.
Sex Chromosomes
 Sex chromosomes determine sex of the individual.
 In humans: XX-female, XY- male
 So all humans have an X and the X chromosome has
important genetic information
 Since these are not homologous chromosomes, this
means that traits carried on the X chromosome do not
follow the same rules.
 Genes on the sex chromosome (X in humans) are sex
linked
Dihybrid cross
 Color in mice is codominant: a CRCW genotype
produces a spotted cat. Long ears is completely
dominant over short ears.
 What is the genotypic and phenotypic ratio of a cross
between 2 heterozygous cats?
 6:3:3:2:1:1
Inheritance in sex linkage
 Because the sex chromosomes are not homologous,
inheritance patterns are different.
 An example of the human genetic disorder that is sexlinked is color blindness
 The condition is recessive- which means you need to
have 2 recessive alleles or if you are male only one
because you do not have the corresponding allele.
 So XbXb or XbY would be color blind
Problem
 A woman with normal sight marries a normal sighted
male and has a son who is color blind. What are the
genotypes of the family? What chance does this family
have of having another son who is colorblind, chance of
having a daughter who is colorblind? Chance that next
child will be colorblind?
 Mother: XbXB, father XBY, son XbY
 50% chance next son will have it
 0% chance daughter will get it (must get a b from mom
and dad
 25%chance next child will have it
Hemophilia
 Follows the same pattern: recessive, X linked.
 A girl is born with hemopilia, her mother is normal, what
are the genotypes and phenotypes of her parents?
 Girl: XhXh, Mother: XhXH, Father XhY
 Females can be heterozygous or homozygous with
respect to sex-linked genes
 Carrier: In human genetics, with reference to a
recessive condition, an individual who is heterozygous
 Female carriers are heterozygous and normal, if
condition is recessive.
More Problems
Arizona genetics problems sex-linked
 A female calico cat mated with a black cat. All the
female cats were black and all the male cats were
calico. Explain.
 Female cat: XcXc, male: XBY, all female offspring:
XcXB, all male XcY
Human Genetics
 Because human inheritance is normally based on
looking back at the families history, we have a different
way of analyzing inheritance patterns called a
pedigree.
 A pedigree is a family tree that describes the traits and
of parents and children across generations.