Download Genetics - Paxon Biology

Genetics
Gregor Mendel:
- Austrian monk who studied the inheritance pattern in pea plants in 1857.
- He is known as the father of genetics.
- Character: detectable feature in an organism
- Trait: variant of an inheritable character
- Ex: eye color is a character, blue and brown are traits of that
character
- Mendel chose characters in pea plants that differed clearly. He
chose 7 characters, with 2 traits each.
- 1. Flower color (purple or white)
- 2. Flower position (axial or terminal)
- 3. Seed color (yellow or green)
- 4. Pod shape (inflated or constricted)
- 5. Pod color (green or yellow)
- 6. Seed shape (round or wrinkled)
- 7. Stem length (tall or short)
-
Mendel chose plants that were true breeding.
True breeding plants only produce the same kind of plant when they are selffertilized.
He started with true breeders, and then performed crosses.
P generation: the parent generation
-
F1: the first generation (P x P)
F2: the second generation (F1x F1)
Mendel’s Laws
- Law of Segregation: two alleles for a character are packaged into separate gametes.
- An allele is an alternative form of a gene.
- Segregation of alleles is a direct result of the separation of homologous
chromosomes during meiosis
- Alternative forms of genes are responsible for variation in inherited characters.
- Ex: the gene for flower color in pea plants exists in two alternative forms;
purple and white.
- For each character, an organism inherits two alleles, one from each parent.
- If the two alleles differ, one is fully expressed (dominant) and the other is
completely hidden (recessive).
- Dominant alleles have a capital letter
- Recessive alleles have a lower case letter
-
The two alleles segregate during gamete production.
Without any knowledge of meiosis, Mendel concluded that each sperm or egg
only has one copy of each allele!
-
Homozygous: having two identical alleles for a given trait.
- Can be homozygous dominant (PP or AA)
- Can be homozygous recessive (bb or tt)
- All gametes are identical
- Homozygotes are true breeding
Heterozygous: Having two different alleles for a trait.
- Ex: Aa Bb etc.
- 50% of gametes carry the dominant allele
- 50% of gametes carry the recessive allele
-
-
-
-
-
- They are not true breeding.
Genotype: what the genes (alleles) look like.
- Ex: If Aa, then the genotype is heterozygous.
- If AA, then the genotype is homozygous dominant.
Phenotype: what the organism physically looks like.
- Ex: Blue, brown, purple etc.
- The genotype may not be apparent from the phenotype.
- If brown eyes are dominant, then having brown eyes could be the genotype
BB or Bb.
In order to determine the genotype of an organism that has the dominant
phenotype, a test cross must be done.
Test cross: where you cross the organism with the dominant phenotype with
another organism you know the genotype of (a homozygous recessive).
Mendel’s law of independent assortment:
- Used dihybrid crosses to show that not only did alleles segregate, they did so
independently of each other.
- He used true breeding plants that differed in two characters.
- He crossed plants homozygous for round and yellow seeds (RRYY) with plants
homozygous for green and wrinkled seeds (rryy).
- The F1 generations were all heterozygotes (RrYy).
- From the F1 generation, he couldn’t tell if the two characters were inherited
together or separate. He then self-pollinated the F1 plants for the F2 generation.
- Hypothesis 1: If the 2 characters segregate together, F1 hybrids can only produce
the same two classes of gametes (RY and ry) that they received from their
parents. Therefore the F2 generation would show a 3:1 phenotypic ratio.
-
-
-
Hypothesis 2: If the two characters segregate independently, F1 will produce 4
types of gametes (RY, Ry, rY and ry). Therefore the F2 generation would show a
9:3:3:1 phenotypic ratio.
After performing a dihybrid cross, he found his 2nd hypothesis to be correct.
Law of Probability:
- Gametes have a probability of the way the will fuse for fertilization.
- Both are random events.
- Examples:
- Tossing heads on a quarter ½ = .50 = 50%
- Tossing tails on a quarter ½ = .50 = 50%
- The sum of all probabilities must add up to 1.0 (or 100%).
- Rolling a 1 on a die: 1/6
- Rolling any # but 1: 5/6
- 1/6 + 5/6 = 1.0
- Random events are independent of each other:
- The outcome of a random event is unaffected by the outcome of previous such
events.
- EX: you toss a coin 2 times and get tails twice.
- What is the probability that the next coin toss will be tails?
- .50 or 50%
- A. The Rule of Multiplication:
-
-
The probability that independent events will occur simultaneously is the
product of their individual probabilities.
- Ex: You have 2 coins. What is the probability that they will both be heads?
- Chance of 1 coin having heads = ½
- Chance of the other coin having heads = ½
- Chance of both having heads when they are tossed at the same time?
- ½ x ½ = ¼ or 25%
- B. The Rule of Addition:
- The probability of an event that can occur in two or more independent ways is
the sum of the separate probabilities of the separate ways.
- Example:
- You have two pea plants: Pp x Pp. What is the probability that the
offspring will be a heterozygote?
- Probability that the dominant allele will be in the egg and the recessive
allele in the sperm: ½ x ½ = ¼
- Probability that the dominant allele will be in the sperm and the recessive
allele in the egg is: ½ x ½ = ¼
- Therefore there are 2 ways to get a heterozygote. ¼ + ¼ = ½ or 50%
- Another example:
- A couple wants to have 3 kids. They want two boys and a girl in that
order. What is the chance of this happening?
- ½ x ½ x ½ = 1/8 = .125
- What is the chance they will have two boys and one girl in any order?
- B (1/2) B (1/2) G (1/2) = 1/8
- B G B = 1/8
- G B B = 1/8
- 1/8 + 1/8 + 1/8 =. 375 or 37.5%
- Ex: A Trihybrid Cross
- What is the probability of getting an offspring with the genotype of aabbcc
when you cross two heterozygotes?
- AaBbCc x AaBbCc
- Because segregation of each allele pair is an independent event, treat this
as three separate monohybrid crosses.
- AaBbCc x AaBbCc
- Aa x Aa: probability of aa = ¼
- Bb x Bb: probability of bb = ¼
- Cc x Cc: probability of cc = ¼
- Probability of these independent events occurring simultaneously:
- ¼ aa x ¼ bb x ¼ cc = 1/64
Incomplete Dominance:
- Where one allele is not completely dominant over the other.
- The heterozygote has a phenotype intermediate between the
phenotypes of the parents.
- Ex: snapdragons
- R = red and r = white
- P gen = RR x rr
-
-
All F1 offspring are heterozygotes (Rr)
But instead of being Red, they are pink.
Both the phenotypic and genotypic ratio is 1:2:1
Codominance:
- Inheritance characterized by full expression of both alleles in the heterozygote.
- Seen in:
- Chickens
- Tay Sacs disease
- Blood Types
- Chickens
- You have a black chicken and a white chicken. You cross them and get
checkered chickens.
- B = black feathers & W = white feathers
- BB = Black chicken
- WW = white chicken
- BW = checkered chicken
- The clue for codominance taking place is that the phenotype is both characters
being expressed from both parents.
- Both alleles are evenly expressed.
- Tay Sacs
- A recessively inherited disorder; only children who are homozygous recessive
have the disease.
- Brain cells lack a crucial lipid-metabolizing enzyme. Lipids accumulate in the
brain and it leads to death (by age 5).
- At the organismal level, heterozygotes are symptom free. It “looks” like a
case of complete dominance.
-
-
-
At the biochemical level it seems to be incomplete dominance because there is
an intermediate phenotype: heterozygotes have an enzyme activity level that
is an intermediate between homozygotes.
- But at the molecular level, the normal allele & the Tay Sacs allele are actually
codominant.
- Heterozygotes produce EQUAL numbers of normal and dysfunctional
enzymes. They lack symptoms, because having ½ the normal amount of
enzymes is sufficient to break down lipids.
Multiple Alleles:
- Some genes have more than 2 alleles controlling the characteristic.
- Ex: ABO blood groups in humans
- There are 3 possible alleles, but each individual can only have 2 alleles.
Blood types:
- There are 4: A, B, AB and O
- A & B refer to 2 genetically inherited A and B antigens on the surface of red
blood cells.
- IA: codes for A
- IB: codes for B
- I: codes for no antigen = type O blood
- IA and IB are codominant to each other.
- IA and IB are both completely dominant to I.
-
Pleiotropy:
- The ability of a single gene to have multiple phenotypic effects.
- Sickle cell anemia
- Fur pigmentation & blue eyes in Siamese cats.
-
Epistasis:
- Different genes can interact to control a single phenotypic character.
- Ex: fur color in rodents
-
Polygenic Inheritance:
- Many genes have an additive effect on one phenotypic characteristic.
- Phenotype varies by degree.
- Skin color
- Height
- Weight
-
Sex-linked genes:
- Some traits are linked to the gender of the organism.
- The gene is located on a sex chromosome rather than an autosome.
- Most sex-linked genes are on the X because:
- The human X is much larger than the Y.
- X linked genes have no homologous loci on the Y.
- Most genes on the Y only encode for traits found in males.
-
-
Hemophilia
Colorblindness
Duchenne Muscular Dystrophy:
- Individuals die ~ 20 yrs.
- All males who have the trait are sterile
- Female carries ~ 5% of population
- Female homozygous are not possible
- ~ 1 in 4,000 males have disease
Testicular feminization syndrome:
- ~ 1 in 65,000 male births
- All individuals are 44 + XY (normal chromosomal males)
-
-
-
Develop as females because there is malfunction in the androgen receptor.
All are sterile, but live as normal behaving females.
Y linked:
- There have been no proven genes that are linked to phenotype in humans.
- “Hairy ear rims” has been suggested.
X-Inactivation in Females:
- “Designed” to compensate for the fact that females have a double dosage of
sex-linked genes while males have only one.
- In female mammals, most diploid cells only have one active X chromosome.
- Proposed by Mary Lyons
- In females, each of the embryonic cells inactivates one of the X chromosomes.
- The inactive X condenses into an object called a Barr Body. It lies on the
nuclear envelope.
-
-
-
Barr bodies are reactivated in a cell that undergoes Meiosis.
Females are a mosaic of two types of cells:
- Those with an active maternal X
- Those with an active paternal X.
Which X is inactivated is random.
Inactivation occurs by methylation.
After an X is inactivated, all mitotic descendants will have the same
inactivated X.
If a female is heterozygous for a sex-linked trait, about ½ the cells will
express one allele and ½ will express the other allele.
Ex: Calico Cats- all are female.
Pedigree:
- A family tree that represents the relationships among parents and offspring
across generations.
- They also show the inheritance pattern of a particular phenotypic character.
- Represents a male who shows a trait
- Represents an unaffected male
- Represents a female who shows a trait
- Represents a female who is unaffected.
- Examples of human traits:
- Widow’s Peak vs. absence
- Free earlobes vs. attached
- Tongue rolling vs. inability to tongue roll
-
-
-
Humans are very difficult to study because:
- Long generation time
- Few offspring in most cases
- Well-planned breeding experiments are impossible.
- Figure 14.16 Large families provide excellent case studies of human genetics
Recessively Inherited Disorders:
- Cystic Fibrosis:
- ~ 4% of Caucasians are carriers
- 1/2500 have the disease
- The dominant allele codes for a membrane protein that controls chloride
traffic into and out of the cell membrane.
- Chloride channels are defective or absent in individuals that are
homozygous recessive.
- Result: accumulation of mucous in the lungs and pancreas.
- Individual usually dead ~ 20 yrs.
- Sickle Cell Anemia (SCA):
- Found more in African Americans, Hispanics from central America.
- One amino acid is different from the normal hemoglobin molecule.
- Results in ‘sickle-shaped’ cells.
- Many phenotypic changes occur.
- Shows Hybrid Vigor
- People who are heterozygotes (hybrids) have an advantage over
homozygotes.
- People with SCA can not get Malaria.
- People who are hybrids have no bad effects from being hybrids, and their
chances of getting Malaria are greatly reduced.
Dominantly Inherited Disorders:
- Achondroplasia affects 1/10,000 people who are heterozygotes.
- Homozygous dominant condition results in spontaneous abortion of the
fetus.
- Homozygote recessives are normal (~99.99% of the population).
-
-
Alzheimer’s
Hypercholesterolemia (1/500 are heterozygous)
Huntington’s Disease:
- Is the degenerative disease of the nervous system
- Near the tip of chromosome # 4.
- Effects do not show up until 25- early 40’s
Aneuploidy:
- Having an abnormal number of chromosomes.
- Results from a process called nondisjunction.
- Sister chromatids do not separate properly in meiosis I or meiosis II.
-
-
Ex:
- Down’s Syndrome (trisomy 21).
- Klinefelter Syndrome
- Super-female
- Super-male
- Turner Syndrome
The way to detect aneuploidy is to perform a karyotype on the fetus.
-
Linked Genes:
- Genes located closely together on the same chromosome are said to be
“linked’ because they do not assort independently.
- Thomas Hunt Morgan performs experiments in the early 1900’s that
proved Mendel’s inheritable factors are located on chromosomes.
- Found that some genes are linked together.
- But the linkage is not a total one; some recombination does occur.
- Morgan performed a testcross.
- b = black body
- b+ = gray body
- vg = vestigial wings
- vg+ = normal wings
-
B + bvg + vg x bb vgvg
Gray, normal wings x black, vestigial wings
If the genes did assort independently, then Mendel’s ratio should be
expected.
If the genes did not assort at all, then there would be 100% linkage.
-
-
Morgan defined one map unit as 1% recombination frequency.
Map units are now called centimorgans.
Hardy-Weinberg Equilibrium
- Hardy and Weinberg developed a simple equation that can be used to
discover the genotype frequencies in a population and to track their
changes from one generation to another.
- This has become known as the "Hardy-Weinberg equilibrium equation."
- Hardy, Weinberg, and the population geneticists who followed them came
to understand that evolution will not occur in a population if seven
conditions are met:
- 7 conditions for HW:
- 1. mutation is not occurring
- 2. Natural selection is not occurring
- 3. the population is infinitely large
- 4. all members of the population breed
- 5. all mating is totally random
- 6. everyone produces the same number of offspring
- 7. there is no migration in or out of the population
- In other words, if no mechanisms that can cause evolution to occur are
acting on a population, evolution will not occur--the gene pool frequencies
will remain unchanged.
- However, since it is highly unlikely that any of these seven conditions, let
alone all of them, will occur in the real world, evolution is the inevitable
result.
- In this equation (p² + 2pq + q² = 1), p is defined as the frequency of the
dominant allele and q as the frequency of the recessive allele for a trait
controlled by a pair of alleles (A and a).
- In other words, p equals all of the alleles in individuals who are
homozygous dominant (AA) and half of the alleles in people who are
heterozygous (Aa) for this trait.
-
-
-
-
Likewise, q equals all of the alleles in individuals who are homozygous
recessive (aa) and the other half of the alleles in people who are
heterozygous (Aa).
Because there are only two alleles in this case, the frequency of one plus
the frequency of the other must equal 100%, which is to say p + q = 1
Since this is logically true, then the following must also be valid: p = 1 - q
q = 1 - p and p² + 2pq + q² = 1
In this equation, p² is the frequency of homozygous dominant (AA)
individuals in a population, 2pq is the frequency of heterozygous (Aa)
individuals, and q² is the frequency of homozygous recessive (aa)
individuals.
From observations of phenotypes, it is usually only possible to know the
frequency of homozygous recessive individuals, or q² in the equation,
since they will not have the dominant trait.
Those who express the trait in their phenotype could be either
homozygous dominant (p²) or heterozygous (2pq).
The Hardy-Weinberg equation allows us to determine which ones they
are. Since p = 1 - q and q is known, it is possible to calculate p as well.
Knowing p and q, it is a simple matter to plug these values into the HardyWeinberg equation (p² + 2pq + q² = 1).
This then provides the frequencies of all three genotypes for the selected
trait within the population.
Example Problem:
- Albinism is a rare genetically inherited trait that is only expressed in
the phenotype of homozygous recessive individuals (aa). The most
characteristic symptom is a marked deficiency in the skin and hair
pigment melanin.
- This condition can occur among any human group as well as among
other animal species. The average human frequency of albinism in
North America is only about 1/20,000.
- What is the frequency of the recessive allele and the dominant allele?
- Referring back to the Hardy-Weinberg equation (p² + 2pq + q² = 1),
the frequency of homozygous recessive individuals (aa) in a
population is q². Therefore, in North America the following must be
true for albinism: q² = 1/20,000 = .00005
- By taking the square root of both sides of this equation, we get: q =
.007
- In other words, the frequency of the recessive albinism allele (a) is
.00707 or about 1/140. Knowing one of the two variables (q) in the
Hardy-Weinberg equation, it is easy to solve for the other (p).
- p=1–q
- p = 1 - .007
- p = .993
- The frequency of the dominant, normal allele (A) is, therefore, .99293
or about 99/100
- What is the frequency of individuals in the population that are:
-
-
-
-
- 1. Homozygous dominant?
- 2. Heterozygous?
- 3. Homozygous recessive?
The next step is to plug the frequencies of p and q into the HardyWeinberg equation:
- p² + 2pq + q² = 1
- (.993)² + 2 (.993)(.007) + (.007)² = 1
- .986 + .014 + .00005 = 1
This gives us the frequencies for each of the three genotypes for this
trait in the population:
- p² = frequency of homozygous dominant individuals = .986
= 98.6% 2pq = frequency of heterozygous individuals = .014
= 1.4% q² = frequency of homozygous recessive individuals (the
albinos) = .00005 = .005%
By comparing genotype frequencies from the next generation with
those of the current generation in a population, one can also learn
whether or not evolution has occurred and in what direction and rate
for the selected trait.
However, the Hardy-Weinberg equation cannot determine which of
the various possible causes of evolution were responsible for the
changes in gene pool frequencies.