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NOTES.2 – Basic Genetic Principles
Unit #9: Mendelian Genetics
Chapters 11 and 14 in textbook
BASIC VOCABULARY Part One:
What is Genetics? • Gene$csisthestudyofheredity.• Heredityisthepassingoftraitsfromparentstooffspring.• Atraitisaphysicalorphysiologicalcharacteris7c.
• Forexample,heightandcolorblindness• Sometraitsareinherited,whileothersarenot.
• Forexample,theshapeofearsvs.piercedears
What makes a trait inheritable? • Thephenotypeisthephysicalappearanceofatrait.• Thephenotypeforanyinheritedtraitisdeterminedbytheproteinsthatyourcells
produce.• Yourcellsknowwhichproteinstomakebasedonwhatyourgenestellthem.• GenesaresequenceofbasesintheDNAfoundwithinyourchromosomes.• Youinherit50%ofyourchromosomesfromeachparent.• Yourownuniquecombina7onofgenesarereferredtoasyourgenotype.• Itisyourgenotypethatdeterminesyourphenotypeofaninheritedtrait.
Protein
Forexample,eyelashlengthisatraitthatisdecidedbyagenefoundonchromosome#1.Ifyouinherited…
Long Long
Chromosome #1 from mom
Chromosome #1 from dad
Your Phenotype: long eyelashes
Short Short Your Phenotype: short eyelashes
Chromosome #1 from mom
Chromosome #1 from dad
There are two versions of the gene: long eyelashes and short eyelashes. A version of a gene is called an allele. If you inherit two identical alleles for a trait, then you have a homozygous genotype and are considered to be purebred or true-breeding.
Ifyouinherittwodifferentallelesforatrait,…
Long
Chromosome #1 from mom
Chromosome #1 from dad
Short Your Phenotype: Long eyelashes
…then you have a heterozygous genotype, and are considered a hybrid or a carrier. Since long eyelashes are phenotypically expressed when only one allele is present, it is considered dominant. The dominant allele is represented by a capital letter (A). Short eyelashes are considered recessive, because there must be two alleles present in order for the recessive phenotype to be expressed. The recessive allele is represented by a lower-case letter (a).
PRACTICE TERMINOLOGY 1. Circle the dominant allele.
A aa Aa AA a 2. Circle the heterozygous genotype.
A aa Aa AA a 3. Circle the homozygous dominant genotype.
A aa Aa AA a 4. Circle the recessive genotype.
A aa Aa AA a 5. Circle the recessive allele.
A aa Aa AA a 6. Circle the phenotype.
AA Aa long eyelashes 7. Circle the trait.
purple flowers round peas plant height
Howdowerepresenttheseallelesonpaper?We use CAPITAL LETTERS to represent the dominant allele. (A = Long) We use lowercase letters to represent the recessive allele. (a = short)
DESCRIPTION: GENOTYPE? PHENOTYPE? Homozygous dominant ____________ _____________ Heterozygous ____________ _____________ Pure recessive ____________ _____________ Hybrid ____________ _____________ True-breeding dominant ____________ _____________ Carrier ____________ _____________
LAWS OF INHERITANCE Part Two:
Gregor Mendel • Austrian monk • Studied the inheritance of traits in pea plants
• His work was not recognized until the turn of the 20th C
• Known today as the “Father of Genetics” • Developed the three basic laws of inheritance
#1 - Law of Dominance • In a cross between parents that are pure for contrasting
phenotypes, only one form of the trait will appear in the next generation.
• All of the offspring will be heterozygous and express only the dominant trait.
• For example in seed color: P: Pure Yellow (YY) x Pure Green (yy) F1: Hybrid Yellow (Yy)
y y
Y
Y
Yy Yy
Yy Yy
#2 - Law of Segregation • During meiosis, the two alleles responsible for a single trait
separate from each other. • Alleles for a trait are then “recombined” at fertilization,
producing the genotype for the traits of the offspring.
#3 - Law of Independent Assortment • Alleles for different traits are distributed to gametes (and
offspring) independently of one another*. *Unless the loci are near each other on the same chromosome - some genes are “linked.”
PUNNETT SQUARES Part Three:
Punnett Squares • A tool used to predict the genotype and
phenotype combinations in a genetic cross. • It does not show actual results. • Each offspring produced is a separate event.
Monohybrid Cross A breeding experiment that tracks the inheritance of ONE trait
• Illustrates Mendel’s Law of Segregation – Each allele for a given trait separates independently during gamete formation (meiosis).
EXAMPLE PROBLEM Purple flower color is dominant over white flower color in pea plants. If a gardener cross-pollinates a homozygous purple flowering plant with a heterozygous purple flowering plant, then what is the genotypic and phenotypic ratios of the F1 offspring?
STEP 1: Identify the trait. Flower Color STEP 2: Create a Key. A – purple a – white STEP 3: Write the genotypes of the Parents. P: _A A_ x _A a_ STEP 4: Fill in the Punnett Square. STEP 5: Determine the genotypic and phenotypic ratios of the offspring.
A A
A
a
A A A A
A a A a
Genotypic Ratio: 2 A A : 2 A a : 0 a a Phenotypic Ratio: 100% purple flowers
PRACTICE PROBLEM Tall pea plants are dominant over short pea plants. If a gardener cross-pollinated two hybrid pea plants, then what is the genotypic and phenotypic ratios of the F1 offspring?
STEP 1: Identify the trait. STEP 2: Create a Key. A – __________________________________ a – __________________________________ STEP 3: Write the genotypes of the Parents. P: _______ x ________ STEP 4: Fill in the Punnett Square. STEP 5: Determine the genotypic and phenotypic ratios of the offspring.
Genotypic Ratio: Phenotypic Ratio:
PRACTICE ANSWER Tall pea plants are dominant over short pea plants. If a gardener cross-pollinated two hybrid pea plants, then what is the genotypic and phenotypic ratios of the F1 offspring?
STEP 1: Identify the trait. Plant Height STEP 2: Create a Key. A – tall a – short STEP 3: Write the genotypes of the Parents. P: Aa x Aa STEP 4: Fill in the Punnett Square. STEP 5: Determine the genotypic and phenotypic ratios of the offspring.
Genotypic Ratio: 1 AA : 2 Aa : 1 aa Phenotypic Ratio: 3 tall : 1 short
A a
A
a
A A A a
A a a a
Dihybrid Cross A breeding experiment that tracks the inheritance of TWO traits
• Illustrates Mendel’s Law of Independent Assortment
– Each pair of alleles segregates independently during gamete formation (meiosis).
Determining Gametes for a Dihybrid Cross
• Trait #1: Seed shape • Alleles:
R - round seed shape r - wrinkled seed shape
• Trait #2: Seed color • Alleles:
Y - yellow seed color y - green seed color
RrYy x RrYy
RY Ry rY ry RY Ry rY ry All possible gamete combinations from each parent.
P: hybrid (RrYy) x hybrid (RrYy)
Setting up & Filling In a Dihybrid
RRYY
RRYy
RrYY
RrYy
RRYy
RRyy
RrYy
Rryy
RrYY
RrYy
rrYY
rrYy
RrYy
Rryy
rrYy
rryy
Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1
RY Ry rY ry
RY
Ry
rY
ry 9:3:3:1 phenotypic ratio
PRACTICE PROBLEM In pea plants, tall plants are dominant over short plants and yellow pea color is dominant over green pea color. If a gardener cross-pollinates a hybrid pea plant with a pea plant that is short with green peas, then what is the phenotypic ratio of the F1 offspring?
STEP #1: Identify Traits. STEP #2: Make a Key. STEP #3: Parent�s Genotypes. P: ___________ x ___________ STEP #4: Fill in Punnett Square STEP#5: Find the Ratio. ______ Tall, Yellow ______ Tall, Green ______ Short, Yellow ______ Short, Green
PRACTICE ANSWER In pea plants, tall plants are dominant over short plants and yellow pea color is dominant over green pea color. If a gardener cross-pollinates a hybrid pea plant with a pea plant that is short with green peas, then what is the phenotypic ratio of the F1 offspring?
STEP #1: Identify Traits. Height & Pea Color STEP #2: Make a Key. A – tall a – short B – yellow b – green STEP #3: Parent�s Genotypes. P: AaBb x aabb STEP #4: Fill in Punnett Square STEP#5: Find the Ratio. ¼ tall, yellow ¼ tall, green ¼ short, yellow ¼ short, green
AB Ab aB ab
ab
ab
ab
ab
AaBb Aabb aaBb aabb
AaBb
AaBb
AaBb
Aabb
Aabb
Aabb
aaBb
aaBb
aaBb
aabb
aabb
aabb
Challenge Question: How many different gametes will be produced for the
following genotypes? Use the Formula:
2n = # of possible gametes, n = # of heterozygotes
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQq
Challenge ANSWERS: How many different gametes will be produced for the
following genotypes? Use the Formula:
2n = # of possible gametes, n = # of heterozygotes
1. RrYy = 22 = 4 gametes
RY Ry rY ry
2. AaBbCCDd = 23 = 8 gametes
ABCD ABCd AbCD AbCd
aBCD aBCd abCD abCD
3. MmNnOoPPQQRrssTtQq = 26 = 64 gametes
LET’S TALK DOMINANCE Part Four:
What is dominance? • It is important to understand that an allele is called “dominant” because it is seen in
the phenotype, not because it somehow subdues a recessive allele. • Alleles are simply variations in a gene’s nucleotide sequence. • When a dominant allele coexists with a recessive allele in a heterozygote, the alleles
do not interact. • It is the pathway from genotype to phenotype that dominance and recessiveness
comes into play. • There are different levels of dominance seen in traits.
Phenotype
Genotype
COMPLETE dominance • In pea plants, round seed shape is
dominant to a wrinkled seed shape.
• The dominant allele (E) codes for the synthesis of an enzyme that helps convert sugar to starch in the seed.
• The recessive allele (e) codes for a defective form of this enzyme. In a recessive homozygote, sugar accumulates in the seed because it is not converted to starch. As the seed develops, the high sugar concentration causes the osmotic uptake of water and the seed swells. When the mature seed dries, it develops wrinkles.
• In contrast, if a dominant allele is present, sugar is converted to starch, and the seeds do not wrinkle when they dry. One dominant allele results in enough of the enzyme to convert sugar to starch, and thus dominant homozygotes and heterozygotes result in the same phenotype: round seeds.
Round Seeds
Round Seeds
Wrinkled Seeds
INCOMPLETE dominance • Neither allele is completely dominant.
• When two purebred varieties cross, the F1 hybrids have an appearance somewhere in between the phenotypes of the two parental varieties.
• For example, when purebred red snapdragons cross with purebred white snapdragons, the resulting F1 hybrids have pink flowers.
• This pink phenotype results from flowers of heterozygotes having less red pigment than the red homozygotes.
CO-dominance • Both alleles are dominant.
• When two purebred varieties cross, the F1 hybrids have an appearance that expresses both phenotypes of the two parental varieties.
• For example, when pure white-feathered chickens cross with pure black feathered chickens, the resulting F1 hybrids have both white and black feathers.
• This speckled phenotype results from the feathers of heterozygotes having both white and black pigments. Gene expression plays a large role.
INHERITANCE CAN BE COMPLICATED
Part Five:
Polyallelic Traits • When a trait is caused by more than
two alleles in a population. • An individual still only inherits two alleles
for the trait – one from each parent. However, they may be two entirely different alleles from its neighbor. – For example: Coat color in Rabbits
• There are four alleles contributing to coat color: C, c, cch, and ch
• Full color: CC, Ccch, Cch, Cc
• Chinchilla: cchch, cchcch, cchc
• Himalayan: chc, chch
• Albino: cc
Polygenic Traits • Traits produced by the inheritance
of several genes in an individual.
– There is an additive effect of two or more genes on a single phenotype, which results in a wide range of phenotypes along a continuum.
– For example, there is evidence that skin pigmentation in humans is controlled by at least 3 separately inherited genes (probably more, but we will simplify).
– There is a dark-skin allele for each gene (A, B, C) contributing one “unit” of darkness to the phenotype and becoming incompletely dominant to the other alleles (a, b, c). An AABBCC person would be very dark, while an aabbcc person would be very light.
– What is another polygenic trait?
Can one gene �stop� another? • Epistasis = from the Greek for �stopping�
• A gene at one locus alters the
phenotypic expression of a gene at a second locus.
• In mice, black coat color (B) is
dominant to brown (b). A second gene determines whether or not pigment will be deposited in the hair. The dominant allele (C) results in the deposition of either black or brown pigment, depending on the genotype at the first locus. But if the mouse is homozygous recessive for pigment (cc), then the coat is white (albino), regardless of the genotype at the black/brown locus. The gene for pigment deposition is said to be epistatic to the gene that codes for black or brown pigment.
Can the environment have an effect on phenotype? • YES! For example, a single tree, locked
into its inherited genotype, has leaves that vary in size, shape, and greenness, depending on exposure to wind and sun.
• The outcome of a genotype lies within its
norm of reaction, a phenotypic range that depends on the environment in which the genotype is expressed.
• Hydrangea flowers of the same genetic
variety range in color from blue-violet to pink, depending on the acidity of the soil.
• Can you think of environmental
influences that affect human traits?
ABO blood groups in humans
• Blood Type Genotype Phenotype 1. type A = IAIA or IAi A 2. type B = IBIB or IBi B 3. type AB = IAIB AB 4. type O = ii O
• There are three alleles that contribute to human blood type:
IA, IB, and i.
• There are four phenotypes for blood type:
A, B, AB, and O.
• An example of a polyallelic trait showing both complete dominance and codominance
• What type of blood can a person with type A receive in a blood transfusion?
A or O • What type of blood can a person with type AB receive in a blood
transfusion? A, B, AB, or O
• Type O is called the Universal Donor.
• Type AB is called the Universal Recipient.
PRACTICE PROBLEM #1: Blood Type • A homozygous male with type B blood mates with a
heterozygous female with type A blood. What is the chance that their child will be type B blood?
STEP #1: Create a Key. STEP #2: Determine Parent�s Genotype
P: __________ x ____________
(male) (female) STEP #3: Fill in the Punnett Square STEP #4: Determine the phenotypes of F1 offspring. STEP #5: Answer the question.
Phenotypic Ratio: ____Type A ____ Type B ____Type AB ____ Type O
ANSWER #1: Blood Type • A homozygous male with type B blood mates with a
heterozygous female with type A blood. What is the chance that their child will be type B blood?
STEP #1: Create a Key. STEP #2: Determine Parent�s Genotype
P: IBIB x IAi
(male) (female) STEP #3: Fill in the Punnett Square STEP #4: Determine the phenotypes of F1 offspring. STEP #5: Answer the question. There is a 50% that their child will have type B blood.
Phenotypic Ratio: 0 Type A ½ Type B ½ Type AB 0 Type O
IA i
IB
IB IAIB
IAIB IBi
IBi
PRACTICE PROBLEM #2: Blood Type
• If a boy has a blood type O and his sister has blood type AB, then what are the genotypes and phenotypes of their parents? – Boy: ____________ – Girl: ____________ – Parent�s must be:
ANSWER #2: Blood Type
• If a boy has a blood type O and his sister has blood type AB, then what are the genotypes and phenotypes of their parents? – Boy: type O (ii) – Girl: type AB (IAIB) – Parent�s must be:
• type A (IAi) and type B (IBi)
IA i
IB
i
IAIB
ii
Sex-linked Traits • Traits (genes) that are located on the sex
chromosomes.
XX chromosome - female Xy chromosome - male
fruit fly eye color
• Why are 99% of sex-linked genes found on the X chromosome rather than the Y chromosome?
• Since sex-linked genes are mostly on the X chromosome, then sons
inherit these genes from which parent – mother or father?
Because the X is larger than the Y.
The mother gives the X to the son; the father gives the Y to the son.
Female Carriers • In a sex-linked trait (such as hemophilia),only females are carriers,
because they have two X chromosomes therefore allowing them to carry two versions of the gene.
• Since males only have one X chromosome, they either express the trait or not. Therefore, sex-linked traits are more commonly expressed in males.
PRACTICE PROBLEM #1: Sex-linked • In fruit flies, eye color is sex-linked. Red eyes are dominant to white eyes. If
a red-eyed male is mated with a white-eyed female, then what percentage of the F1 offspring will have white eyes?
STEP #1: Make a Key. R – red eye color ; r – white eye color XX – female ; XY – male STEP #2: Determine the Parent�s Genotypes. P: _____________ x _____________
(male) (female) STEP #3: Fill in the Punnett Square STEP #4: Determine the Phenotypic Ratio of the F1 offspring. STEP #5: Answer the question.
Phenotypic Ratio: _____ red-eyed male _____ white-eyed male _____ red-eyed female _____ white-eyed female
ANSWER #1: Sex-linked • In fruit flies, eye color is sex-linked. Red eyes are dominant to white eyes. If
a red-eyed male is mated with a white-eyed female, then what percentage of the F1 offspring will have white eyes?
STEP #1: Make a Key. R – red eye color ; r – white eye color XX – female ; XY – male STEP #2: Determine the Parent�s Genotypes. P: XRY x XrXr (male) (female) STEP #3: Fill in the Punnett Square STEP #4: Determine the Phenotypic Ratio of the F1 offspring. STEP #5: Answer the question. There is a 50% chance of a F1 offspring having white eyes.
Phenotypic Ratio: 0 red-eyed male ½ white-eyed male ½ red-eyed female 0 white-eyed female
Xr Xr
XR
Y
XRXr XRXr
XrY XrY
PRACTICE PROBLEM #2: Sex-linked • Hemophilia in humans is due to an X-chromosome mutation. It
is a recessive trait. What will be the results of a mating between a normal (non-carrier) female and a hemophiliac male?
– Use the letter N to represent the non-hemophiliac and n to represent the hemophiliac.
KEY: P: ________ x ________ Phenotypic Ratio of F1 offspring:
ANSWER #2: Sex-linked • Hemophilia in humans is due to an X-chromosome mutation. It
is a recessive trait. What will be the results of a mating between a normal (non-carrier) female and a hemophiliac male?
– Use the letter N to represent the non-hemophiliac and n to represent the hemophiliac.
KEY: XN – normal ; Xn – hemophilia XX – female ; XY – male P: XN XN x Xn Y Phenotypic Ratio of F1 offspring: 100% normal offspring (The females are carriers.)
XN XN
Xn
Y
XNXn XNXn
XNY XNY
X-inactivation in Female Mammals • Female mammals inherit two X chromosomes, but one X in each cell
becomes almost completely inactivated during embryonic development. • Females consist of a mosaic of two types of cells, because X inactivation
occurs randomly. • The inactive X in each cell condenses into a compact object called a Barr
body. Most of the genes of the Barr body are not expressed. • In the ovaries, Barr body chromosomes are reactivated in the cells that give
rise to eggs, so every female gamete has an active X.
