unit vi - mendelian genetics baby campbell – ch 9 big campbell – ch 14, 15
TRANSCRIPT
I. MENDEL• Mendel’s Experiments
o Worked with ______________o Eliminated
________________________ and controlled ____________________
o P Generation - True-breeding pea plants with one trait X true-breeding pea plants with another trait
o Produced hybrids also known as F1
___________________________F1 Phenotype = F1 Genotype =
o F1 X F1 → F2
___________________________F2 Phenotype Ratio = F2 Genotype Ratio =
pea plants
self-pollinationcross-pollination
TT x tt
4 tall : 0 short
0 TT : 4 Tt : 0 tt
Tt x Tt
3 tall : 1 short
1 TT : 2 Tt : 1 tt
I. MENDEL, cont• Mendel’s Principles
1) Alternative versions of genes known as ______________ account for variations in inherited characters.
2) Organisms inherit _____ alleles for each trait
3) If alleles at a locus differ; that is; if the genotype is _______________, the allele that shows is known as the ________________ allele.
4) Law of Segregation – two alleles for a heritable character segregate during meiosis
5) Law of Independent Assortment – each pair of alleles segregates independently of each other pair of alleles during meiosis
alleles
2
heterozygous
dominant
II. ANALYZING PROBABILITY OF TRAIT INHERITANCE
• Test Crosso Organisms with dominant phenotype crossed with
_____________________________ to determine genotype• Punnett Square• Multiplication Rule
o States the probability of 2 or more independent events occurring together can calculated by multiplying individual probabilities
o For example, Determine the probability of a homozygous recessive short plant
produced from F1 X F1
Cross = Tt x Tt
Probability of egg carrying t = ½ Probability of sperm carrying t = ½ Probability of tt offspring = ¼
homozygous recessive
II. ANALYZING PROBABILITIES, cont• Addition Rule
o States that the probability of 2 or more mutually exclusive events occurring can be calculated by adding together their individual probabilities
o For example, Determine the probability of a heterozygous plant produced from
F1 X F1
Tt x Tt Chance of egg carrying T = ½ Chance of sperm carrying t = ½
Chance of sperm carrying T = ½ Chance of egg carrying t = ½
Probability of Tt offspring = ¼ + ¼ = ½
II. ANALYZING PROBABILITIES, cont• Crosses Involving Multiple Characters
o Determine the genotype ratios of the offspring for the cross BbDD X BBDd
II. ANALYZING PROBABILITIES, cont
• Crosses Involving Multiple Characterso Determine the genotype ratios of the offspring for the
cross YyRr X YyRr
II. ANALYZING PROBABILITIES, cont• Crosses Involving Multiple Characters
o In the cross, PpYyRr X Ppyyrr, what is the probability of offspring that are purple, green, & round?
P= purple, p = white Y = yellow, y = green R = round, r = wrinkled
Probability Practice Makes Perfect! In pea plants, • long stems are dominant to short stems• purple flowers are dominant to white, and • round seeds are dominant to wrinkled. A plant that is heterozygous for all three loci self-pollinates and 2048 progeny are
examined. How many of the resulting plants would you expect to be long-stemmed with purple flowers, producing wrinkled seeds?
III. VARIATIONS IN INHERITANCE• Co-Dominance
o Both alleles affect phenotype in separate & distinguishable ways
o Often designated with 2 different “big letters”
• Incomplete Dominanceo Neither allele is dominant; heterozygotes show a blend of
two homozygous phenotypeso One allele designated with “big letter’, the other with “big
letter prime”; for example T T’
III. VARIATIONS IN INHERITANCE, cont
• Multiple Alleleso Many genes have more than 2
alleleso Example, ABO blood groups in
humanso Three alleles
Phenotype Genotype
A
B
AB
O
• A woman with O blood has a child with Type A blood. The man she claims is the father has AB blood. Is it possible that he is the father of this child?
III. VARIATIONS IN INHERITANCE, cont• Polygenic Inheritance
o For example, AABBCC = very dark skin; aabbcc = very light skin.o Intensity based on units; in other words, AaBbCc and AABbcc
individuals would have the same pigmentation
III. VARIATIONS IN INHERITANCE, cont• Epistasis
o Gene at one locus alters phenotypic expression of a gene at a second locus
o For example,A dominant allele, P causes the production of purple pigment; pp individuals are white. A dominant allele C is also required for color production; cc individuals are white. What proportion of offspring will be purple from a ppCc x PpCc cross?
IV. SEX-LINKED INHERITANCEo First recognized by Thomas Hunt Morgan
Drosophila melanogasterFruit fliesExcellent organism for genetic studies
Prolific breeding habits Simple genetic make-up; 4 pairs of chromosomes → 3 pairs of
autosomes, 1 pair of sex chromosomes Crossed true-breeding wild-type females with true-breeding
mutant males Mutant trait showed up in ½ male F2 offspring ; was not seen in F2
femaleso Determined mutant allele was on X-chromosome; thus
inherited differently in males versus females In females, In males,
IV. SEX-LINKED INHERITANCE, cont• Red-green colorblindness is caused by a sex-linked recessive allele. A
color-blind man marries a woman with normal vision whose father was colorblind. What is the probability she will have a colorblind daughter?
IV. SEX-LINKED INHERITANCE, cont• The gene for amber body color in Drosophila is sex-linked recessive. The
dominant allele produces wild type body color. The gene for black eyes is autosomal recessive; the wild type red eyes are dominant. If males with amber bodies, heterozygous for eye color are crossed with females heterozygous for eye color and body color, calculate the expected phenotype ratios in the offspring.
IV. SEX-LINKED INHERITANCE, cont• X Inactivation in Females
o During embryonic development, one X chromosome in female cells is inactivated due to addition of methyl group to its DNA
o Dosage compensationo Inactive X chromosome condenses; known as Barr body
IV. SEX-LINKED IN INHERITANCE, cont
o Occurs randomlyFemales will have some cells where “Dad’s copy” of X is
inactivated, some where “Mom’s copy” is inactiveTherefore, females are a mosaic of cellsPreserved in mitosisIn ovaries, Barr body chromosome is reactivated for
meiosis and oogenesis
V. GENE MUTATIONSo Change in the nucleotide sequenceo May be spontaneous mistakes that
occur during replication, repair, or recombination
o May be caused by mutagens; for example, x-rays, UV light, carcinogens
o If changes involve long stretches of DNA, known as chromosomal mutations
o Point mutations – change in a gene involving a single nucleotide pair; 2 types Substitution Frameshift – due to addition or
deletion of nucleotide pairs
V. GENE MUTATIONS, cont• Classification of Gene Mutations
o Traits may be described as dominant, recessive, etc . based on the effect of the abnormal allele on the organism’s phenotype
o Instruction encoded by genes carried out through protein synthesiso Vast majority of proteins are enzymeso Abnormal allele → Defective enzyme
If the enzyme produced by the normal allele is present in sufficient quantities to catalyze necessary reactions,
No noticeable effect on phenotypeDefective allele is classified as recessive
If the lack of normal enzyme production by defective allele cannot be overcome by normal allele,
Organism’s phenotype is affectedDefective allele is classified as dominant
VI. INHERITED GENETIC DISORDERS
• Due to gene mutations• Classified as autosomal or sex-linked, depending on
chromosome location of affected gene
• Autosomal Disorders – Grouped according to path of inheritanceo Autosomal Recessive Disorders
VII. TESTING FOR INHERITED GENETIC DISORDERS
• Identification of Carriers/Genetic Counselingo Tests are available for Tay-Sachs, sickle cell, cystic fibrosis,
Huntington’s, PKU, & many others• PGD – Preimplantation Genetic Diagnosis
VII. GENETIC TESTING, cont• Fetal Testing
o Amniocentesis Performed between 14th-16th weeks of pregnancyCells collected, tested
VII. GENETIC TESTING, cont
o Chorionic Villus Sampling (CVS)Narrow tube inserted
through mother’s vagina, cervix
Small tissue sample suctioned from placenta (organ that transmits nutrients, removes wastes from fetus)
Testing may be done earlier in pregnancy but not suitable for all types of testing
• Newborn Screeningo PKU
VIII. CHROMOSOMAL BASIS OF INHERITANCE• Chromosomal Theory of
Inheritanceo States genes occupy specific loci
on chromosomeso During meiosis, chromosomes
undergo segregation & independent assortment
• Linked Geneso During Thomas Morgan’s work
with Drosophila, he recognized Two genes located on same
chromosome were linked; that is, inherited together
However, offspring phenotypes showed this wasn’t always true
VIII. CHROMOSOMAL BASIS OF INHERITANCE, cont
• Linked Genes, conto In fruit flies, normal wild-type phenotype is gray body, normal
wings – both genes are located on same chromosomeG = wild-type (gray) body; g = black bodyW = wild-type wings; w = mutant wings
o True-breeding wild type flies X true-breeding mutantso F1 showed all
o F1 X test crossCounted 2300 offspringShould have counted
VIII. CHROMOSOMAL BASIS OF INHERITANCE, cont
965 GWgw 944 gwgw 206 Gwgw 185 gWgw
Morgan realized variation in probabilities due to
VIII. CHROMOSOMAL BASIS OF INHERITANCE, cont
• Linkage Mapso In crossing over, the further apart two genes are, the higher
the probability that a crossover will occur between them and therefore, the higher the recombination frequency.
VIII. CHROMOSOMAL BASIS OF INHERITANCE, cont
o Recombination Frequency = # recombinants__ X 100 total # offspringo One map unit = 1% recombination frequency……………………………………………………………………………………..965 GWgw 944 gwgw 206 Gwgw 185
gWgw
VIII. CHROMOSOMAL BASIS OF INHERITANCE, cont• The genes for vestigial wings, black body color, and cinnabar eyes are
linked genes.• In controlled crosses . ..
The gene for vestigial (vg) wings and body color (b) have a 17% crossover rate.
The gene for eye color (cn) and body color (b) have a 9% crossover rate. The gene for eye color (cn) and vestigial wings (vg) have a 9 ½%
crossover rate.• Draw the chromosome.
IX. CHROMOSOMAL DISORDERS• Alterations in Chromosome Number
o Most commonly due to nondisjunction
o Results in aneuploid gametes
IX. CHROMOSOMAL DISORDERS, cont• Cri du Chat – Caused by deletion in chromosome 5
• Chronic Myelogenous Leukemia – Caused by reciprocal translocation. Large piece of chromosome 22 exchanges places with tip of chromosome 9. Resulting chromosome 22 easily recognizable; known as Philadelphia chromosome.