BIO 202 Midterm: Exam 3 Study Guide
16 Jun 2018
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Midterm 3 Objectives:
Chapter 15:
1. Understand how nondisjunction occurs and gives rise to aneuploidy.
Aneuploidy: incorrect number of chromosomes. Most common chromosomal abnormality in humans.
Nondisjunction (not coming apart): Failure to separate homologous chromosomes in meiosis I or the
failure of separating sister chromatids in meiosis II.
-Gametes will have more than one, or too little chromosomes.
2. Predict the outcome of meiotic products due to nondisjunction during meiosis I or meiosis II.
Meiosis I: two gametes will have an extra chromosome(trisomic); two gametes have one less
chromosome(monosomic).
Meiosis II: One gamete has extra chromosome, one has one less, and two normal gametes.
3. Be able to predict the genotype that results from fertilization of abnormal gametes.
4. Define trisomy, triploidy, and polyploidy. Explain how these major chromosomal changes occur and
describe the consequences.
Trisomy: Three copies of a chromosome.
Triploidy: Having extra SET of chromosome.
Polyploidy: Contains more than two complete SETS of
chromosomes.
-Can occur when a normal gamete fertilizes another gamete
that has nondisjunction for all its chromosomes.
-Common in plants. Uncommon in animals. No known
mammals are polyploidy.
5. Describe the type of chromosomal alterations implicated in the following human disorders: Down
syndrome, Klinefelter's syndrome, extra Y, triple-X syndrome, Turner's syndrome.
Down syndrome: Caused by an extra chromosome 21. AKA trisomy 21. Chromosome 21 is the smallest
chromosome (1.5% of total DNA) so it makes sense this is the only chromosome aneuploidy that is not
lethal.
Klinefelter's syndrome: (XXY) Two X chromosomes, 1 Y chromosome.
Extra Y: (XYY) Extra Y chromosome that can result in taller than average males.
triple-X syndrome: (XXX)Produces healthy females
Turner's syndrome: (X0) Only one X chromosome. Only known human monosomy.
6. Distinguish among deletions, duplications, inversions, and translocations and understand their
phenotypic consequences.
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Deletions: One chromosome segment is deleted. Information lost
Duplications: A segment of chromosome is repeated. Information gained.
Inversions: A segment of chromosome is inverted. Can alter phenotype because gene expression may
be influenced by the location.
Translocations: A segment of a chromosome is moved to another chromosome. Can alter phenotype
because gene expression may be influenced by the location.
Ex: Philadelphia Chromosome: Translocation between chromosome 9 and 22 causes CML
(chronic myelogenous leukemia). This alters tyrosine kinase which causes overexpression of
white blood cells.
Chapter 17:
1. Desie Beadle ad Tatu’s epeients with Neurospora and explain how they contributed to our
understanding of how genes control metabolism.
-Beadle and Tatum developed the one gene-one, one enzyme hypothesis. They did this through
experimenting with neurospora (bread mold).
-Neurospora usually has all the enzymes required to grow in a minimally nutrient rich environment
making it ideal for experimentation on gene encoding.
-The scientists mutated the spores using radiation, the mold could no longer grow in that medium. It
showed that depending on the type of radiation (x-ray, UV, or neutron) the neurospora would not be
able to create an essential molecule due to gene mutation.
-To pipoit the eat utiet that ould’t e eated, the ultiated the utated euospoa i
different amino acids, the amino acid in which the neurospora grows in is the amino acid the
2. Distiguish etee the oe gee-oe eze hpothesis ad the oe gee-oe polpeptide
hypothesis and explain why the original hypothesis was changed.
one gene-one enzyme: Idea that each gene encodes an enzyme.
one gene-one polypeptide: Ideology that some genes code for a subunit, or polypeptide of a protein,
not a whole protein, and proteins besides enzymes can be made from a gene.
3. Understand the basis of the early techniques used to crack the genetic code and identify amino
acids it specifies.
4. Explain the significance of the reading frame during translation (can identify an open reading frame)
-Reading Frame is important in translation because it determines what amino acids will be created due
to the reading of codons. If an insertion or deletion of 1 or 2 bases occurs, the reading frame will be
messed up, and the polypeptide will be entirely different.
-Open reading frames start with the start codon AUG (which codes for methionine amino acid) the
reading from ends at a stop codon (either UAA, UAG, or UGA). There is usually a span of about 33
codons in between the start and stop codon.
5. Be able to identify regions with predicted protein-encoding genes, and provide information about
which strand of DNA is the template, and the approximate length (in amino acids) of the protein.
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-Template Strand (non-coding strand,
atisese stad: 3’→5’ dietio stad.
RNA polymerase copies tis strand uses it as a
TEMPLATE to create mRNA.
-Coding Strand (non-template strand, sense
stad: 5’→3’ direction strand. mRNA has
the same base sequence as this strand as it is
the same direction as mRNA. The CODE on
this strand is the same as the copied mRNA.
-Length of amino acids=# of codons
6. Understand the process of RNA synthesis. When provided with a double standed DNA molecule, be
able to predict the sequence of RNA product or DNA template.
Process of mRNA synthesis is transcription.
Genes provide instructions for making proteins. RNA is a copy of these instructions.
RNA polymerase creates the mRNA fo 5’ to 3’.
Usually only one strand is copied. VERY RARELY are both strands being transcribed.
Does not require a primer, uses start sites. Does not use any proofreading.
3 steps:
Initiation: Transcription factors recognize and bind to promotor region (ex: a TATA box). This
recruits RNA polymerase and causes the creation of a transcription initiation complex (which
consists of many proteins).
Elongation: DNA is untwisted as RNA polymerase moves along the template strand to create
mRNA strand. The RNA is eated fo the 5’→3’ dietio. The DNA is closed as the RNA
polymerase passes it and RNA molecule peels away.
Termination: When RNA polymerase encounters a terminator sequence in the DNA. This causes
the RNA polymerase to stop transcription. Differs in prokaryotes and eukaryotes. In eukaryotes,
RNA polymerase is not cleaved right away. 10-35 nucleotides after termination sequence mRNA
is cut.
7. Be able to describe the function of the following in terms of transcription: mRNA, 5' end, 3' end,
RNA polymerase, phosphodiester bond, promoter, start site, coding sequence, transcription
terminator.
mRNA, 5' end, 3' end: RNA is eated i this dietio 5’→3’.
RNA polymerase: RNA polymerase makes the mRNA strand from the DNA strand template. They add
complimentary nucleotides and create phosphodiester linkage to bond the nucleotides together.
phosphodiester bond:
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Document Summary
Chapter 15: understand how nondisjunction occurs and gives rise to aneuploidy. Nondisjunction (not coming apart): failure to separate homologous chromosomes in meiosis i or the failure of separating sister chromatids in meiosis ii. Gametes will have more than one, or too little chromosomes: predict the outcome of meiotic products due to nondisjunction during meiosis i or meiosis ii. Meiosis i: two gametes will have an extra chromosome(trisomic); two gametes have one less chromosome(monosomic). Meiosis ii: one gamete has extra chromosome, one has one less, and two normal gametes: be able to predict the genotype that results from fertilization of abnormal gametes, define trisomy, triploidy, and polyploidy. Explain how these major chromosomal changes occur and describe the consequences. Polyploidy: contains more than two complete sets of chromosomes. Can occur when a normal gamete fertilizes another gamete that has nondisjunction for all its chromosomes.
