exam 1 & 2 learning objectives

Question 1

nondisjunction

Answer 1

The failure of homolog or sister chromatid separation during cell division. Results in nuclei with the wrong number of chromosomes.

Question 2

aneuploidy

Answer 2

An uneven number of chromosomes. Usually the result of the gain or loss of a chromosome—that is, (trisomy) or (monosomy).

Question 3

monosomy

Answer 3

The presence of a single chromosome instead of a homologous pair, resulting in a chromosome number that is 2n-1

Question 4

trisomy

Answer 4

The presence in a genome of three copies of a chromosome rather than a homologous pair of chromosomes and resulting in a number of chromosomes that is 2n-1

Question 5

haploinsufficiency

Answer 5

A wild-type allele that is unable to support wild-type function in a heterozygous genotype. Classified as a recessive wild-type allele.

Question 6

gene dosage

Answer 6

The number of copies of a gene.

Question 7

euploidy

Answer 7

A number of chromosomes that is an exact multiple of the haploid number.

Question 8

polyploidy

Answer 8

The presence of more than two complete sets of chromosomes in a genome.

Question 9

autopolyploidy

Answer 9

A pattern of polyploidy produced by the duplication of chromosomes from a single genome.

Question 10

allopolyploidy

Answer 10

A polyploidy organism arising through the union of chromosome sets from different species.

Question 11

complete dominance

Answer 11

The complete dominance of one allele also results in the exclusive expression of the dominant phenotype among the heterozygous progeny of a cross between pure-breeding homozygous parents, while the progeny display a 3:1 ratio of dominant to recessive phenotypes.
(purple & white flowers)

Question 12

incomplete dominance

Answer 12

the phenotype of the heterozygous organism is distinctive
( red+white=pink)

Question 13

codominance

Answer 13

leads to a heterozygous phenotype different
identified when the protein products of both alleles are detectable in heterozygous organisms
ex: ABO blood type

Question 14

multiple alleles

Answer 14

the presence of more than two alternative forms (alleles) of a gene that can occupy the same locus on a chromosome.

Question 15

recessive lethal allele

Answer 15

type of allele that can cause the death of an organism when present in a homozygous recessive state

Question 16

dominant lethal allele

Answer 16

an allele that causes the death of an organism when present in just one copy (heterozygous state).

Question 17

gene interaction

Answer 17

phenomenon where different genes influence each other’s expression and contribute collectively to a particular phenotype.

Question 19

complementary gene interaction

Answer 19

Two or more genes interact to produce a specific phenotype that neither can produce alone.

Question 20

recessive epistasis

Answer 20

when the presence of two recessive alleles at one gene locus masks or suppresses the expression of alleles at a different locus.

Question 21

dominant epistasis

Answer 21

occurs when a dominant allele at one gene locus masks or suppresses the expression of alleles at a different locus.

Question 22

pleiotropy

Answer 22

phenomenon where a single gene influences multiple, seemingly unrelated phenotypic traits.

Question 23

autosomal dominant

Answer 23

a pattern of inheritance in which a single copy of a dominant allele on an autosome (a non-sex chromosome) is sufficient to cause the expression of a trait or disorder.

Question 24

autosomal recessive

Answer 24

a trait or disorder is determined by two copies of a recessive gene on a non-sex chromosome (autosome)

Question 25

x-linked recessive

Answer 25

refers to a mode of inheritance in which a mutation in a gene on the X chromosome causes the phenotype to be expressed in males (who have only one X chromosome) and in females who are homozygous for the gene mutation (having two copies of the mutated gene).

Question 26

x-linked dominant

Answer 26

a mode of genetic inheritance by which a dominant gene is carried on the X chromosome. This pattern of inheritance is characterized by the expression of the trait or disorder in individuals who have at least one copy of the dominant allele on their X chromosome.

Question 27

y-linked

Answer 27

the transmission of genes located on the Y chromosome. These genes are passed exclusively from father to son, as only males possess a Y chromosome.

Question 28

sex-limited traits

Answer 28

characteristics that are expressed in only one sex, even though the genes for these traits are present in both sexes. These traits are typically influenced by the hormonal environment of the individual, which determines whether the trait will be expressed.

Question 29

sex-influenced traits

Answer 29

autosomal traits that are expressed differently in males and females due to hormonal differences. These traits can appear in both sexes but are influenced by the individual’s sex.

Question 30

Explain how eukaryotic DNA undergoes compaction to form condensed chromatin using the following terms: histone, H1, nucleosome, solenoid, supercoiling

Question 31

genetic drift

Answer 31

A change in the gene pool of a population due to chance (more common in a small population)

Question 32

founder effect

Answer 32

Occurs when a few individuals colonize an isolated habitat – The alleles present are not a good representation of the whole population * The British colony of Tristan da Cunha – Disproportionately high rate of hereditary blindness * Galápagos Islands – Organisms that arrived as strays diverged from the populations from which they came (Amish)

Question 33

bottleneck effect

Answer 33

Results from a drastic reduction in population size – Fires, floods, earthquakes, hunting By chance certain alleles may be – Overrepresented – Underrepresented – Eliminated – Unchanged Can decrease genetic variability in a population – Loss of individual variation results in a loss of adaptability

Question 34

gene flow

Answer 34

Genetic exchange with another population – When fertile individuals move into or out of a population – When gametes are transferred between populations * Pollen * Results in – Gain or loss of alleles – Reduces differences between populations

Question 35

natural selection

Answer 35

is a process in which organisms with certain inherited characteristics are more likely to survive and reproduce than are individuals with other characteristics * i.e. differential success and reproduction * Result: A population changes over generations 25

Question 36

directional

Answer 36

Shifts the overall makeup of a population by selecting in favor of one extreme phenotype

Question 37

disruptive

Answer 37

Favors variants at opposite extremes over intermediate individuals – Can lead to two new species

Question 38

stabilizing

Answer 38

Favors intermediate phenotypes (heterozygotes) – Most common

Question 39

Explain the Hardy-Weinberg equilibrium and describe the “ideal” population that it models

Answer 39

Models an “ideal” population * Large population * No migration * No mutations * No natural selection * Random mating * Can only be used when there is no evolution

Question 40

linked genes

Answer 40

Genes located on the same chromosome cannot undergo independent assortment

Question 41

unlinked genes

Question 42

parental/noncrossover gametes

Answer 42

Without crossing over (complete linkage) largest proportion

Question 43

recombiant/crossover gametes

Answer 43

Occurs between two nonsister chromatids – Both parental and recombinant (crossover) gametes are produced

Question 44

single crossover

Question 45

double crossover

Answer 45

smallest proportion

Question 46

conjugation

Answer 46

The short-term union of two bacterial cells for the unidirectional transfer of DNA from the “donor” to the “recipient.” The transferred material may be plasmid DNA or donor bacterial chromosome DNA.

Question 47

transformation

Answer 47

The bacterial process of gene transfer in which donated DNA fragments originating in a dead donor cell, or plasmid DNA, are taken up across the cell wall and membrane of a recipient cell and recombined into the transformant genome. (2) More generally refers to the process by which exogenous DNA is directly taken up by a cell resulting in a genetic alteration of the cell. (3) The conversion of animal cells to an abnormal unregulated state by an oncogenic virus or by transforming DNA

Question 48

transduction

Answer 48

in bacterial systems, the process of transfer of DNA from a donor bacterial cell to a recipient cell using a bacteriophage as a vector. More generally can refer to the process by which foreign DNA is introduced into another cell via a viral vector.

Question 49

cotransformation

Answer 49

Simultaneous transformation of two or more genes carried on a donor DNA fragment into a recipient.

Question 50

cotransduction

Answer 50

The simultaneous transduction of two or more genes contained on a donor DNA fragment into a recipient cell, where it undergoes homologous recombination to be spliced into the transductant chromosome.

Question 51

vertical gene transfer

Answer 51

the passage of genetic material from a parent organism to its offspring, primarily through sexual or asexual reproduction.

Question 52

horizontal gene transfer

Answer 52

the process where an organism acquires genetic material from another organism that is not its parent or offspring.

Question 53

bacteriophage

Answer 53

A virus whose host is a bacterium.

Question 54

partial diploid +

Answer 54

is a bacterial cell that carries two copies of some of its genes

Question 55

Describe in detail the process of transcription in bacteria

Answer 55

1. Initiation a. Promoter Recognition Transcription starts at a specific DNA sequence called a promoter, usually found just upstream (before) the gene. Bacterial promoters typically contain two key consensus sequences: −10 region (TATAAT) – also called the Pribnow box −35 region (TTGACA) b. RNA Polymerase Binding Bacteria use a single type of RNA polymerase made up of several subunits (core enzyme: α₂ββ′ω). To begin transcription, it needs a σ (sigma) factor, which helps it recognize and bind to the promoter. ✅ Together, the RNA polymerase core enzyme and sigma factor form the holoenzyme. c. Open Complex Formation After binding, the DNA strands are unwound around the −10 region, creating an open complex. The template strand (3′→5′) will be used to make RNA. d. Abortive Initiation and Promoter Escape RNA polymerase starts making short RNA sequences (abortive transcripts). Once it synthesizes ~10 nucleotides successfully, it releases the sigma factor and moves forward into elongation. 2. Elongation The core RNA polymerase continues without the sigma factor. It adds ribonucleotides (A, U, G, C) one by one, complementary to the DNA template strand. RNA is built in the 5′ to 3′ direction. A short RNA-DNA hybrid (about 8–9 base pairs) forms within the enzyme. 🔁 RNA polymerase has no proofreading ability like DNA polymerase, so errors may occur but are usually not harmful. 3. Termination There are two types of termination in bacteria: a. Rho-Independent Termination (Intrinsic) The RNA forms a GC-rich hairpin loop followed by a stretch of uracils (U’s). This structure causes RNA polymerase to stall and detach from the DNA. b. Rho-Dependent Termination A protein called Rho factor binds to a specific site on the newly made RNA. It moves along the RNA toward RNA polymerase. When RNA polymerase pauses, Rho catches up and causes it to release the RNA and DNA. 🧪 Result: A single mRNA molecule is produced. In bacteria, this mRNA is often polycistronic, meaning it can code for multiple proteins from a single transcript (especially in operons)

Question 56

Describe in detail the process of translation in bacteria

Answer 56

1. 🧭 Initiation a. mRNA Binding to the Ribosome The bacterial ribosome is made of two subunits: 30S (small) subunit 50S (large) subunit A ribosome binding site on the mRNA, called the Shine-Dalgarno sequence, base-pairs with a complementary sequence on the 16S rRNA of the 30S subunit. This helps position the start codon (AUG) correctly in the P site of the ribosome. b. Initiator tRNA Binding The initiator tRNA, carrying formyl-methionine (fMet), binds to the start codon (AUG) via its anticodon. This tRNA is called tRNA^fMet in bacteria. c. Assembly of the Complete Ribosome Once the initiator tRNA is in place, the 50S subunit joins, forming the 70S initiation complex. This step requires initiation factors (IF-1, IF-2, IF-3) and GTP for energy. 2. 🔁 Elongation This stage builds the polypeptide chain one amino acid at a time. a. Aminoacyl-tRNA Entry A charged tRNA (with an amino acid) matching the next codon enters the A site of the ribosome. This step is helped by elongation factor EF-Tu and requires GTP. b. Peptide Bond Formation The enzyme peptidyl transferase (part of the 50S subunit) forms a peptide bond between: The amino acid in the P site (fMet or growing chain) The new amino acid in the A site c. Translocation The ribosome shifts three nucleotides (one codon) along the mRNA. The tRNA in the P site moves to the E site (Exit site) and is released. The tRNA in the A site moves to the P site, making room for the next tRNA. This step is powered by EF-G and GTP. 🔁 This cycle repeats, adding one amino acid at a time to the growing chain. 3. 🛑 Termination a. Stop Codon Recognition When the ribosome reaches a stop codon (UAA, UAG, or UGA), there is no matching tRNA. Instead, release factors (RF-1, RF-2) bind to the stop codon. b. Polypeptide Release The ribosome releases the finished polypeptide chain from the tRNA in the P site. c. Ribosome Disassembly RF-3 and GTP help release the ribosome subunits, tRNA, and mRNA so they can be reused. 🧬 Final Product: A polypeptide chain (protein), which may then fold into its functional 3D shape or undergo further modification.

Question 57

summarize the process of eukaryotic gene expression

Answer 57

Transcription (in the nucleus) RNA polymerase II binds to a gene's promoter with help from transcription factors. It makes a pre-mRNA copy from the DNA template strand. The pre-mRNA is complementary to the DNA and built in the 5′ to 3′ direction. 🧪 2. RNA Processing (also in the nucleus) Before leaving the nucleus, pre-mRNA is modified: 5′ Cap added – protects mRNA and helps it bind to ribosomes. Poly-A tail added – increases stability. Splicing – removes non-coding sequences (introns) and joins coding parts (exons) together. ✅ Result: Mature mRNA 🚪 3. mRNA Export The mature mRNA exits the nucleus through a nuclear pore into the cytoplasm. 🧫 4. Translation (in the cytoplasm) Ribosomes read the mRNA codons (3-letter sequences). tRNA molecules bring matching amino acids. The ribosome links amino acids into a polypeptide chain (protein). 🧬 5. Protein Folding and Modification The polypeptide folds into its final shape. It may be modified (e.g. phosphorylation, glycosylation) in the ER or Golgi apparatus before becoming fully functional.

Question 58

F+

Answer 58

A donor bacterium containing an extrachromosomal fertility plasmid.

Question 59

F-

Answer 59

A bacterial recipient cell; does not contain an F (fertility) plasmid.

Question 60

F'

Answer 60

A bacterial donor cell harboring an F (fertility) plasmid.

Question 61

Hfr

Answer 61

Pertaining to Hfr chromosomes or to Hfr donors in bacterial conjugation.