Question 1

The two chromosomes in each of the 22 homologous pairs in our cells ...

Accepted Answer

The homologs are derived from both parents, normally have the same set of loci, and show the same banding pattern. However, the sequences are not expected to be identical.

Question 2

Indicate true (T) and false (F) statements below about the human genome. Your answer would be a six-letter string composed of letters T and F only, e.g. FTFFFT.
( )	Only about 1.5% of the human genome is highly conserved.
( )	Almost half of our genome is composed of repetitive sequences.
( )	Genes occupy almost a quarter of the genome.
( )	There are roughly as many pseudogenes in the human genome as functional genes.
( )	Transposable elements occupy almost 10% of our genome.
( )	On average, exons comprise 1.5% of our genes.

Accepted Answer

Only about 5% of our genome is highly conserved. Our relatively long genes (including their exons, introns, and some regulatory sequences) cover almost a quarter of the genome, but only about 1.5% of our genome is composed of exonic sequences. These exons constitute roughly 6% of our genes. In contrast, a whopping 50% of our genome is made of various repeated sequences, most notably the transposable DNA elements. Please refer to Table 4-1 for the data.

Question 3

In assembling a nucleosome, normally the …(1) histone dimers first combine to form a tetramer, which then further combines with two … (2) histone dimers to form the octamer.

Accepted Answer

The H3-H4 and the H2A-H2B dimers appear to be stable intermediates in histone assembly and exchange. The H3-H4 tetramers are also stable, and are thought to be assembled (and inherited) mostly as a single unit.

Question 4

The acetylation of lysines on the histone tails …

Accepted Answer

Histone acetylation on lysine residues is generally an activating mark favoring more open chromatin structures (in contrast to what HP1 does), but this effect is mediated mostly through the recruited non-histone proteins that recognize this mark, not the chemical modification itself. The acetylation eliminates the positive charge on lysine, which helps-but is not sufficient alone-to loosen the nucleosome packaging. All known natural covalent modifications on the histone tails are reversible.

Question 5

The chromatin remodeling complexes play an important role in chromatin regulation in the nucleus. They …

Accepted Answer

The ATP-dependent chromatin remodeling complexes can move the nucleosomes on DNA, interact with histone chaperones, and exchange histones or remove them from DNA.

Question 6

Indicate true (T) and false (F) statements below regarding histones. Your answer would be a six-letter string composed of letters T and F only, e.g. TTFFFF.
( )	The histones are highly acidic proteins.
( )	The histone fold consists of three ? helices.
( )	The core histones are much more conserved than the H1 histone.
( )	The N-terminal tails of the core histones undergo a variety of reversible post-translational modifications.
( )	Every nucleosome core is made up of three polypeptide chains.
( )	The H1 histone is absent in the 30-nm fibers.

Accepted Answer

The highly basic histone proteins are made of the histone fold that consists of three ? helices connected by two loops. Additionally, each core histone has an N-terminal tail which, along with the rest of the protein, can be modified post-translationally. Each nucleosome core particle contains eight histone proteins, two copies of each type. The less well conserved histone H1 is not part of the nucleosome core, but is present in the 30-nm fibers.

Question 7

Chromosome 3 contains nearly 200 million nucleotide pairs of our genome. If this DNA molecule could be laid end to end, how long would it be? The distance between neighboring base pairs in DNA is typically around 0.34 nm.

Accepted Answer

Multiplying the size of the chromosome by the distance between consecutive bases gives the end-to-end distance as follows: (200 × 10⁶ nucleotide pairs [np]) × (0.34 × 10-⁹ m/np) = ~0.068 m = ~7 cm.

Question 8

Which of the following is true regarding heterochromatin in a typical mammalian cell?

Accepted Answer

Heterochromatin covers a significant fraction of the mammalian genome including the centromere and telomere neighborhoods, but contains very few, mostly inactive, genes. Even though heterochromatin has a very high degree of compaction, its genes can still become active under appropriate conditions through the remodeling of the chromatin into more open arrangements. There are many genes that are not packaged in heterochromatin but are nevertheless transcriptionally inactive.

Question 9

Compared to the human genome, the genome of yeast typically has …

Accepted Answer

The yeast has a concise genome with a much higher ratio of coding to noncoding DNA.

Question 10

Complete the DNA sequence below such that the final sequence is identical to that of the complementary strand. Your answer would be a seven-letter string composed of letters A, C, T, and G only, e.g. TTCTCAG.

Accepted Answer

The final sequence should be 5?-CTTTAGATCTAAAG-3?. The complementary strand would then have the exact same sequence. This is an example of a palindromic sequence.

Question 11

In a double-stranded DNA molecule, one of the chains has the sequence CCCATTCTA when read from the 5? to the 3? end. Indicate true (T) and false (F) statements below regarding this chain. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT.
( )	The other chain is heavier, i.e. it has a greater mass.
( )	There are no C residues in the other chain.
( )	The 5?-terminal residue of the other chain is G.
( ) The other chain is pyrimidine-rich.

Accepted Answer

The sequence of the other chain is 5?-TAGAATGGG-3?, which makes it slightly heavier because it is mainly composed of the bulkier purine bases.

Question 12

Indicate whether each of the following descriptions better applies to a centromere (C), a telomere (T), or an origin of replication (O). Your answer would be a seven-letter string composed of letters C, T, and O only, e.g. TTTCCTO.
( )	It contains repeated sequences at the ends of the chromosomes.
( )	It is NOT generally longer in higher organisms compared to yeast.
( )	Each eukaryotic chromosome has many such sequences.
( )	There are normally two such sequences in each eukaryotic chromosomal DNA molecule.
( )	There is normally one such sequence per eukaryotic chromosomal DNA molecule.
( )	It is where DNA duplication starts in S phase.
( )	It attaches the chromosome to the mitotic spindle via the kinetochore structure.

Accepted Answer

There is normally one centromere per chromosomal DNA molecule, and two per mitotic chromosome. The mitotic kinetochore structure forms at the centromere. In contrast, the telomeres are at the two ends of each linear chromosome. There are usually many replication origins per eukaryotic chromosome. Replication origins and centromeres are both generally much longer in higher eukaryotes compared to yeast.

Question 13

Which of the following correlates the best with biological complexity in eukaryotes?

Accepted Answer

Biological complexity correlates better with the number of genes than it does with genome size, number of chromosomes, or number of genes per chromosome.

Question 14

For the Human Genome Project, cloning of large segments of our genome was first made possible by the development of yeast artificial chromosomes, which are capable of propagating in the yeast Saccharomyces cerevisiae just like any of the organism's 16 natural chromosomes. In addition to the cloned human DNA, these artificial vectors were made to contain three elements that are necessary for them to function as a chromosome. What are these elements? Write down the names of the elements in alphabetical order, and separate them with commas, e.g. gene, histone, nucleosome.

Accepted Answer

These three elements are required in a functional chromosome.

Question 15

The position effect variegation (PEV) phenotype described in this chapter can be used to identify new genes that regulate heterochromatin formation. For instance, strains of Drosophila melanogaster with the White variegation phenotype have been subjected to mutagenesis to screen for dominant mutations (in other genes) that either enhance or suppress PEV, meaning the mutations result in either lower or higher red pigment production, respectively. Which of the following mutations is expected to be an enhancer of variegation?

Accepted Answer

An enhancer of variegation would facilitate the spread of heterochromatin into the nearby White gene, and consequently decrease red pigmentation. For example, the production of a hyperactive methyl transferase enzyme would result in elevated levels of H3 lysine 9 methylation, leading to enhanced heterochromatin expansion. The other described mutations bring about suppression of variegation.

Question 16

Indicate which feature (1 to 4) in the schematic drawing below of a chromatin fiber corresponds to each of the following. Your answer would be a four-digit number composed of digits 1 to 4 only, e.g. 2431.
  
( )	Nucleosome core particle
( )	Linker DNA
( )	Histone octamer
( )	Non-histone protein

Accepted Answer

The nucleosome core particle is composed of a histone octamer wrapped by 147 nucleotide pairs of DNA, and is connected to its neighbors via a linker DNA of variable length. Non-histone proteins are also abundant in the chromatin.

Question 17

The eukaryotic chromosomes are organized inside the nucleus with a huge compaction ratio of several-thousand-fold. What is responsible for such a tight packaging?

Accepted Answer

The various chromatin proteins, including histone and non-histone proteins, wrap and fold the DNA to achieve an astonishing compaction ratio. The nuclear envelope and the nuclear matrix are dispensable for this effect, as evident by the high compaction seen in mitotic chromosomes when the nucleus is disassembled.

Question 18

Indicate which numbered feature (1 to 5) in the schematic drawing below of the DNA double helix corresponds to each of the following. Your answer would be a five-digit number composed of digits 1 to 5 only, e.g. 52431.
  
( )	Hydrogen-bonding
( )	Covalent linkage
( )	Phosphate group
( )	Nitrogen-containing base
( )	Deoxyribose sugar

Accepted Answer

In the double-stranded DNA, the sugar-phosphate backbones form two covalently continuous chains, while the nitrogenous bases from one chain are hydrogen-bonded to those of the other chain according to the Watson-Crick model.

Question 19

A DNA nucleotide pair has an average mass of approximately 660 daltons. Knowing the number of nucleotides in the human genome, how many picograms of DNA are there in a diploid human nucleus? Avogadro's number is 6 × 10²³. Write down the picogram amount without decimals (round the number to the closest integer), e.g. 23 pg.

Accepted Answer

There are about 6.4 × 10? nucleotide pairs (np) in a diploid nucleus. The average mass of 660 daltons is equivalent to 660 grams per mole (6 × 10²³) of the nucleotide pairs. Thus, the total mass of nuclear DNA is:
(6.4 × 10? np) × (660 g/mole) / (6 × 10²³ np/mole) = ~7 × 10-¹² g = ~7 pg

Question 20

Which of the following features of DNA underlies its simple replication procedure?

Accepted Answer

In principle, replication would have been conceptually as simple without any of the other features.

Quiz 4: DNA, Chromosomes, and Genomes