PUBH 1145 Lecture Notes - Lecture 5: Zygosity, Karyotype, Meiosis

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12 Sep 2020

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Dna : double stranded helix of genetic material: adenine-thymine, guanine-cytosine. Genes : differing lengths and sequences of dna. Humans have 46 chromosomes (23 pairs: 44 autosomes, 2 sex chromosomes. Karyotype is the complete genetic makeup of a cell: able to be examined under the microscope for abnormalities. Sex cells (egg and sperm) only have 23, and are called haploid: as a result of meiosis, when fertilization occurs, new cell will have full 46 chromosomes. Each paired chromosome pair has a variety of the specific gene called an allele. Genes occupy a specific locus on a chromosome. Differences in alleles results in how that trait will manifest. Homozygous: both alleles are the same. Dominant alleles will always express when inherited: whether homozygous or heterozygous. Recessive alleles will only express when the individual is homozygous. Co-dominant alleles will both express to varying degrees when inherited. Other genes, environmental factors, gender can influence penetrance of expression.

Related Questions

___1. Nationally, the most common cause of Down Syndrome is
a. nondisjunction of Chromosome 21 during meiosis
b. translocation of Chromosome 21 onto another chromosome
c. inversion of a large section of Chromosome 21
d. deletion of the Down Syndrome gene
___2. In humans, sex is determined by
a. The presence or absence of the Y chromosome
b. The ratio of X chromosomes to sets of autosomes
c. The temperature during fetal development
d. Which parent is stronger at the time of conception
___3. A translocation is
a. a chromosomal abnormality in which a piece of one chromosome breaks off and attaches to a different chromosome
b. an abnormality in which the chromosomes are translocated out of the nucleus
c. a condition in which the chromosome has broken in two places and come back together with the middle segment backward
d. an abnormality in which there is an extra copy of one chromosome
___ 4. In snapdragons, a red-flowered plant crossed with a white-flowered plant produces all pink-flowered offspring. But crossing two pink-flowered plants produces red, pink, and white in a 1:2:1 ratio. This is a good example of
a. Incomplete dominance
b. Dominance of the red allele over the white allele
c. Lethal effects of the pink allele
d. Sex-linkage of the flower-color alleles
___ 5. If a person with type AB blood marries a person with Type O blood, what blood types are possible in their children? (Assume this trait is a single-gene trait, not complicated by interaction with other genes).
a. A, B, AB, or O
b. A or B only
c. AB or O only
d. AB only 2
___6. In what phase of meiosis are haploid nuclei first formed? a. Anaphase I b. Metaphase II c. Telophase II d. Metaphase I e. Telophase I
___7. What is the phenotype of a heterozygous tall pea plant? a. tall b. Tt c. TT d. tt e. dominant
___ 8. If genes A and B are far apart on a chromosome, and A and C are close together, which of the following observations is expected?
a. Crossing over will occur more often between A and C than between A and B
b. Crossing over will occur at equal frequencies between A and C, and between A and B
c. Crossing over will occur more often between A and B than between A and C
___9. Sometimes an organismâs genotype for one gene masks the effect of a different gene. This phenomenon is called
a. Dominance
b. Incomplete dominance
c. Co-dominance
d. Epistasis
e. Variable expressivity
___10. The dominant allele of a gene is
a. The one that is expressed most often in a population
b. The one that is expressed most often in any particular cross
c. The one that is expressed in an individual that has two different alleles
d. The one that is most healthy to have, that gives the greatest fitness
___11. In human cells, DNA replication occurs
a. In S phase of interphase
b. In G1 phase of interphase
c. In G2 phase of interphase
d. In prophase of mitosis
e. Whenever the cell needs to read the DNA to make proteins
___ 12. A process in which recipient cells take up genes from DNA molecules floating free in the surrounding medium is called
a. conjugation
b. specialized transduction
c. generalized transduction
d. transformation 3
___ 13. Most calico cats are female. However, a few rare males are observed. What is the genotype of a male calico cat?
a. X BX 0
b. X OY
c. X BY
d. X BYY
e. X BX OY
___ 14. A type of red-green color-blindness is a sex-linked (X-linked) recessive trait in humans. Can a woman have this type of color-blindness?
a. Yes, a woman could be color-blind if her mother and father were both heterozygous.
b. Yes, a woman could be color-blind but only if she has an abnormal number of sex chromosomes.
c. No, women can't show sex-linked traits because they lack a Y chromosome.
d. No, women can't show sex-linked traits because they have two X chromosomes.
e. Yes, a woman could be color-blind if her father was color-blind and her mother was a carrier (heterozygous).

In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that

	DNA is a double helix.
	DNA may be taken up by bacterial cells and be active.
	DNA is present in bacteriophage T2.
	eukaryotic cells may be genetically transformed.
	DNA binds to the cell wall of the bacterium.

In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.

	temperatures high enough to kill cells
	enzymes that hydrolyze proteins
	enzymes that hydrolyze DNA
	enzymes that hydrolyze polysaccharides
	enzymes that hydrolyze RNA

Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?

	Pure RNA from S cells
	Cell-membrane extracts from S cells
	A mixture of pure RNA and protein from S cells
	Cell-free extracts from S cells
	Pure protein from S cells

A-T base pairs in a DNA double helix

	form three hydrogen bonds with each other.
	are more heat-stable than G-C base pairs.
	are of a substantially different length than G-C base pairs.
	are not accessible to DNA binding proteins.
	are chemically distinct from G-C base pairs.

If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?

	27
	50
	0
	73
	23

The uniform diameter of the DNA structure provides evidence for

	antiparallel DNA strands.
	a right-handed helix.
	a double-stranded helix.
	3â² end phosphorylation.
	purineâpyrimidine pairing.

If a sequence of one strand of DNA is 5â²-TGACTATC-3â², what is the complementary strand?

	5â²-ACTGATAG-3â²
	3â²-TGACTATC-5â²
	3â²-ACTGATAG-5â²
	3â²-ACTGATAG-3â²
	5â²-TGACTATC-3â²

What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?

	3â² end phosphorylation
	Purineâpyrimidine pairing
	Antiparallel DNA strands
	Covalent bonds between sugars and phosphate groups
	Hydrogen bonding between paired bases

If the MeselsonâStahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only ¹⁵N and the second only ¹⁴N, this result would have indicated that replication was

	unidirectional.
	dispersive.
	conservative.
	bidirectional.
	semiconservative.

10.

The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3â² hydroxyl group (âOH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.

	The replication complex will no longer bind to origins of replication.
	The DNA polymerase will only incorporate acyclovir molecules.
	Acyclovir will bind single-strand binding proteins.
	DNA polymerase will not be able to link a successive nucleotide.
	Acyclovir will block the activity of the DNA helicase.

11.

Which of the following does not demonstrate the stability of the DNA double helix?

	Single-strand binding proteins are required to keep the strands apart.
	Paired bases form hydrogen bonds.
	High temperatures are required to denature it.
	DNA synthesis requires energy.
	DNA helicases require ATP.

12.

What effect would a primase inhibitor have on DNA replication?

	DNA polymerase would not be able to add bases to the DNA.
	Primase would be blocked from joining the DNA replication complex.
	Primase would not be able to provide primers for DNA polymerases.
	DNA polymerase would not be able to use DNA as a template.
	There would be no effect on DNA replication.

13.

To replicate their DNA in a timely manner, most eukaryotic chromosomes

	normally have uncoiled DNA.
	have multiple origins of replication.
	contain linear molecules of single-stranded DNA.
	have two replication forks that move in the same direction.
	are composed entirely of DNA.

14.

Which statement about DNA replication is false?

	Okazaki fragments are synthesized as parts of the leading strand.
	Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
	The sliding clamp increases the rate of DNA synthesis.
	Replication forks represent areas of active DNA synthesis on the chromosomes.
	Ligases and polymerases function in the vicinity of replication forks.

15.

In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.

	leading; telomerase
	lagging; telomerase
	lagging; DNA polymerase I
	leading; DNA polymerase I
	leading; replicase

16.

A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?

	DNA polymerase is constantly active.
	Primase is able to extend the telomeres.
	The mismatch repair system is extra efficient.
	The telomerase gene is being expressed.
	The cells do not need the ends of their chromosomes.

17.

If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?

	The replication complex excises the incorrect nucleotide.
	The excision repair proteins remove the incorrect nucleotide.
	The mismatch repair system scans for base-pairing mismatches.
	There is no correction, because nucleotide errors by DNA polymerases are not repaired.
	DNA ligase connects the correct nucleotide.

18.

Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.

	1, 2, 3, 4
	2, 1, 3, 4
	3, 4, 2, 1
	2, 4, 1, 3
	4, 2, 3, 1

19.

Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.

	64
	32
	12
	128
	6

20.

When double-stranded DNA is heated to temperatures above 90Â°C, it denatures. Denaturation is a process that

	breaks covalent bonds.
	hydrolyzes DNA.
	separates double-stranded DNA into two single strands.
	causes new hydrogen bonds to form.
	irreversibly destroys the double helical structure.

Messenger RNA (mRNA)...

a) Is a structural part of a ribosome

b) Brings amino acids to DNA

c) Carries genetic information from the nucleus to the cytoplasm

d) Is a structural part of a chromosome

e) Initiates and directs chemical reactions

Mr. Potato is homozygous dominant (KK), his son, Potato Jr. is heterozygous (Kk). Based on this information, what are the possible genotypes for Mrs. Potato? (Hint: use a Punnet Square).

a) Her genotype is definitely kk

b) Her genotype is either Kk or kk

c) Her genotype is either KK or Kk

d) Her genotype is either KK, or Kk, or kk

e) Mr. Potato cannot be Potato Jrâs biological father

12. Homologous chromosomes:

a) Are members of different chromosome pairs

b) Are genetically identical

c) Carry genetic information that influences the same traits (same genes but probably different alleles)

d) Are only inherited from the mother

e) Are only inherited from the father.

âFitnessâ in an evolutionary sense, refers to an individualâs:

a) Strength

b) Reproductive success

c) Aggressiveness

d) Size

e) Age at death

The fact that individuals who possess favorable traits are more likely to survive and reproduce than those who possess less favorable traits is the basis for the theory of...

a) uniformitarianism.

b) natural selection.

c) the inheritance of acquired characteristics.

d) catastrophism.

e) the fixity of species.

Even with strong selection, genetic variation of a population can be maintained because:

a) Selection only affects one gene at a time.

b) The effects of many alleles are hidden from selection by the effects of other alleles

c) Mutation constantly adds new alleles at an extremely fast rate.

d) Populations always go extinct before their gene frequencies are dramatically altered.

e) Only a and c

Gene Flow is defined as the:
a) Production of new alleles
b) New musical group in which Adam Levine will be the leading vocalist
c) Exchange of genes between populations
d) differential reproductive success of individuals
e) Process by which a DNA base gets extracted out of the nucleus to initiate the process of protein synthesis.

The basic unit of DNA is a molecule called a: a) Nucleotide

b) Base
c) Thymine
d) Prokaryoticcell e) mRNA

How many bases code for one amino acid? a) one

b) two
c) it depends on what amino acid it is d) three
e) four

Hemophilia is a condition that affects: a) more females than males
b) more males than females
c) females and males equally

d) only children born prematurely
e) individuals with a missing portion of chromosome number 5

A clear example of directional selection is:
a) Darwin finches with deeper and thicker beaks associated to a severe drought b) Sickle cell anemia in regions in which Malaria is endemic
c) Achondroplasticdwarfism
d) Fly halteres becoming fully developed wings after modification of a hox gene e) all of the above

Meiosis is evolutionarily significant because:
a) New combination of alleles can arise through independent assortment and crossing over

22.

23.

24.

25.

26.

b) It is the mechanism by which body tissues are repaired
c) It creates somatic cells that are crucial for our body organs to maintain us alive d) It allows forensics to determine the cause of death of unidentified people
e) none of the above

Homologous chromosomes:

a) Are exact copies of each other and contain exactly the same genetic material
b) Are members of different chromosome pairs
c) Are only inherited from the father
d) carry genetic information that influences the same traits (the same genes), however, each of the homologous chromosomes can contain different versions of the same gene (different alleles).

e) carry genetic information that influences different traits because each of the homologous chromosomes contains completely different genes at very different loci

How many chromosome pairs occur in a normal human somatic cell?

a) 46 b) 23 c) 22 d) 24 e) 45

What is the scientific method? Use bullet points to describe the 4 major components.

How is a hypothesis different from a theory?

List the four major components (described during class) necessary for natural selection to occur [AKA Darwinâs Four Postulates] (PDF1-3).

27. Who was Thomas Malthus? In what way(s) Malthus was influential for Darwinâs Theory of Natural Selection? Hint: populations vs resources.

29. What amino acids are being coded for by the gene segment provided below? Nucleotide triplets and their corresponding amino acids are:
ACA= Cysteine; AAA=Phenylalanine; TGG=Threonine; CCT=Glycine; GAT=Leucine; ATA=Tyrosine; AGG=Serine; TCT=Arginine

a) Threonine, serine, glycine, tyrosine, leucine, phenylalanine, arginine, cysteine.

b) Tyrosine, leucine, glycine, threonine, arginine, threonine, serine, cysteine.

c) Arginine, cysteine, serine, glycine, threonine, leucine, phenylalanine, tyrosine.

d) Phenylalanine, leucine, tyrosine, glycine, serine, threonine, cysteine, arginine.

e) Phenylalanine, leucine, tyrosine, glycine, threonine, phenylalanine, arginine, serine.

PUBH 1145 Lecture Notes - Lecture 5: Zygosity, Karyotype, Meiosis

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