Name______
Period ______
STAAR Biology EOC Practice Test
1. Prokaryotes are considered by the scientific community as the most ancient life-forms on Earth. Yet, these primitive cells share many common characteristics with the more modern eukaryotes. However, one significant difference between these two cell types is that only eukaryotes contain
A membrane-bound compartments to carry out specialized functions
B a selectively-permeable cell membrane to maintain homeostasis
C deoxyribonucleic acid to serve as a template to produce proteins
D a rigid cell wall which provides structure and support
Use the information below to answer the following two questions.
Identical lengths of potato cores were cut and massed. One potato core was then placed in 100 ml of each of the following solutions; 0M sucrose, 0.2M sucrose, 0.4M sucrose, 0.6M sucrose, 0.8M sucrose, and 1M sucrose. These were left to equilibrate overnight after which the potato cores were then removed, patted dry, and massed again. Data is recorded in the table below:
Solution / Initial Mass / Final Mass0M sucrose / 2.2 g / 2.4 g
0.2M sucrose / 2.4 g / 2.4 g
0.4M sucrose / 1.9 g / 1.8 g
0.6M sucrose / 2.0 g / 1.8 g
0.8M sucrose / 2.2 g / 1.8 g
1M sucrose / 2.3 g / 1.8 g
2 What is the most likely cause of the potato core gaining mass when placed in the 0M sucrose solution?
A Sucrose moved into the potato core from the solution.
B Sucrose moved from a region of higher concentration to a region of lower concentration.
C Water moved down its concentration gradient.
D Water concentrations stayed the same inside and outside the potato core.
3. The isotonic point in this potato core experiment is closest to
A 0M sucrose because sucrose moved into the potato core causing it to gain mass
B 0.2M sucrose because water didn’t move into or out of the potato core
C 0.2M sucrose because equal amounts of water moved into and out of the potato core
D 1M sucrose because the potato core lost the most mass due to water loss
4 Viruses are considered non-living because they lack the machinery to reproduce on their own and must rely on a cellular host cell to survive. Structures common to both cells and viruses include
A protein capsid
B nucleic acids
C cell wall
D nucleus
5Differences between prokaryotes and eukaryotes which are readily observable through the light microscope include
- prokaryotes are much smaller than eukaryotes
- ribosomes are smaller in prokaryotes
- prokaryotes lack nuclei
AI only
BII only
C I and II only
D I and III only
6Cells often need to take in materials from their environment, many of which are found in lower concentrations outside the cell compared to inside the cell. In order to do this, cells must use the energy from ATP to move these materials
A against the concentration gradient through active transport
B with the concentration gradient through active transport
C against the concentration gradient through passive transport
D with the concentration gradient through passive transport
7 The cellular organelle most directly involved in maintaining homeostasis is the
A plasma membrane
B cell wall
C chloroplast
D mitochondria
8 Viruses are able to infect cells because they share a common genetic code and are able to use host cell enzymes to carry out protein synthesis. There are some structures which are unique to viruses and not found in cells. These include
I. Protein capsid
II. Nucleic acids
III. Cell wall
A I only
B II only
C I and II only
D II and III only
9 During an infection, some viruses remain inside cell but do not cause symptoms of disease. The graphic above depicts the
A lytic cycle and the host cell is destroyed by rapidly reproducing viral particles
B lytic cycle and the viral nucleic acid inserts into the host cell chromosome
C lysogenic cycle and the viral nucleic acid inserts into the host cell chromosome
D lysogenic cycle and the viral nucleic acid replicates independently of the host cell chromosome
10 The stage of the cell cycle pictured above may be best described as
A Cytokinesis: the cytoplasm of the cell is divided
B S phase: DNA is copied, resulting in identical sister chromatids
C G2 phase: the cell grows in size and prepares to divide the nucleus
D Mitotic phase: the nucleus is divided
11 Leaves are plant organs specialized for photosynthesis. Through the light microscope, which organelle would be expected to be seen in greater numbers?
A B C D
12Each of the following statements describes the importance of the cell cycle and Mitosis to an organism EXCEPT
A.The cell cycle and Mitosis are important to produce identical new cells so that a multi-cellular eukaryote can grow larger.
B The cell cycle and Mitosis are important to produce identical new cells so that a unicellular eukaryote can reproduce.
C The cell cycle and Mitosis are important to produce identical new cells so that a multi-cellular eukaryote can repair damaged tissue.
D The cell cycle and Mitosis are important to produce identical new cells so that a multi-cellular eukaryote can reproduce.
13According to the illustration, adult stem cells extracted from bone marrow tissue may be cultured to produce different types of differentiated cells, such as liver cells, nerve cells, and heart muscle cells. Which of the following statements can best explain this result?
A There are different genes in each of the individual stem cells that are cultured.
B Environmental conditions influence which genes are expressed.
C Each stem cell contains only the DNA required to make one cell type.
D DNA can move from one stem cell to another through the culture media.
14 When harmful mutations occur in genes that contain the information to make cell cycle regulatory proteins, this may result in accelerated cell division. The result of these mutations is most likely
A cancer
B slower cell growth
C a halt in mitosis
D entry into G0 phase
15 Which of the following illustrations depicts a nucleotide, the building block of DNA?
16Which statement most accurately describes how deoxyribonucleic acid (DNA) carries the genetic code for building an organism?
A The order of nitrogen bases provides the genetic code
B The genetic code is based on the number of hydrogen bonds found between nitrogen bases
C The number of deoxyribose sugars along the sides of the DNA ladder is used to determine the genetic code
D The shape of the DNA molecule provides the genetic code
17Which of the following structures would pair with the nitrogen base Adenine in a complementary strand of DNA?
18Restriction enzymes isolated from various bacterial species have been extensively used in the biotechnology laboratory for cutting human genes from DNA samples to insert into bacterial plasmids. This is made possible because
A Nitrogen bases form a common genetic code for all organisms.
B DNA is maintained in both bacterial and human cells within nuclear membranes.
C Small, circular pieces of DNA called plasmids are common to both humans and bacteria.
D Restriction enzymes cut at random sequences so it does not matter that humans and bacteria use a different genetic code.
19The model in the graphic above illustrates translation, the latter part of the process of protein synthesis. Which other necessary components for translation are absent from the graphic?
A messenger RNA and transfer RNA
B ER membranes and ribosomes
C ribosomal RNA and DNA
D transfer RNA and ribosome
20 Which of the following statements is NOT correct regarding the process in the illustration above?
A I must be complementary to an mRNA codon.
B This process occurs in the nucleus of a cell.
C III shows the first two amino acids of a growing polypeptide chain.
D The process shown is translation.
21 Bacterial species’ genomes are arranged somewhat differently than those of eukaryotes. Their circular chromosome contains clusters of genes which code for proteins that work together to accomplish a series of related tasks. In this way, groups of genes can be turned on when the gene product(s) are needed, and turned off when they aren’t needed. This confirms the fact that
A Gene expression is a regulated process.
B Genes are arranged similarly in prokaryotes and eukaryotes.
C Eukaryotes have circular chromosomes.
D Bacterial genomes are much larger than those of eukaryotes.
Use the illustration that follows to answer the next two questions:
22 Changes in DNA sequences are called mutations. Mutations may affect only one nitrogen base in the sequence, or may affect large regions of a chromosome. The particular type of gene mutation that is illustrated above is called a
A point mutation
B insertion mutation
C frameshift mutation
D nonsense mutation
23 The significance of this gene mutation is that
A A different amino acid sequence results which is likely to affect the functioning of the final protein product.
B A different amino acid sequence results which is not likely to affect the functioning of the final protein product.
C No protein product will be produced.
D The protein product is unaffected.
24 Maize is a variety of corn that is often used to illustrate genetic crosses. Since each kernel of corn on a cob is an offspring, much data can be obtained from observing a single corn cob. Kernels are found as one of two colors; either purple (P) which is dominant, or yellow (p) which is recessive. If two heterozygous purple plants were crossed, what would be the predicted phenotype ratio of the resulting offspring?
A 1PP:2Pp:1pp
B all PP
C 3 purple:1 yellow
D all purple
......
….. ….
SPOTTED SEED DOTTED SEED
25Lentil seeds can be found in two varieties based on the different sizes of spots which may be observed on their outer surface. Seeds with larger spots are called spotted, while seeds with smaller spots are called dotted. The different-sized spots are inherited in a co-dominant pattern. Predict the resulting phenotype of offspring if a lentil with spotted seeds were crossed with a lentil that has dotted seeds.
A Seeds are both spotted and dotted.
B Seeds have intermediate size spots.
C Seeds have no spots.
D Seeds are all spotted.
26In pea plants, tall (T) is dominant over short (t), and purple flowers (P) are dominant to white flowers (p). Given two parent plants which are heterozygous for both of these traits, what is the probability of producing tall offspring with white flowers?
A 9/16
B 3/16
C 1/16
D 0
27Variation is the raw material for change in organisms, providing a mechanism by which populations may evolve and adapt to changing environmental conditions. Meiosis provides a number of opportunities for increased variation in sexually reproducing populations through each of the following events EXCEPT
A crossing over
B independent assortment of alleles
C random distribution of homologous chromosomes
D fertilization
28Gel electrophoresis and DNA fingerprinting is NOT a useful method for confirming
A the order of nitrogen bases in DNA
B family relationships such as paternity
C a person’s identity
D genes that cause genetic disorders
29 The process of meiosis occurs in sexually reproducing eukaryotes and is important for the continuation of these species because of all of the following EXCEPT
A Chromosome number is halved in the daughter cells.
B Crossing over during Prophase I provides increased variation.
C Diploid chromosome number is maintained in the zygote.
D Gametes produced are 2n.
30 A baby is born with phenotypic characteristics of Down’s syndrome; a rounded face, small chin, almond-shaped eyes, and shorter limbs. Evidence from which of these techniques would best confirm the child has an extra chromosome 21?
A gel electrophoresis
B DNA fingerprinting
C karyotype analysis
D gene therapy
31 The complete amino acid sequence was determined of a protein shared by all primate groups. The data provided in the graph above shows the percentage of amino acids each primate protein shared with the human protein. According to this molecular homology data,
A The gorilla is least related to humans.
B The gorilla and chimpanzee are most closely related.
C The chimpanzee is most closely related to humans.
D The chimpanzee is least related to the gibbon.
32 The rock strata were excavated in an area as illustrated above and the fossil remains examined. According to this information, the most likely scenario is that
A Layer I is the oldest, followed by Layer II, III, and IV.
B Fish suddenly appeared on Earth at the time of Layer III.
C A marine environment gradually filled with sediment and became a terrestrial environment.
D A terrestrial environment was covered by a shallow sea for a period of time.
33 How does competition for limited resources within and between species drive evolutionary change?
A Individuals with favorable adaptations will out-compete others and produce more offspring.
B Competition makes some individuals stronger than others, and more of the stronger individuals survive.
C Stronger individuals develop different adaptations than weaker individuals.
D Competition produces variation between populations.
34 Each of the finch varieties pictured evolved from a common ancestral finch species. This diversity of species likely developed due to
A individuals evolving different traits during their lifetime living longer than others
B finches being born with beneficial variations reproducing more successfully
C immigration of finch species providing a greater variety of traits in the population
D selective predation of one particular variety of finch
35 Cheetahs are the fastest land mammals, yet can only maintain their top speed of about 70 mph for a few seconds. Since all cheetahs are nearly genetically identical due to a past bottleneck event, development of future diversity in this species is most likely to occur by
A gene flow
B immigration
C emigration
D mutation
36 The importance of a uniform taxonomic system to the scientific community includes all of the following EXCEPT
A Binomial nomenclature provides common language.
B Species names are descriptive words.
C Ambiguity of common names is introduced.
D Two or more different kinds of organisms may not have the same name
37Which is the best description of how the fossil record provides evidence of common ancestry?
A Analogous structures can be identified between related groups of organisms.
B Homologous structures can be identified between related groups of organisms.
C Molecular homologies can be identified through trace remains.
D Developmental homologies can be identified through fossilized soft tissues
38Duck feet are especially well-adapted for life in the water. Sharp claws allow them to better navigate the wet, slippery environment and webbed toes aid in their swimming ability. These adaptations probably came about as a result of
A one duck developing these favorable traits over its lifetime and passing them to its offspring
B variation in the duck population; ducks born with one or more of these traits were more reproductively successful
C nature selecting against these adaptations
D selective breeding of ducks with these favorable traits over many generations
39The finch beak adaptations in the graphic most likely evolved through the mechanism of
A stabilizing selection
B adaptive radiation
C genetic drift
D non-random mating
40 Species become more diverse to
I.Occupy available niches
II.Reduce competition
III.Adapt to changing environmental conditions
A I and II only
B I and III only
C II and III only
D I, II, and III
Characteristics of speciesUnicellular
Nucleated cells
Microscopic
Motile
Inhabits aquatic environments
41Given the characteristics described in the information box above, this species belongs in kingdom
A Archaebacteria
B Fungi
C Eubacteria
D Protista
42The organism pictured above, Escherichia coli, is a common symbiont in mammalian digestive systems. This organism has a circular chromosome located in a region of the cell called the nucleoid (A), a rigid outer cell wall (B), and a semi-permeable cell membrane (C). To which kingdom does this organism belong?
A Archaebacteria
B Eubacteria
C Protista
D Fungi
43 Fungi are unique among eukaryotes in that they are
A unicellular
B multicellular
C autotrophic
D decomposers
Use the information in the illustration and table below to answer the next question.
Amphibia(amphibians) / Cold- blooded / Smooth skin / Gills in young, adults respire through lungs, skin
Aves
(birds) / Warm-blooded / Body covered in feathers / lungs
Mammalia
(mammals) / Warm-blooded / Body covered in hair / lungs
Osteichthyes
(bony fish) / Cold-blooded / Body covered in scales / gills
Chondrichthyes
(cartilaginous fish) / Cold-blooded / Body covered in scales / gills
Reptilia
(reptiles) / Cold-blooded / Body covered in scales / lungs
44 The animal species in the illustration has the following characteristics; it breathes with lungs, maintains a constant body temperature, has hair, gives live birth, and nurses young with milk. These characteristics place the animal in the class
A Amphibia
B Aves
C Mammalia
D Osteichthyes
45 The simple sugar glucose, C6H12O6 , is formed through the process of photosynthesis as depicted in the chemical equation above. Simple sugars are unique among bio-molecule monomers in that
A They are used by the body for long-term energy storage.
B They do not dissolve readily in water.
C Carbon, hydrogen, and oxygen atoms are always found in a 1:2:1 ratio.
D They are the building blocks of proteins.
46 When comparing the reactants and products of photosynthesis and cellular respiration, which statement is true?
A Carbon dioxide is a product of photosynthesis.
B Oxygen is a product of cellular respiration.
C Solar energy is converted to chemical energy in the process of photosynthesis.
D Energy is stored in the bonds of glucose in the process of cellular respiration.
47When compared to the process of photosynthesis, cellular respiration produces more
- ATP
- CO2
- O2
A I only