Praxis 5236
Biology: Content Knowledge
Maryland Passing Score
155
MSDE Score Code
7403
Retake Wait
28 days
Score Valid
10 years
10 Free Practice Questions
Question 1 · Molecular and Cellular Biology
During DNA replication in eukaryotes, which enzyme is responsible for synthesizing short RNA primers that are necessary to initiate new strand synthesis?
- A
- B
- C
- D
Explanation
Primase is the enzyme responsible for synthesizing short RNA primers during DNA replication. These primers provide a free 3'-OH group that DNA polymerase requires to begin adding nucleotides. DNA polymerase I (choice A) in prokaryotes removes RNA primers and replaces them with DNA, but does not synthesize the primers. Helicase (choice C) unwinds the double helix by breaking hydrogen bonds between base pairs, but does not synthesize primers. Ligase (choice D) joins Okazaki fragments on the lagging strand by forming phosphodiester bonds, but plays no role in primer synthesis.
Question 2 · Diversity of Life
A biologist examines an organism and notes the following characteristics: prokaryotic cell organization, cell walls composed of peptidoglycan, and the ability to perform nitrogen fixation in root nodules of legumes. To which domain does this organism belong?
- A
- B
- C
- D
Explanation
The correct answer is B. Bacteria are prokaryotic organisms whose cell walls contain peptidoglycan, distinguishing them from Archaea, which lack peptidoglycan. Nitrogen-fixing bacteria such as Rhizobium form mutualistic associations with legume root nodules, a hallmark characteristic of certain bacterial lineages. A is incorrect because Archaea, while prokaryotic, have cell walls made of pseudopeptidoglycan or other polymers—never true peptidoglycan—and do not form nitrogen-fixing root nodule symbioses with plants. C is incorrect because Eukarya are eukaryotic organisms with membrane-bound nuclei and do not have peptidoglycan cell walls. D is incorrect because Protista is a kingdom within domain Eukarya, so it cannot apply to a prokaryotic organism with peptidoglycan cell walls.
Question 3 · Ecology
Which of the following BEST describes the competitive exclusion principle as formulated by G.F. Gause?
- A
- B
- C
- D
Explanation
Gause's competitive exclusion principle states that two species occupying the exact same ecological niche — competing for the same resources in the same way — cannot stably coexist; the superior competitor will eventually drive the other to local extinction or cause it to shift its realized niche. This was demonstrated experimentally with Paramecium species. Option B is incorrect because the principle is not a quantitative relationship between species richness and competition intensity; it is a binary principle about niche overlap and coexistence. Option C is incorrect because the principle involves interspecific competition between similar species, not predator-prey dynamics; predators removing prey is predation, not competitive exclusion. Option D is incorrect because the principle addresses coexistence of species already sharing a habitat, not immigration patterns from surrounding habitats, which relates more to island biogeography theory.
Question 4 · Ecology
Which of the following best describes the role of nitrogen-fixing bacteria such as Rhizobium in the nitrogen cycle?
- A
- B
- C
- D
Explanation
The correct answer is B. Nitrogen fixation is the conversion of atmospheric dinitrogen gas (N₂) — which is largely unavailable to most organisms — into ammonia (NH₃) or ammonium (NH₄⁺) by prokaryotes such as Rhizobium (which lives symbiotically in legume root nodules) and free-living bacteria like Azotobacter. This process makes nitrogen bioavailable and is the primary entry point of nitrogen into most ecosystems. Option A describes nitrification, which is carried out by a different group of bacteria (e.g., Nitrosomonas and Nitrobacter) and involves the oxidation of ammonia to nitrite and then to nitrate — not the function of Rhizobium. Option C describes ammonification (mineralization), the decomposition of organic nitrogen by decomposers (fungi and bacteria), returning nitrogen to inorganic form — again, not the role of Rhizobium. Option D describes denitrification, performed by anaerobic bacteria (e.g., Pseudomonas denitrificans), which reduces nitrate back to N₂ gas and completes the cycle — this is the opposite of nitrogen fixation.
Question 5 · Classical Genetics and Evolution
In a diploid organism, a gene locus has two alleles: T (dominant) and t (recessive). A cross between two heterozygous individuals (Tt × Tt) is performed. What is the probability that any given offspring will be homozygous recessive (tt)?
- A
- B
- C
- D
Explanation
When two heterozygous individuals (Tt × Tt) are crossed, the Punnett square yields four equally probable genotypic outcomes: TT, Tt, Tt, and tt. Only one of these four outcomes is homozygous recessive (tt), giving a probability of 1/4 (25%). Option B (1/2) represents the probability of being heterozygous (Tt). Option C (3/4) represents the probability of displaying the dominant phenotype (TT + Tt combined). Option D (1/3) is incorrect and would only be relevant if one considered only the three genotypic classes (TT, Tt, tt) without accounting for their unequal frequencies.
Question 6 · Molecular and Cellular Biology
The sodium-potassium ATPase pump maintains a resting membrane potential in neurons by actively transporting ions across the plasma membrane. What is the stoichiometry of ion movement per ATP molecule hydrolyzed?
- A
- B
- C
- D
Explanation
The sodium-potassium ATPase (Na⁺/K⁺-ATPase) pumps three sodium ions out of the cell and two potassium ions into the cell for each ATP molecule hydrolyzed. This unequal exchange contributes to the net negative charge inside the cell, helping establish the resting membrane potential. A is incorrect because it reverses the numbers — it is Na⁺ that is exported in greater quantity (3), not K⁺. C is incorrect because a 1:1 ratio would not accurately reflect the actual mechanism of the pump and would not produce a net electrochemical gradient. D is incorrect because the 2:2 ratio is not the correct stoichiometry; the asymmetry of 3 Na⁺ out to 2 K⁺ in is biochemically significant and experimentally well established.
Question 7 · Molecular and Cellular Biology
During DNA replication in eukaryotic cells, which enzyme is responsible for synthesizing the RNA primer that initiates synthesis of each new DNA strand?
- A
- B
- C
- D
Explanation
Primase is the enzyme responsible for synthesizing the short RNA primers needed to initiate DNA synthesis, because DNA polymerases cannot begin a new strand de novo — they can only add nucleotides to an existing 3'-OH group. The RNA primer provides this starting point. A is incorrect because DNA polymerase I (in prokaryotes) removes and replaces RNA primers with DNA but does not synthesize the initial primer. C is incorrect because helicase unwinds and separates the double-stranded DNA helix at the replication fork but does not synthesize primers. D is incorrect because ligase joins Okazaki fragments together by forming phosphodiester bonds between adjacent DNA segments after primer replacement.
Question 8 · Diversity of Life
In the five-kingdom classification system proposed by Whittaker (1969), organisms are grouped partly based on their mode of nutrition. Which of the following correctly matches a kingdom to its primary mode of nutrition?
- A
- B
- C
- D
Explanation
The correct answer is C. In Whittaker's five-kingdom system, Fungi are classified as heterotrophs that obtain nutrients by absorbing digested organic compounds secreted externally by their hyphae (absorptive heterotrophy). Fungi secrete digestive enzymes into their substrate and absorb the resulting small molecules. A is incorrect because Fungi are not photosynthetic autotrophs; they lack chloroplasts and cannot perform photosynthesis. Photosynthetic autotrophy is characteristic of Plantae and some Protista. B is incorrect because Plantae are photoautotrophs, not heterotrophs; they produce their own food through photosynthesis using sunlight, carbon dioxide, and water. D is incorrect because Monera (prokaryotes) display diverse nutritional modes including photoautotrophy (e.g., cyanobacteria), chemoautotrophy (e.g., nitrifying bacteria), and various forms of heterotrophy including absorption and decomposition. Restricting Monera to absorptive heterotrophy only is inaccurate.
Question 9 · Molecular and Cellular Biology
During DNA replication in eukaryotes, which enzyme is responsible for synthesizing a short RNA sequence that provides a free 3'-OH group for DNA polymerase to begin nucleotide addition?
- A
- B
- C
- D
Explanation
Primase is the enzyme that synthesizes a short RNA primer complementary to the template strand, providing the free 3'-OH group that DNA polymerase III requires to begin adding deoxyribonucleotides. DNA polymerase cannot initiate a new strand de novo. (A) DNA ligase seals nicks between Okazaki fragments by forming phosphodiester bonds but does not synthesize primers. (C) Helicase unwinds the double helix by breaking hydrogen bonds between base pairs, but does not synthesize primers. (D) Topoisomerase relieves torsional strain ahead of the replication fork by cutting and rejoining DNA strands, but is not involved in primer synthesis.
Question 10 · Diversity of Life
A researcher is studying two species of flowering plants. Species X produces seeds enclosed within a fruit derived from the ovary wall, while Species Y produces seeds that are exposed on the surface of cone scales with no enclosing fruit tissue. Which of the following correctly classifies these two species?
- A
- B
- C
- D
Explanation
The correct answer is A. The defining feature of angiosperms is the production of seeds enclosed within a fruit, which develops from the ovary wall after fertilization. Gymnosperms, by contrast, bear 'naked' seeds that sit exposed on cone scales or similar structures, lacking an enclosing fruit. Species X matches the angiosperm description and Species Y matches the gymnosperm description. B is incorrect because it reverses the characteristics—angiosperms have enclosed seeds and gymnosperms have exposed seeds, not the other way around. C is incorrect because the production of naked seeds on cone scales is exclusively a gymnosperm trait; no angiosperm produces seeds in this manner. D is incorrect because bryophytes (mosses, liverworts) and pteridophytes (ferns) are non-seed-producing vascular or non-vascular plants and would not be described by seed-bearing characteristics at all.
Frequently Asked Questions
What is the Maryland passing score for Praxis 5236?
The Maryland passing score for the Praxis 5236 (Biology: Content Knowledge) is 155. This is set by MSDE and differs from other states. Always verify current requirements at msde.maryland.gov.
What is the MSDE score recipient code for Maryland?
The Maryland State Department of Education (MSDE) score recipient code is 7403. Select this code at every Praxis registration to have your scores sent directly to MSDE for licensure processing.
How long do I have to wait to retake the Praxis 5236?
Maryland requires a 28-day wait between Praxis 5236 attempts. This wait applies regardless of your score. Plan your test dates accordingly.
How many questions are on the Praxis 5236?
The Praxis 5236 contains 120 selected-response questions plus 2 constructed-response items (for PLT exams). You have 2 hours to complete the exam.
What domains does the Praxis 5236 cover?
The Praxis 5236 covers content knowledge specific to Biology: Content Knowledge. See the official ETS test framework for the complete domain breakdown.
How long are Praxis 5236 scores valid in Maryland?
Praxis scores are valid for 10 years from the test date in Maryland. Scores do not expire for the purposes of Maryland teacher certification within this window.
Can I use a calculator on the Praxis 5236?
No on-screen calculator is provided for the Praxis 5236.
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PraxisPass offers a permanent free tier that includes your Exam Readiness Score diagnostic, Pass Probability Score baseline, and your first complete 25-minute study mission for the 5236. The Individual plan at $19/month unlocks unlimited study sessions across all 50+ Maryland Praxis exams.