DAT Question of the Day

Correct Answer: B. mycorrhizae

The mycorrhizae inhabit the roots of vascular plants and help to increase the surface area of the roots leading to higher rates of nutrient and water absorption. Rhizomes are underground stems that can sprout to produce new shoots and roots for the plant. Likewise, the apical meristem is the portion on the stem that enables upward growth of the stem. The vascular cambium produces secondary xylem and phloem that help sustain the plant as it grows laterally each year. In addition, the cork cambium forms the “woody” part of the plant helping to provide the plant with structure and support as it grows laterally.

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Correct Answer: B. recombinant DNA

The phenotype of the offspring is a mixture of the two parents’ phenotype. This would be due to recombinant DNA, specifically crossing over that likely occurred during meiosis I (prophase I). Restriction enzymes cut the DNA at specific sequences and would not be the direct cause of the change in phenotype observed. Genetic drift describes the change in allele frequency over time not in one generation. Genetic mutations can lead to different features in an offspring than what was present in the parents, however the offspring resemble the parents, with one phenotypic characteristic from each parent expressed.

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Correct Answer: C. CUG ACG UUA

The tRNA has the anticodon to the codon on the mRNA. To solve this problem, you must first transcribe the mRNA from the given DNA strand, which would result in: GAC UGC AAU. Note: For the mRNA, A (Adenine) pairs with U (Uracil) not T (Thymine). Next, you have to find the complementary bases to the mRNA to determine the anticodon. So, the anticodon would be: CUG ACG UUA.

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Correct Answer: C. animal pole

The animal pole does not have yolk and can divide through mitosis at a quicker rate than the vegetal pole (the region with the yolk). The placenta is not known as having high rates of mitosis and is best known as the structure for which the embryo can release waste, perform gas exchange and nutrient absorption.

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Correct Answer: B. Secondary Succession

Succession is the process that ecological areas go through to repopulate a community. Following a forest fire, the environment is beginning from the producers level and must first grow small grasses, then small shrubs, before consumers and higher-level consumers will repopulate the area. In the event of a volcano, this would be an example of primary succession, as there is no soil to sustain plants. Instead, bacteria populate the area, and eventually give rise to soil and plants and animals. Homeostasis refers to the internal regulation of an organism as opposed to the process an external environment goes through. Biological regeneration is a distractor answer choice testing your true knowledge of the topic at hand.

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Correct Answer: D. DNA polymerases require a primer base-paired to the template and with a free 3′-end.

DNA polymerase can add dNTPs (deoxynucleoside triphosphates) only to a pre-existing oligonucleotide chain that is annealed to the template DNA. The enzyme adds dNTPs on a free 3′-OH- group. Primers in replication are RNA based. These primers are then cut out via endonuclease activity and replaced with DNA nucleotides. Finally, DNA ligase fuses the gap between the fragments of DNA.

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Correct Answer: C. prevents glucose from leaking through the plasma membrane.

Glycolysis takes place in the cytosol. It is a characteristic of eukaryotic and prokaryotic cells. The first reaction of glycolysis is irreversible since it requires an energy investment. Glucose-6-phosphate cannot pass back through the plasma membrane.

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Correct Answer: C. Lambdaphage viral vector.

The lambdaphage viral vector can be used injecting a specific sequence of DNA into a target cell, but not for encouraging the cell to uptake plasmid DNA. Bacteria can utilize a pillus for conjugation, in order to exchange genetic material through plasmids. Electroporation involves using a small charge to cause the cell membrane to become more porous and uptake a plasmid. Heat shock involves heating up the target cells to ~42ºC after keeping them on ice in the presence of CaCl2 – the cell membranes will become more porous and uptake the plasmid as with electroporation.

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Correct Answer: C. a homozygous red haired cow

In genetics, a test cross is used to determine the genotype of an organism that expresses the dominant phenotype. The black cow could be BB or Bb for the black hair color. If you were to mate the cow with a homozygous black cow BB, all of the offspring would be black haired and this would not help the farmer determine the genotype of the original black cow. Likewise using a heterozygous black cow (Bb), could result in all black offspring, since there is only a 25% chance of getting a red haired cow if the original cow was heterozygous (Bb). A white cow nor an albino cow are not mentioned in the original question and would not be relevant in determining the genotype of the black cow. A test cross, uses a homozygous recessive individual, since this genotype is easily physically identified. If the 50% of the offspring are black and 50% are red, then the black cow was heterozygous (Bb). If all black offspring result, then the black cow was homozygous dominant (BB).

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Correct Answer: C. a nonsense mutation

A silent mutation often occurs in the wobble site of the codon on the mRNA strand. For most codons, if the third base pair is changed, the correct amino acid can still be made. A missense mutation results in the addition of an incorrect amino acid, but can still lead to a functional protein. Nonsense mutations result in a polypeptide chain that is shorter and is non-functional. For nonsense mutations, an insertion or deletion of a nucleotide base creates a premature stop codon (UGA, UAG, or UAA), terminating translation before the polypeptide can be completed. Transfer mutation is not an accurate term in biology.

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