AP Bio Unit 6 AP CR Questions

Eukaryotes transcribe RNA from DNA that contains introns and exons. Alternative splicing is one posttranscriptional modification that can create distinct mature mRNA molecules that lead to the production of different proteins from the same gene. Figure 1 shows a gene and the RNA produced after transcription and after alternative splicing. The figure presents a diagram with 4 levels and a representation of a substrate below the fourth level. From top to bottom the levels are labeled as follows, level 1: D N A, level 2: Pre m R N A, level 3: m R N A, level 4: Enzyme. The D N A in level 1 has 5 separate exons labeled from left to right as Exon 1, Exon 2, Exon 3, Exon 4, and Exon 5. A vertical arrow labeled Transcription points down to level 2, the pre m R N A level. The Pre m R N A strand has 5 separate exons labeled from left to right as Exon 1, Exon 2, Exon 3, Exon 4, and Exon 5. Under the Pre m R N A is the label After Alternative Splicing, from which 3 arrows each point down to a different m R N A in level 3. The first m R N A is labeled A and includes Exon 1, Exon 2, Exon 3, Exon 4, and Exon 5. The second m R N A is labeled B and includes Exon 1, Exon 2, Exon 4, and Exon 5. The third m R N A is labeled C and includes Exon 1, Exon 2, Exon 3, and Exon 5. Following each m R N A is an arrow pointing down to the corresponding enzyme. Enzyme A is represented by a rectangle with a small rectangular notch cut into the top, enzyme B is represented by a rectangle with a small triangular notch cut into the top, and enzyme C is represented by a rectangle with a small semicircular notch cut into the top. The Substrate shown below the three enzymes is a represented by a rectangle with a small rectangular extension from its bottom surface. Figure 1. Model of posttranscriptional alternate splicing of mRNA A cell needs to metabolize the substrate illustrated in Figure 1 for a vital cellular function. Which of the following best explains the long-term effect on the cell of splicing that yields only enzyme C mRNA? A The cell will die because it is unable to metabolize the substrate without enzyme A, which is structurally specific for the substrate shown. B The cell will remain healthy because all three of the above enzymes can metabolize the substrate, as they are from the same gene. C The cell will remain healthy because the enzyme C mRNA will undergo alternative splicing again until it transformed into enzyme A mRNA. D The cell will remain healthy because enzyme-substrate interactions are nonspecific and enzyme C will eventually metabolize the substrate

A

Figure 1 represents a portion of a process that occurs during protein synthesis. The figure presents a segment of D N A that has a Binding Site followed by a Structural Gene in the central portion of the segment. The binding site is located between 2 tick marks, and the base sequence A C A G T G A is shown within the tick marks. An oval labeled Transcriptional Repressor Protein is positioned above the binding site. Moving slightly to the right along the segment, the Structural Gene is also located between 2 tick marks. Below the gene is a line labeled m R N A that starts near the left end of the gene and curves down and away from the D N A. Figure 1. Model of selected features of DNA transcription Which claim is most consistent with the information provided by the diagram and current scientific understanding of gene regulation and expression? A Reversible changes in the DNA sequence may influence how a gene is expressed in a cell. B Some sequences of DNA can interact with regulatory proteins that control transcription. C This is an inducible operon controlled by several regulatory factors. D The transcription factor may produce mutations in the binding site at the promoter sequence inhibiting the synthesis of the protein

B

Huntington's disease, an autosomal dominant disorder, is caused by a mutation in the HTT gene. The HTT gene contains multiple repeats of the nucleotide sequence CAG. A person with fewer than 35 CAG repeats in the HTT gene is unlikely to show the neurological symptoms of Huntington's disease. A person with 40 or more CAG repeats almost always becomes symptomatic. Due to errors in meiosis, an individual without symptoms of Huntington's disease can produce gametes with a larger number of CAG repeats than there are in their somatic cells. A woman develops Huntington's disease. Her father had the disorder. Her mother did not, and there is no history of the disorder in the mother's family. Which of the following best explains how the woman inherited Huntington's disease? A She inherited an allele with fewer than 40 CAG repeats in the HTT gene because her mother did not have Huntington's disease. B She inherited an allele with more than 40 CAG repeats in the HTT gene from her father. C Her mother produced eggs that all have more than 40 repeats in the HTT gene. D Her mother produced eggs that all have fewer than 40 CAG repeats in the HTT gene

B

Lynch syndrome is an inherited condition associated with an increased risk for colon cancer, as well as certain other cancers. Mutations in one of several genes involved in DNA repair during DNA replication have been associated with Lynch syndrome. DNA sequencing was performed for an individual. The results indicated that the individual carries one of the dominant alleles that has been associated with Lynch syndrome. Which of the following best explains how the results should be interpreted? A The individual does not have an increased risk of developing cancer because one dominant allele is insufficient to cause the disease. B The individual has an increased risk of developing colon cancer. C Because the person's DNA has the mutation, other family members must have cancer. D Results cannot be interpreted until testing determines if additional mutated alleles are present.

B

All cells must transcribe rRNA in order to construct a functioning ribosome. Scientists have isolated and identified rRNA genes that contribute to ribosomal structure for both prokaryotes and eukaryotes. Figure 1 compares the transcription and processing of prokaryotic and eukaryotic rRNA. The figure presents a diagram with two halves that together compare Prokaryotic with Eukaryotic r R N A processing. Each half of the diagram has 2 rows; row 1 is labeled Pre r R N A, and row 2 is labeled Mature r R N A. For each type of organism, Row 1 consists of a rectangle that is divided into alternating white and shaded segments. For the Prokaryotic half of the diagram, the pre r R N A rectangle consists of a white segment, followed by a shaded segment labeled 16 S, followed by a white segment, followed by a shaded segment labeled 23 S, followed by a white segment, followed by a shaded segment labeled 5 S, followed by a white segment. An arrow labeled r R N A Processing points from the pre r R N A to three individual shaded segments of mature r R N A that are labeled 16 S, 23 S, and 5 S. For the Eukaryotic half of the diagram, the pre r R N A rectangle consists of a white segment, followed by a shaded segment labeled 18 S, followed by a white segment, followed by a shaded segment labeled 5.8 S, followed by a white segment, followed by a shaded segment labeled 28 S, followed by a white segment. An arrow labeled r R N A Processing points from the pre r R N A to three individual shaded segments of mature r R N A that are labeled 18 S, 5.8 S, and 23 S. Figure 1. Comparison of rRNA processing in prokaryotes and eukaryotes Which of the following statements provides the best explanation of the processes illustrated in Figure 1 ? A Introns are removed from the pre-rRNA, and the mature rRNA molecules are joined and then translated to produce the protein portion of the ribosome. B Introns are removed from the pre-rRNA, and each mature rRNA molecule is translated to produce the proteins that make up the ribosomal subunits. C Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to combine with proteins to form the ribosomal subunits. D Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to bring different amino acids to the ribosome

C

An evolutionary biologist hypothesizes that two morphologically similar plant species are not closely related. To test the hypothesis, the biologist collects DNA samples from each of the two plant species and then uses restriction enzymes to cut the DNA samples into fragments, which are then subjected to gel electrophoresis. The results are shown in Figure 1. The figure shows a rectangle representing a gel with 2 lanes, labeled Species A and Species B. On the right side of the rectangle is the label, Direction of Movement, with a vertical arrow pointing downward. An open rectangle at the top of each lane represents the sample loading well. The data from the representation of the gel are as follows. Lane A contains 12 dark narrow rectangles, representing bands in the gel. The first band is near the top, just below the sample loading well. The next nine bands are fairly evenly spaced from close to the top band to about three fourths of the way along the lane. Two bands below these top ten are spaced a bit further apart and are in the next one eighth of the lane. No bands are in the final eighth of the lane. Lane B contains 9 bands. The first band of lane B is in a position that is just below the second band of lane A. Two more bands are below the first band, then there is a space with no bands that is approximately at the second one fourth of the lane. Five more bands are in the second half of the lane, the first two a bit spread out and the next three closer together. The last band in lane B is in a position that is lower than that of the last band in lane A. Figure 1. DNA analysis of two species Given the results shown in Figure 1, which of the following correctly describes a relationship between the two species? A Species B is the ancestor of species A because it has fewer bands. B Species A is more complex than species B because it has more bands. C Species B has more short fragments of DNA than species A does. D Species A has more short fragments of DNA than species B does.

C

Cystic fibrosis (CF) is a progressive genetic disease that causes persistent lung infections and affects the ability to breathe. CF is inherited in an autosomal recessive manner, caused by the presence of mutations in both copies of the gene for the cystic fibrosis transmembrane conductance regulator (CFTR) protein. Partial nucleotide sequences and the corresponding amino acid sequences for an unaffected individual and an affected individual are modeled in Figure 1. The figure presents 2 diagrams of partial nucleotide sequences and the corresponding amino acids for an unaffected individual and an affected individual. For an unaffected individual the nucleotide sequence, in triplets, and the corresponding amino acids are as follows. From left to right, nucleotides A C A and amino acid T h r, nucleotides T G G and amino acid T r p, nucleotides T A T and amino acid T y r, nucleotides G A C and amino acid A s p, nucleotides T C T and amino acid S e r, nucleotides C T T and amino acid L e u, nucleotides G G A and amino acid G l y. There is a G after the last nucleotide triplet. For an affected individual the nucleotide sequence, in triplets, and the corresponding amino acids are as follows. From left to right, nucleotides A C A and amino acid T h r, nucleotides G G T and amino acid G l y, nucleotides A T G and amino acid M e t, nucleotides A C T and amino acid T h r, nucleotides C T C and amino acid L e u, nucleotides T T G and amino acid L e u, nucleotides G A G and amino acid G l u. Figure 1. CFTR protein sequences in unaffected and affected individuals Based on the information in Figure 1, which type of mutation explains the nature of the change in DNA that resulted in cystic fibrosis in the affected individual? A Substitution, because the amino acid tryptophan is replaced with glycine. B Insertion, because an extra guanine is present, which changes the reading frame. C Deletion, because a thymine is missing, which changes the reading frame. D Duplication, because the amino acid leucine occurs twice, which changes the reading frame

C

Figure 1 depicts a simplified model of a replication bubble. The figure represents a simplified model of a horizontally presented D N A replication bubble. The origins of replication are indicated in the center of each template strand of the bubble. Within the bubble each nucleotide base of the template strand is paired with the nucleotide base of the new complementary strand. Outside the bubble the template strands are shown as separated single strands. Arrows indicate that the direction of fork movement is to the left of the left end of the replication bubble and to the right of the right end of the replication bubble. The left end of the upper template strand is labeled 5 prime, and the right end is labeled 3 prime. The left end of the lower template strand is labeled 3 prime, and the right end is labeled 5 prime. Within the replication bubble, arrows indicate the direction of DNA synthesis. A single arrow representing synthesis of DNA complementary to the upper strand at the left replication fork is labeled Roman numeral 1 and points from the origin of replication toward the left. The left end of the arrow is labeled 3 prime. Four short sequential arrows representing synthesis of DNA complementary to the upper strand at the right replication fork are labeled Roman numeral 2 and point from the fork toward the origin of replication. The right end of the rightmost arrow is labeled 5 prime. Four short sequential arrows representing synthesis of DNA complementary to the lower strand at the left replication fork are labeled Roman numeral 3 and point from the fork toward the origin of replication. The left end of the leftmost arrow is labeled 5 prime. A single arrow representing synthesis of DNA complementary to the lower strand at the right replication fork is labeled Roman numeral 4 and points from the origin of replication toward the right. The right end of the arrow is labeled 3 prime. Figure 1. DNA replication bubble. Which of the following best explains how this model illustrates DNA replication of both strands as a replication fork moves? A I and IV are synthesized continuously in the 5′ to 3′ direction. B II and III are synthesized in segments in the 3' to 5' direction. C I is synthesized continuously in the 5′ to 3′ direction, and III is synthesized in segments in the 5′ to 3′ direction. D II is synthesized in segments after DNA polymerase is released from synthesizing strand IV

C

Nucleotide base pairing in DNA is universal across organisms. Each pair (T−A; C−G) consists of a purine and a pyrimidine. Which of the following best explains how the base pairs form? A Ionic bonds join two double-ringed structures in each pair. B Hydrogen bonds join two single-ringed structures in each pair. C Hydrogen bonds join a double-ringed structure to a single-ringed structure in each pair. D Covalent bonds join a double-ringed structure to a single-ringed structure in each pair

C

Retroviruses such as HIV and hepatitis B virus use RNA as their genetic material rather than DNA. In addition, they contain molecules of reverse transcriptase, an enzyme that uses an RNA template to synthesize complementary DNA. Which of the following best predicts what will happen when a normal cell is exposed to a retrovirus? A The reverse transcriptase will cut the host DNA into fragments, destroying the host cell. B The reverse transcriptase will insert the viral RNA into the host's genome so it can be transcribed and translated. C The reverse transcriptase will produce DNA from the viral RNA, which can be incorporated into the host's genome and then transcribed and translated. D The reverse transcriptase will force the host ribosomes to translate the viral RNA prior to polypeptide assembly.

C

Scientists conducted a transformation experiment using E. coli bacteria and the pTru plasmid. Samples of the pTru plasmid (lane A) and the chromosomal DNA from two different E. coli strains that the scientists attempted to transform (lane B and lane C) were compared using gel electrophoresis. The results are shown in Figure 1. The figure shows a rectangle representing a gel with 3 lanes, labeled A, B, and C. An open rectangle at the top of each lane represents the sample loading well. The samples are as follows. Lane A contains the p T r u plasmid, Lane B contains E. coli strain 1, and Lane C contains E. coli strain 2. The data from the representation of the gel are as follows. Lane A contains a single filled in rectangle at the very bottom of the lane. Lane B contains a single filled in rectangle at the top of the lane, just a bit below the loading well rectangle. Lane C contains two filled in rectangles, one that is a bit below the loading well rectangle and another at the very bottom of the lane. Figure 1. Results of E. coli transformation with pTru plasmid Which of the following statements best explains the experimental results observed in Figure 1 ? A E. coli in both lanes B and C have been successfully transformed and contain additional genetic information. B E. coli in lane B have been successfully transformed and contain additional genetic information. C E. coli in lane C have been successfully transformed and contain additional genetic information. D Which E. coli have been transformed cannot be determined from this gel.

C

A simplified model of a DNA replication fork is represented in Figure 1. The protein labeled Enzyme 1 carries out a specific role in the DNA replication process. The figure presents a model of a DNA replication fork. The DNA is double stranded and coiled on the left side of the figure, and the two strands are separated on the right side of the figure. The 3 prime and 5 prime ends of the two strands are labeled at each end. The right end of the upper uncoiled strand has a 3 prime label, and a long arrow that points from right to left is shown immediately under this strand. The right end of the arrow is labeled 5 prime, and the left end is labeled 3 prime. A shaded circle overlaps the separated upper strand and the 3 prime end of the arrow. The right side of the lower uncoiled strand has a 5 prime label, and two short arrows that point from left to right are shown immediately above this strand. The left end of one arrow is labeled 5 prime, and the right end is labeled 3 prime. Another shaded circle overlaps the coiled portion of the helix, and is labeled Enzyme 1. Figure 1. A simplified model of a DNA replication fork Which of the following statements best explains the role of Enzyme 1 in the DNA replication process? A Enzyme 1 is a DNA ligase that joins together the DNA fragments at a replication fork to form continuous strands. B Enzyme 1 is a DNA primase that catalyzes the synthesis of RNA primers on the lagging strand of a replication fork. C Enzyme 1 is a DNA polymerase that synthesizes new DNA by using the leading and lagging strands of a replication fork as templates. D Enzyme 1 is a topoisomerase that relieves tension in the overwound DNA in front of a replication fork.

D

Cycloheximide (CHX) is a eukaryote protein synthesis inhibitor. It is used in biomedical research to inhibit protein synthesis in eukaryotic cells studied in vitro. Its effects are rapidly reversed by simply removing it from the culture medium. In a translation experiment using a fungus culture, radiolabeled amino acids were added to the culture, allowing the researchers to measure the growth of a single polypeptide chain by measuring counts per minute (CPM). As the chain grew, the CPM increased. After a certain amount of time, CHX was added to the mixture, and the experiment continued. After an additional amount of time, the CHX was removed from the culture medium. Which of the following graphs best predicts the data collected during the experiment? A. up then straight B. up down up C. up straight down D. up straight up

D

Exposure to ultraviolet (UV) radiation is the leading cause of skin cancer in humans. Figure 1 shows a model of how UV exposure damages DNA. The figure presents a diagram of two D N A molecules. The first D N A molecule is labeled Before, and an arrow labeled Incoming U V Photon points to one strand in the central portion of the molecule. The second D N A helix is labeled After and displays an expanded area of the helix where the photon made contact. In this area, some bases of one strand are separated from their complementary bases of the other strand, and some of the separated bases appear to be binding to other bases on the same strand. Figure 1. Model of damage to DNA caused by UV exposure Which of the following statements best explains what is shown in Figure 1 ? A UV exposure triggers DNA replication, which results in rapid cell proliferation. B Naturally occurring dimers are removed by the UV photons, causing misshapen DNA, which results in replication errors. C The hydrogen bonds between base pairs absorb the UV photons, separating the two DNA strands, which results in rapid DNA replication. D UV photons cause dimers to form, leading to misshapen DNA, which results in replication and transcription errors

D

Labeled nucleotides were supplied to a cell culture before the cells began DNA replication. A simplified representation of the process for a short segment of DNA is shown in Figure 1. Labeled DNA bases are indicated with an asterisk ( The figure presents a diagram that starts with an original double stranded D N A molecule composed of five base pairs. The strand on the left side has a 5 prime label at the top and a 3 prime label at the bottom. The strand on the right has a 3 prime label at the top and a 5 prime label at the bottom. Arrows point from this molecule to the next part of the diagram in which the two strands are shown separately, with their 5 prime and 3 prime ends in the identical positions as in the original molecule. An arrow that is labeled DNA Synthesis points from each of the two single strands to a new double stranded DNA molecule. The two new molecules are labeled Copies of the Original DNA Molecule. Asterisks are shown by all bases on the strand of each new molecule that is complementary to the single strand. The 5 prime and 3 prime ends of each strand are shown, and the two strands are in opposite orientations, as in the original molecule. Figure 1. A simplified representation of the DNA replication process Which of the following best helps explain how the process represented in Figure 1 produces DNA molecules that are hybrids of the original and the newly synthesized strands? A Each template strand is broken down into nucleotides, which are then used to synthesize both strands of a new DNA molecule. B Each template strand is broken into multiple fragments, which are randomly assembled into two different DNA molecules. C Each newly synthesized strand is associated with another newly synthesized strand to form a new DNA molecule. D Each newly synthesized strand remains associated with its template strand to form two copies of the original DNA molecule.*

D

Small single-stranded RNA molecules called microRNAs (miRNAs) are capable of base pairing with specific binding sites in the 3′ untranslated region of many mRNA transcripts. Transcription of gene Q yields an mRNA transcript that contains such an miRNA binding site, which can associate with miRNA‑delta, a specific miRNA molecule. Which of the following best supports the claim that binding of miRNA‑delta to the miRNA binding site inhibits translation of gene Q mRNA? A When the promoter for gene Q is altered, transcription is inhibited. B Translation of Q mRNA is inhibited regardless of whether the miRNA binding site sequence is altered. C Translation of Q mRNA is inhibited in the absence of miRNA‑delta. D When the miRNA binding site sequence is altered, translation of Q mRNA occurs in the presence of miRNA-delta

D

The trp operon in E. coli is an example of a repressible operon that consists of genes coding for enzymes used to synthesize tryptophan. When tryptophan levels are high, the operon is turned off and these genes are not transcribed. However, it is also known that tryptophan does not bind directly to the operator DNA sequence. A regulatory gene called trpR has also been discovered although it is not part of the trp operon. The proposed model of how tryptophan acts as a corepressor is shown in Figure 1. The figure presents a stretch of D N A that includes a t lowercase r p uppercase R regulatory gene and then, further along, a separate area labeled as the t lowercase r p Operon. The operon includes a Promoter followed by Functional Genes. An R N A Polymerase is shown binding to the first part of the promoter, and the second part of the promoter is labeled as the Operator. Arrows above the Operator and the Functional Genes point to the right. The figure indicates that an m R N A is produced from the t lowercase r p uppercase R gene, and another m R N A is produced from the t lowercase r p Operon. The figure also indicates that a t lowercase r p Repressor Protein is produced from the t lowercase r p uppercase R m R N A. The protein is shown in one shape when tryptophan is absent, where it is labeled as Inactive and in a different shape when tryptophan is present, where it is labeled as Active. Arrows pointing in both directions are present between the inactive and active representations of the repressor protein. Figure 1. Model of proposed regulation of the trp operon by corepressors trp repressor and tryptophan Which of the following evidence best supports a claim that tryptophan functions as a corepressor? A Normal expression of trpR causes the trp operon to be transcribed regardless of tryptophan levels. B When the operator sequence is mutated, the trp operon is not transcribed. C The trpR gene codes for a repressor protein that has a DNA binding domain. D When trpR is mutated, the trp operon is transcribed regardless of tryptophan levels.

D

Which of the following best explains how the pattern of DNA arrangement in chromosomes could be used, in most cases, to determine if an organism was a prokaryote or a eukaryote? A Prokaryotic DNA Eukaryotic DNA Single circular chromosome Multiple circular chromosomes B Prokaryotic DNA Eukaryotic DNA Multiple chromosomes Single chromosome C Prokaryotic DNA Eukaryotic DNA Single linear chromosome Multiple linear chromosomes D Prokaryotic DNA Eukaryotic DNA Single circular chromosome Multiple linear chromosomes

D

Arsenic is a toxic element found in both aquatic and terrestrial environments. Scientists have found genes that allow bacteria to remove arsenic from their cytoplasm. Arsenic enters cells as arsenate that must be converted to arsenite to leave cells. Figure 1 provides a summary of the arsenic resistance genes found in the operons of three different bacteria. E. coli R773R773 is found in environments with low arsenic levels. Herminiimonas arsenicoxydans and Ochrobactrum tritici are both found in arsenic‑rich environments. Researchers claim that bacteria that live in environments heavily contaminated with arsenic are more efficient at processing arsenic into arsenite and removing this toxin from their cells. Justify this claim based on the evidence shown in Figure 1. a. There are multiple operons controlling the production of proteins that process and remove arsenite from cells in both H. arsenicoxydans and O. tritici. In contrast, E. coli has only one operon devoted to arsenic removal. b. Both H. arsenicoxydans and O. tritici contain the arsRarsR gene that codes for a repressor that turns on the operon to eliminate arsenite from the cell. c. Both O. tritici and E. coli contain the arsDarsD gene, which codes for a protein that helps remove arsenite from the cell. d. Both H. arsenicoxydans and O. tritici. have more arsenic resistance genes than has E. coli.

a

Both liver cells and lens cells have the genes for making the proteins albumin and crystalline. However, only liver cells express the blood protein albumin and only lens cells express crystalline, the main protein in the lens of the eye. Both of these genes have enhancer sequences associated with them. The claim that gene regulation results in differential gene expression and influences cellular products (albumin or crystalline) is best supported by evidence in which of the following statements? a.Liver cells possess transcriptional activators that are different from those of lens cells. b.Liver cells and lens cells use different RNARNA polymerase enzymes to transcribe DNADNA. c.Liver cells and lens cells possess the same transcriptional activators. d.Liver cells and lens cells possess different general transcription factors.

a

Antibiotics interfere with prokaryotic cell functions. Streptomycin is an antibiotic that affects the small ribosomal subunit in prokaryotes. Specifically, streptomycin interferes with the proper binding of tRNAtRNA to mRNAmRNA in prokaryotic ribosomes. Which of the following best predicts the most direct effect of exposing prokaryotic cells to streptomycin? a. Amino acid synthesis will be inhibited. b. No m R N A will be transcribed from D N A . c. Posttranslational modifications will be prevented. d. synthesis of polypeptides will be inhibited.

d

Nondisjunction during meiosis can negatively affect gamete formation. A model showing a possible nondisjunction event and its impact on gamete formation is shown in Figure 1. Which of the following best describes the most likely impact on an individual produced from fertilization between one of the daughter cells shown and a normal gamete? a. Because nondisjunction occurred in anaphase II, all gametes will be normal and the resulting individual will be phenotypically normal. b.Because nondisjunction occurred in anaphase II, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event. c. Because nondisjunction occurred in anaphase IIII, all gametes will be normal and the resulting individual will be phenotypically normal. d. Because nondisjunction occurred in anaphase IIII, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event.

d