A.P. Biology 2nd Trimester Exam (MC)

Probability that the genotype ccdd will be produced by the parents CcDd x CcDd (A)0 (B)1/16 (C)1/4 (D)1/2 (E)3/4

1/16

Probability that the genotype Rr will be produced by the parents Rr x rr (A) 0 (B) 1/16 (C) 1/4 (D) 1/2 (E) 3/4

1/2

In which of the following would there NOT be a change in the amino acid sequence of the peptide coded for by this DNA? 3' TAG TTC AAA CCG CGT AAC ATT 5' (A)Changing 3' AAA 5' to read 3' AAG 5' (B)Changing 3' TTC 5' to read 3' ATC 5' (C)Changing 3' CCG 5' to read 3' GGC 5' (D)Deleting the first A from 3' AAA 5' (E)Deleting the last triplet

Changing 3' AAA 5' to read 3' AAG 5'

Which of the following is most likely to create genetic variation in a population? (A)RNA polymerase errors during transcription (B)Helicase failure to unwind DNA during DNA replication (C)DNA polymerase errors during replication (D)Misincorporation of amino acids by tRNA during translation

DNA polymerase errors during replication

Gregor Mendel's pioneering genetic experiments with pea plants occurred before the discovery of the structure and function of chromosomes. Which of the following observations about inheritance in pea plants could be explained only after the discovery that genes may be linked on a chromosome? (A)Pea color and pea shape display independent inheritance patterns. (B)Offspring of a given cross show all possible combinations of traits in equal proportions. (C)Most offspring of a given cross have a combination of traits that is identical to that of either one parent or the other. (D)Recessive phenotypes can skip a generation, showing up only in the parental and F2 generations.

Most offspring of a given cross have a combination of traits that is identical to that of either one parent or the other.

Enzyme used in the synthesis of mRNA (A)DNA ligase (B)DNA polymerase (C)RNA polymerase (D)Restriction enzyme (E)Reverse transcriptase

RNA polymerase

Achondroplastic dwarfism is a dominant genetic trait that causes severe malformation of the skeleton. Homozygotes for this condition are spontaneously aborted (hence, the homozygous condition is lethal) but heterozygotes will develop to be dwarfed. Matthew has a family history of the condition, although he does not express the trait. Jane is an achondroplastic dwarf. Matthew and Jane are planning a family of several children and want to know the chances of producing a child with achondroplastic dwarfism. If three children are born to Matthew and Jane, what are the chances that the first two children will not express the trait but that the third child will be an achondroplastic dwarf? (A)5/8 (B)4/8 (C)3/8 (D)1/8 (E)1/16

1/8

The probability that the genotype Aa will be produced by the parents Aa x Aa (A)0 (B)1/16 (C)1/4 (D)1/2 (E)3/4

1/2

Probability that the genotype TTSs will be produced by the parents TTSs x TtSS (A) 0 (B) 1/16 (C) 1/4 (D) 1/2 (E) 3/4

1/4

In sheep, eye color is controlled by a single gene with two alleles. When a homozygous brown-eyed sheep is crossed with a homozygous green-eyed sheep, blue-eyed offspring are produced. If the blue-eyed sheep are mated with each other, what percent of their offspring will most likely have brown eyes? (A)0% (B)25% (C)50% (D)75% (E)100%

25%

3' TAG TTC AAA CCG CGT AAC ATT 5' The mRNA transcribed from the DNA would read (A)5' TAG TTC AAA CCG CGT AAC AAT 3' (B)5' ATC AAG TTT GGC GCA TTG TAA 3' (C)5' AUC AAG UUU GGC GCA UUG UAA 3' (D)5' AAU CAA UGC GCC AAA CUU GAU 3' (E)5' AUU GUU ACG CGG UUU GAA CUA 3'

5' AUC AAG UUU GGC GCA UUG UAA 3'

Achondroplastic dwarfism is a dominant genetic trait that causes severe malformation of the skeleton. Homozygotes for this condition are spontaneously aborted (hence, the homozygous condition is lethal) but heterozygotes will develop to be dwarfed. Matthew has a family history of the condition, although he does not express the trait. Jane is an achondroplastic dwarf. Matthew and Jane are planning a family of several children and want to know the chances of producing a child with achondroplastic dwarfism. The probability that Matthew and Jane's first child will be an achondroplastic dwarf is (A)0% (B)25% (C)50% (D)75% (E)100%

50%

One of the affected males from the third generation has a child with a female who is a carrier. For the pedigree shown above, which of the following best expresses the probability that the couple's first son will be affected with the disorder? (A)25% (B)50% (C)75% (D)100%

50%

The diagram above shows a developing worm embryo at the four-cell stage. Experiments have shown that when cell 3 divides, the anterior daughter cell gives rise to muscle and gonads and the posterior daughter cell gives rise to the intestine. However, if the cells of the embryo are separated from one another early during the four-cell stage, no intestine will form. Other experiments have shown that if cell 3 and cell 4 are recombined after the initial separation, the posterior daughter cell of cell 3 will once again give rise to normal intestine. Which of the following is the most plausible explanation for these findings? (A)A cell surface protein on cell 4 signals cell 3 to induce the formation of the worm's intestine. (B)The plasma membrane of cell 4 interacts with the plasma membrane of the posterior portion of cell 3, causing invaginations that become microvilli. (C)Cell 3 passes an electrical signal to cell 4, which induces differentiation in cell 4. (D)Cell 4 transfers genetic material to cell 3, which directs the development of intestinal cells.

A cell surface protein on cell 4 signals cell 3 to induce the formation of the worm's intestine.

The vertebrate forelimb initially develops in the embryo as a solid mass of tissue. As development progresses, the solid mass near the end of the forelimb is remodeled into individual digits. Which of the following best explains the role of apoptosis in the remodeling of the forelimb? (A)Apoptosis replaces old cells with new ones that are less likely to contain mutations. (B)Apoptosis involves the regulated activation of proteins in specific cells of the developing forelimb that leads to the death of those cells. (C)Apoptosis involves the destruction of extra cells in the developing forelimb, which provides nutrients for phagocytic cells. (D)Apoptosis in the developing forelimb triggers the differentiation of cells whose fate was not already determined.

Apoptosis involves the regulated activation of proteins in specific cells of the developing forelimb that leads to the death of those cells.

In dogs, one pair of alleles determines coat color (dark and albino). Another pair of alleles determines hair length (short and long). Thus, each gamete will contain one of the coat-color alleles, C or c, and one of the hair-length alleles, B or b. In repeated crosses of a specific dark, short-haired dog with an albino, long-haired dog, all the offspring were dark with short hair, as shown in cross I. However, in subsequent crosses of another dark, short-haired dog with a dark, long-haired dog, the ratios shown in cross II below were obtained. Cross 1. (dark, short hair) x (albino, long hair): all dark, short hair Cross 2. (dark, short hair) x (dark, long hair): (3 dark, short), (3 dark, long), (1 albino, short), (1 albino, long) Which of the following is probably the genotype of the dark, short-haired parent in cross I? (A)CcBb (B)ccbb (C)CCBB (D)CCbb (E)ccBB

CCBB

In dogs, one pair of alleles determines coat color (dark and albino). Another pair of alleles determines hair length (short and long). Thus, each gamete will contain one of the coat-color alleles, C or c and one of the hair-length alleles, B or b. In repeated crosses of a specific dark, short-haired dog with an albino, long-haired dog, all the offspring were dark with short hair, as shown in cross I. However, in subsequent crosses of another dark, short-haired dog with a dark, long-haired dog, the ratios shown in cross II below were obtained. Cross 1. (dark, short hair) x (albino, long hair): all dark, short hair Cross 2. (dark, short hair) x (dark, long hair): (3 dark, short), (3 dark, long), (1 albino, short), (1 albino, long) In cross II, the genotype of the dark, short-haired parent is (A)CcBb (B)ccbb (C)CCBB (D)CCbb (E)ccBB

CcBb

Enzyme used during replication to attach Okazaki fragments to each other (A)DNA ligase (B)DNA polymerase (C)RNA polymerase (D)Restriction enzyme (E)Reverse transcriptase

DNA ligase

Enzyme used to position nucleotides during DNA replication (A)DNA ligase (B)DNA polymerase (C)RNA polymerase (D)Restriction enzyme (E)Reverse transcriptase

DNA polymerase

Figure 1 illustrates a model of the molecules involved in DNA replication and their placement relative to each other. Which of the following correctly explains where DNADNA replication will begin on the strand oriented 5'→3'5′→3′, reading from left to right? (A)DNA replication will be randomly initiated along the unwound portion of the DNA strand since base pairing will occur. (B)DNA replication cannot occur since there is already an RNA base pairing with the template strand. (C)DNA replication will be initiated immediately to the left of the RNA since DNA polymerase requires an RNA primer. (D)DNA replication will be initiated at the site of the topoisomerase since that is where DNA begins to uncoil.

DNA replication will be initiated immediately to the left of the RNA since DNA polymerase requires an RNA primer.

In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (ampr). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria. I) No AMP- lawn III) No AMP/Plasmid- lawn II) AMP- none IV) AMP/Plasmid- speckled Plates that have only ampicillin-resistant bacteria growing include which of the following? (A)I only (B)III only (C)IV only (D)I and II

IV only

The epinephrine signaling pathway plays a role in regulating glucose homeostasis in muscle cells. The signaling pathway is activated by the binding of epinephrine to the beta-2 adrenergic receptor. A simplified model of the epinephrine signaling pathway is represented in Figure 1. (A) It involves the opening and closing of ion channels. (B) It involves enzymes activating other enzymes. (C)It involves changes in the expression of target genes. (D)It involves protons moving down a concentration gradient.

In involves enzymes activating other enzymes. Based on Figure 1, the epinephrine signaling pathway involves enzymes activating other enzymes. For example, the pathway includes several protein kinases, enzymes that catalyze the transfer of a phosphate group from ATPATP to a protein substrate. As represented in Figure 1, protein kinase A catalyzes the transfer of a phosphate group from ATPATP to phosphorylase kinase, which results in the activation of phosphorylase kinase. The activated phosphorylase kinase activates glycogen phosphorylase in a similar manner.

The Trp operon is a coordinately regulated group of genes (trpA-trpE) that are required for tryptophan biosynthesis in E. coli. Based on the figure above, which of the following correctly describes the regulation of the Trp operon? (A)In the absence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon. (B)In the presence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon. (C)In the absence of tryptophan, the trpR gene is inactive, preventing the production of the repressor that blocks expression of the operon. (D)In the presence of tryptophan, the trpR gene is inactive, preventing the production of the repressor that blocks expression of the operon.

In the presence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon.

Australian dragon lizards have a ZW sex-determination system. The male genotype is homogametic (ZZ), and the female genotype is heterogametic (ZW). However, all eggs incubated at temperatures above 32°C tend to develop into females. Which of the following best explains how the development of phenotypic female Australian dragon lizards with a ZZ genotype occurs when incubation temperatures are above 32°C? (A)Lizard embryos with a ZZZZ genotype cannot develop at temperatures above 32°C32°C. (B)At incubation temperatures above 32°C32°C, ZZ chromosomes are mutated into WW chromosomes. (C)At incubation temperatures above 32°C32°C, crossing over transfers genes from the WW chromosome to the ZZ chromosome, producing females. (D)Incubation temperatures above 32°C32°C inhibit the genes on the ZZ chromosome that produce proteins necessary for male development.

Incubation temperatures above 32°C32°C inhibit the genes on the ZZ chromosome that produce proteins necessary for male development. Similar in function to other SRYSRY genes found on the YY chromosome in mammals, the ZZ chromosome contains genes that need to actively produce proteins during development of the embryo to trigger production of male characteristics. Above 32°C32°C, these genes are inhibited, so the default female phenotype develops.

Which of the following correctly explains the process shown in Figure 1 ? ATGATCTCGTAA ----------------> AUGAUCU / / / / / / / TACTAGAGCATT (A)DNA replication is occurring because replication is semi-conservative and the new strand is a copy of the template strand. (B)Initiation of transcription is occurring because a strand of RNA is being produced from a DNA template strand. (C)Translation is occurring because the two strands have separated and a new strand is being produced. (D)Alternative splicing of mRNAmRNA is occurring because the mRNAmRNA strand is being synthesized from only one strand of DNADNA.

Initiation of transcription is occurring because a strand of RNARNA is being produced from a DNADNA template strand. Transcription is occurring. Since uracil (U) is present in the new strand being formed, RNARNA is being formed from a DNADNA template.

Achondroplastic dwarfism is a dominant genetic trait that causes severe malformation of the skeleton. Homozygotes for this condition are spontaneously aborted (hence, the homozygous condition is lethal) but heterozygotes will develop to be dwarfed. Matthew has a family history of the condition, although he does not express the trait. Jane is an achondroplastic dwarf. Matthew and Jane are planning a family of several children and want to know the chances of producing a child with achondroplastic dwarfism. The genotypes of Matthew and Jane are best represented as (A) Matthew- AA Jane- Aa (B)Matthew- Aa Jane- aa (C)Matthew- aa Jane- aa (D)Matthew- aa Jane- Aa (E)Matthew- Aa Jane- Aa

Matthew- aa Jane- Aa

In a certain signal transduction pathway, the binding of an extracellular molecule to a cell-surface protein results in a rapid increase in the concentration of cyclic AMP inside the cell. The cyclic AMP binds to and activates cytosolic enzymes that then activate other enzymes in the cell. Which of the following statements best describes the role of cyclic AMP in the signal transduction pathway? (A)It acts as a signaling molecule that passes the signal from the cell to other cells. (B)It acts as a receptor that carries the signal from outside the cell to inside the cell. (C)It acts as a second messenger that helps relay and amplify the signal within the cell. (D)It acts as a channel protein that transmits the signal across the cell's nuclear membrane.

It acts as a second messenger that helps relay and amplify the signal within the cell. In the signal transduction pathway, cyclic AMPAMP acts as a second messenger by relaying the signal from the plasma membrane to cytosolic enzymes. The signal is amplified inside the cell by the production of many cyclic AMPAMP molecules, which then activate enzyme molecules that are capable of catalyzing a specific reaction repeatedly.

In dogs, one pair of alleles determines coat color (dark and albino). Another pair of alleles determines hair length (short and long). Thus, each gamete will contain one of the coat-color alleles, C or c, and one of the hair-length alleles, B or b. In repeated crosses of a specific dark, short-haired dog with an albino, long-haired dog, all the offspring were dark with short hair, as shown in cross I. However, in subsequent crosses of another dark, short-haired dog with a dark, long-haired dog, the ratios shown in cross II below were obtained. Cross 1. (dark, short hair) x (albino, long hair): all dark, short hair Cross 2. (dark, short hair) x (dark, long hair): (3 dark, short), (3 dark, long), (1 albino, short), (1 albino, long) Which of the following correctly describes the relationship of the dark-coat-color allele to the albino condition? (A)It is dominant. (B)It is recessive. (C)It is codominant. (D)It is a polygenic inheritance pattern. (E)The alleles are linked.

It is dominant.

Both mitosis and meiosis begin with a parent cell that is diploid. Which of the following best describes how mitosis and meiosis result in daughter cells with different numbers of chromosomes? (A)In mitosis, the chromosomes consist of a single chromatid, which is passed to two haploid daughter cells. In meiosis, the chromosomes consist of two chromatids during the first round of division and one chromatid during the second round of division, resulting in two haploid daughter cells. (B)In mitosis, synapsis of homologous chromosomes results in four haploid daughter cells after one division. In meiosis, synapsis of homologous chromosomes occurs during the second division and results in four diploid daughter cells. (C)Mitosis produces one identical daughter cell after one round of division. Meiosis has two rounds of division and doubles the number of chromosomes in the second round of division, producing four diploid cells. (D)Mitosis produces two identical diploid daughter cells after one round of division. Meiosis produces four haploid daughter cells after two rounds of division.

Mitosis produces two identical diploid daughter cells after one round of division. Meiosis produces four haploid daughter cells after two rounds of division.

Enzyme found in retroviruses that produce DNA from an RNA template (A)DNA ligase (B)DNA polymerase (C)RNA polymerase (D)Restriction enzyme (E)Reverse transcriptase

Reverse transcriptase

In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (ampr). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria. I) No AMP- lawn III) No AMP/Plasmid- lawn II) AMP- none IV) AMP/Plasmid- speckled Which of the following statements best explains why there are fewer colonies on plate IV than on plate III? (A)Plate IV is the positive control. (B)Not all E. coli cells are successfully transformed. (C)The bacteria on plate III did not mutate. (D)The plasmid inhibits E. coli growth.

Not all E. coli cells are successfully transformed.

Which of the following statements best explains the structure and importance of plasmids to prokaryotes? (A)Plasmids are circular, single-stranded RNARNA molecules that transfer information from the prokaryotic chromosome to the ribosomes during protein synthesis. (B)Plasmids are circular, double-stranded DNADNA molecules that provide genes that may aid in survival of the prokaryotic cell. (C)Plasmids are single-stranded DNADNA molecules, which are replicated from the prokaryotic chromosome, that prevent viral reproduction within the prokaryotic cell. (D)Plasmids are double-stranded RNARNA molecules that are transmitted by conjugation that enable other prokaryotic cells to acquire useful genes.

Plasmids are circular, double-stranded DNADNA molecules that provide genes that may aid in survival of the prokaryotic cell.

For the following group of questions first, study the description of the situation or data and then choose the one best answer to each question following it and fill in the corresponding oval on the answer sheet. The diagram below illustrates the results of electrophoresis of DNA sequences obtained from a family of two adults and three children, and amplified using PCR. The bands represent short repeating sequences of variable length. Results for another female (X) are included for comparison. Father- 1, 2, 3 Mother- 1, 2, X Child 1- father, mother, 2, X Child 2- mother, father, 1, 3 Child 3- 2, father X- mother, 1, 2 Which of the following is the best explanation for the fragment pattern for individual X ? (A)She has only one member of this chromosome pair. (B)She has only one living parent. (C)She is homozygous for this particular DNA fragment. (D)She is the mother's child from another marriage. (E)She is not related to any member of the family being tested.

She is the mother's child from another marriage.

Most cells that have become transformed into cancer cells have which of the following characteristics when compared to normal, healthy cells? (A)Shorter cell cycle (B)More carefully regulated rates of cell division (C)Lower rates of mitosis (D)Higher rates of protein translation (E)Identical DNA

Shorter cell cycle

Erwin Chargaff investigated the nucleotide composition of DNA. He analyzed DNA from various organisms and measured the relative amounts of adenine (A), guanine (G), cytosine (C), and thymine (T) present in the DNA of each organism. Table 1 contains a selected data set of his results. Table 1. Nucleotide composition of sample DNA from selected organisms Which of the following statements best explains the data set? (A)Since the %A%A and the %G%G add up to approximately 50 percent in each sample, adenine and guanine molecules must pair up in a double-stranded DNADNA molecule. (B)Since the %A%A and the %T%T are approximately the same in each sample, adenine and thymine molecules must pair up in a double-stranded DNADNA molecule. (C)Since the %(A+T)%(A+T) is greater than the %(G+C)%(G+C) in each sample, DNADNA molecules must have a poly-AA tail at one end. (D)Since the %C%C and the %T%T add up to approximately 50 percent in each sample, cytosine and thymine molecules must both contain a single ring.

Since the %A and the %T are approximately the same in each sample, adenine and thymine molecules must pair up in a double-stranded DNADNA molecule. The percentages of adenine and thymine are approximately the same because adenine aligns with thymine in double-stranded DNADNA and they are therefore present in each organism in the same amounts.

Cross 1: blue male X blue female- 60 blue male, 124 blue female, 52 white male, 0 white female Cross 2: blue male X white female- 0 blue male, 115 blue female, 108 white male, 0 white female Cross 3: white male X blue female- 48 blue males, 42 blue females, 53 white males, 49 white female The data above represent the results of three different crosses involving the inheritance of a gene that determines whether a certain organism is blue or white. Which of the following best explains the mechanism of inheritance of the gene? (A)The allele for white is an autosomal dominant allele because a 1:1 phenotype ratio of blue to white among both sexes is observed in cross 3. (B)The allele for blue is an autosomal dominant allele because an approximate 3:1 phenotype ratio of blue to white is observed in cross 1. (C)The allele for white is an X-linked dominant allele because no white females are produced in cross 1. (D)The allele for blue is an X-linked dominant allele because there are no blue male offspring in cross 2.

The allele for blue is an X-linked dominant allele because there are no blue male offspring in cross 2.

In fruit flies, purple eyes and ebony body are traits that display autosomal recessive patterns of inheritance. In a genetics experiment, students cross wild-type flies with flies that have purple eyes and ebony bodies. The students observe that all the flies in the F1 generation have normal eyes and a normal body color. The students then allow the F1 flies to mate and produce an F2 generation. The students record observations about the flies in the F2 generation and use the data to perform a chi-square goodness-of-fit test for a model of independent assortment. The setup for the students' chi-square goodness-of-fit test is presented in Table 1. Phenotype, Observed, Expected, Normal eyes/normal body, 187, 171 Normal eyes/ebony body, 49, 57 Purple eyes/normal body, 41, 57 Purple eyes/ebony body, 27, 19\ The students choose a significance level of p=0.01. Which of the following statements best completes the next step of the chi-square goodness-of-fit test? (A)The calculated chi-square value is 2.11, and the critical value is 7.82. (B)The calculated chi-square value is 2.11, and the critical value is 11.35. (C)The calculated chi-square value is 10.48, and the critical value is 7.82. (D)The calculated chi-square value is 10.48, and the critical value is 11.35.

The calculated chi-square value is 10.48, and the critical value is 11.35. The calculated chi-square value is 10.48, and the critical value for 3 degrees of freedom and a significance level of p=0.01p=0.01 is 11.35. Because the calculated chi-square value is less than the critical value, the null hypothesis cannot be rejected. Consequently, the students can conclude that the data fit a model of independent assortment.

Based on the model of eukaryotic cell cycle regulation shown in the figure, which of the following best describes the effect of a drug that blocks the production of the mitotic cyclin? (A)The cell cycle would proceed uncontrollably, and the cell would become cancerous. (B)The G1 cyclin would functionally replace mitotic cyclin, and the cell would continue dividing normally. (C)DNA synthesis would be prevented, and the cell would stop dividing. (D)The cell would be prevented from entering mitosis, and the cell would stop dividing.

The cell would be prevented from entering mitosis, and the cell would stop dividing.

Antibiotics can be used to kill the specific pathogenic bacterium, Mycobacterium tuberculosis, that causes tuberculosis. The appearance of antibiotic-resistant strains has made it more difficult to cure M. tuberculosis infections. These antibiotic-resistant bacteria survive and pass on the genes to their offspring, making the resistant phenotype more common in the population. DNA analysis indicates that the genes for antibiotic resistance are not normally present in bacterial chromosomal DNA. Which of the following statements best explains how the genes for antibiotic resistance can be transmitted between bacteria without the exchange of bacterial chromosomal DNA? (A)The antibiotic-resistant bacteria release a hormone that signals neighboring bacteria to become resistant. (B)The genes for antibiotic resistance are located on a plasmid that can be passed to neighboring bacteria. (C)The antibiotic-resistant bacteria are the result of bacteria that specifically modify their own chromosomal DNADNA to neutralize the antibiotics. (D)The antibiotic alters the bacterial genome of each bacterium, which results in an antibiotic-resistant population.

The genes for antibiotic resistance are located on a plasmid that can be passed to neighboring bacteria. Bacteria can exchange and express the genes found on plasmids, which are foreign, extrachromosomal loops of DNADNA that they pick up. Genes for antibiotic resistance are located on these plasmids.

Which of the following statements best describes how a growth factor stimulates cell division from outside a cell? (A)The growth factor binds to other cells in the same area and holds them together to form a large, multicellular structure. (B)The growth factor binds to receptors on the cell surface, initiating a signal transduction pathway that activates specific target genes. (C)The growth factor binds to sugar molecules in the extracellular fluid and provides them to the cell as a source of energy. (D)The growth factor binds to phospholipids in the plasma membrane, creating a channel through which substances enter the cell.

The growth factor binds to receptors on the cell surface, initiating a signal transduction pathway that activates specific target genes.

In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (ampr). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria. I) No AMP- lawn III) No AMP/Plasmid- lawn II) AMP- none IV) AMP/Plasmid- speckled Which of the following best explains why there is no growth on plate II? (A)The initial E. coli culture was not ampicillin-resistant. (B)The transformation procedure killed the bacteria. (C)Nutrient agar inhibits E. coli growth. (D)The bacteria on the plate were transformed.

The initial E. coli culture was not ampicillin-resistant.

A gene that influences coat color in domestic cats is located on the X chromosome. A female cat that is heterozygous for the gene (XBXO) has a calico-colored coat. In a genetics experiment, researchers mate a calico-colored female cat (XBXO) with an orange-colored male cat (XOY) to produce an F1 generation. The researchers record observations for the cats in the F1 generation and plan to use the data to perform a chi-square goodness-of-fit test for a model of X-linked inheritance. The data for the chi-square goodness-of-fit test are presented in Table 1. Phenotype, genotype, observed, expected Calico-colored female, XBXO, 15, 10 Orange-colored female, XOXO, 6, 10 Black-colored male, XBY, 11, 10 Orange-colored male, XOY, 8, 10 The researchers calculate a chi-square value of 4.6 and choose a significance level of p=0.05. Which of the following statements best completes the chi-square goodness-of-fit test? (A)The null hypothesis can be rejected because the chi-square value is greater than the critical value. (B)The null hypothesis can be rejected because the chi-square value is less than the critical value. (C)The null hypothesis cannot be rejected because the chi-square value is greater than the critical value. (D)The null hypothesis cannot be rejected because the chi-square value is less than the critical value.

The null hypothesis cannot be rejected because the chi-square value is less than the critical value. There are four potential phenotypic categories, so there are three degrees of freedom. At the p=0.05 significance level, the critical value is 7.82. Because the calculated chi-square value is less than the critical value, the null hypothesis cannot be rejected.

The following the DNA sequence is a small part of the coding (nontemplate) strand from the open reading frame of β-hemoglobin gene. Given the codon chart listed below, what would be the effect of a mutation that deletes the G at the beginning of the DNA sequence? 5'- GTT TGT CTG TGG TAC CAC GTG GAC TGA - 3' (A)The mutation precedes the gene, so no changes would occur. (B)Lysine (lys) would replace glutamine (gln), but there would be no other changes. (C)The first amino acid would be missing, but there would be no other change to the protein. (D)The reading frame of the sequence would shift, causing a change in the amino acid sequence after that point.

The reading frame of the sequence would shift, causing a change in the amino acid sequence after that point.

A series of crosses is performed with fruit flies (Drosophila melanogaster) to examine inheritance of the genes vestigial (vg) and cinnabar (cn). The recessive vg allele causes small, malformed wings called vestigial wings. The recessive cn allele causes bright-red eyes called cinnabar eyes. In the first cross, a female with wild-type wings and eyes is mated with a male with vestigial wings and cinnabar eyes. All the F1 individuals have wild-type wings and eyes. In the second cross, female F1 flies are mated with males with vestigial wings and cinnabar eyes. The phenotypes of 500 F2 individuals are shown in the table wildtype wings, wild type eyes- 226 wildtype wings, cinnabar eyes- 25 vestigial wings, wildtype eyes- 26 vestigial wings, cinnabar eyes- 223 (A)The two genes are located on two different chromosomes. (B)The two genes are sex-linked. (C)The two genes are located on mitochondrial DNA. (D)The two genes are linked on an autosome.

The two genes are linked on an autosome.

Researchers studying cell cycle regulation in budding yeast have observed that a mutation in the CDC15 gene causes cell cycle arrest in telophase when the yeast cells are incubated at an elevated temperature. Which of the following statements best predicts the effect of the cell cycle arrest on proliferating yeast cells? (A)The yeast cells will transition out of G0G0 but will fail to complete the G1G1 phase. (B)The yeast cells will initiate mitosis but will fail to complete the G2G2 phase. (C)The yeast cells will replicate their chromosomes but will fail to complete cytokinesis. (D)The yeast cells will replicate their organelles but will fail to complete the S phase.

The yeast cells will replicate their chromosomes but will fail to complete cytokinesis. Because the cell cycle arrest occurs in telophase, the yeast cells will progress through the cell cycle, including replicating their chromosomes in the S phase, and will enter the M phase but will fail to complete mitosis and cytokinesis.

A genetic counselor is consulted by a young man who is worried about developing Huntington's disease, an inherited disorder caused by a dominant allele of a single gene. The young man explains that his cousin was recently diagnosed with Huntington's disease, and the news has caused him to consider his own risk of developing the disorder. Which of the following questions will best help the genetic counselor to evaluate the risk of the young man developing Huntington's disease and transmitting it to his children? (A)Were you and your cousin born in the same geographical area? (B)Were your parents or grandparents ever diagnosed with Huntington's disease? (C)Were you in physical contact with a person diagnosed with Huntington's disease? (D)Were you ever exposed to substances that are suspected of being mutagens?

Were your parents or grandparents ever diagnosed with Huntington's disease?

Figure 1 represents a metabolic process involving the regulation of lactose metabolism by E. coli bacteria. Lactose is utilized for energy by E. coli when glucose is not present. Allolactose is an isomer of lactose that is in the environment of these bacteria when lactose is present. The CAP site prevents the binding of RNA polymerase when glucose is present in the environment. The lacZ, lacY, and lacA genes code for proteins needed for lactose metabolism. Figure 1. Model of lac operon, comparing repressed and active states Which is a scientific claim that is consistent with the information provided and Figure 1 ? (A)The presence of excess lactose blocks the functioning of RNARNA polymerase in this operon. (B)When bound to the operator, the repressor protein prevents lactose metabolism in E. coli. (C)The binding of the repressor protein to the operator enables E. coli to metabolize lactose. (D)Allolactose acts as an inducer that binds to the operator, allowing E. coli to metabolize lactose.

When bound to the operator, the repressor protein prevents lactose metabolism in E. coli. When the repressor protein is bound to the operator, the lac operon is turned off and the genes for lactose metabolism cannot be transcribed.

When DNA replicates, each strand of the original DNA molecule is used as a template for the synthesis of a second, complementary strand. Which of the following figures most accurately illustrates enzyme-mediated synthesis of new DNA at a replication fork?

straight line one way, dotted line the other way

DNA replication occurs (A)during the S phase of the cell cycle (B)as the nuclear envelope breaks down in early mitosis (C)during mitosis but not during meiosis (D)in animal cells but not in plant cells (E)only in cells destined to become gametes

during the S phase of the cell cycle

For the following group of questions first, study the description of the situation or data and then choose the one best answer to each question following it and fill in the corresponding oval on the answer sheet. The diagram below illustrates the results of electrophoresis of DNA sequences obtained from a family of two adults and three children, and amplified using PCR. The bands represent short repeating sequences of variable length. Results for another female (X) are included for comparison. Father- 1, 2, 3 Mother- 1, 2, X Child 1- father, mother, 2, X Child 2- mother, father, 1, 3 Child 3- 2, father X- mother, 1, 2 The banding patterns of the DNA fragments reveal that (A)child 1 and child 2 cannot be biological siblings (B)child 1 and child 3 probably look like the mother (C)the mother cannot be the biological parent of all three children (D)the mother's DNA has the same DNA sequence as the father's DNA (E)child 2 and child 3 inherited all of their DNA from the father

the mother cannot be the biological parent of all three children