1. DNA was the primary source of heritable information.
2. RNA was the primary source of heritable information.
3. Protein was the primary source of heritable information.
4. Phages were the primary source of heritable information.

1. The precipitate contained sulfur.
2. The precipitate contained oxygen.
3. The precipitate contained phosphorus.
4. The precipitate contained protein.

1. Comparison of DNA from a known source or individual with analysis of the sequence of an unknown sample of DNA allows scientists to find out if both of them are similar or not.
2. DNA from the unknown sample is sequenced and analyzed. The result of the analysis is then matched with any random population. The matching individual then helps in forensics.
3. Comparison of DNA from a known source or individual with analysis of the sequence of bases in strands of an unknown sample of RNA allows scientists to find out if both of them are similar or not.
4. Comparison of DNA from a known source or individual with analysis of the sugars and phosphates in strands of an unknown sample of DNA allows scientists to find out if both of them are similar or not.

1. Rosalind Franklin used X-ray diffraction methods to demonstrate the helical nature of DNA, while Watson and Crick formulated the double stranded structural model of DNA.
2. Rosalind Franklin, Watson and Crick first employed the technique of X-ray diffraction to understand the storage of DNA. Since it did not work out, Watson and Crick then ran experiments to ascertain the DNA structure.
3. Rosalind Franklin, Watson and Crick used X-ray diffraction methods to demonstrate the helical nature of DNA, while Rosalind Franklin formulated the double stranded structural model of DNA.
4. Watson and Crick used X-ray diffraction methods to demonstrate the helical nature of DNA, while Rosalind Franklin formulated the double stranded structural model of DNA.

1. Both contain four different nucleotides.
2. Both are usually double-stranded molecules.
3. Both contain adenine and uracil.
4. Both contain ribose.

1. to make copies of DNA
2. to isolate a change in a single nucleotide
3. to separate DNA fragments
4. to sequence DNA

Explain how the components of DNA fit together.

1. DNA is composed of nucleotides, consisting of a five-carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with thymine and guanine pairs with cytosine. Adenine and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other.
2. DNA is composed of nucleotides, consisting of a five-carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form two hydrogen bonds and guanine and thymine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other.
3. DNA is composed of nucleotides, consisting of a five-carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other.
4. DNA is composed of nucleotides, consisting of a five-carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run parallel to each other.

Describe the Sanger DNA sequencing method used for the human genome sequencing project.

1. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is added to the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPs.
2. A DNA sample is denatured by heating and then put into four tubes. A primer, RNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is added to the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPs.
3. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is removed from the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPs.
4. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four deoxynucleotides (dNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent dNTP is added the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different dNTPs.

What process is illustrated in the figure?

1. transcription
2. mutation
3. excision
4. translation

1. Topoisomerase relieves the pressure that results from supercoiling by breaking and reforming DNA’s phosphate backbone ahead of the replication fork.
2. Topoisomerase increases the pressure to increase supercoiling by breaking and reforming DNA’s phosphate backbone ahead of the replication fork.
3. Topoisomerase relieves the pressure that results from supercoiling by breaking and reforming DNA’s nucleotide base pairs ahead of the replication fork.
4. Topoisomerase relieves the pressure that results from separation of DNA strands by breaking and reforming DNA’s phosphate backbone ahead of the replication fork .

1. weather variations
2. dietary changes
3. DNA mutations
4. translation failure

What can be the result of DNA failing to undergo repair after too much UV exposure?

1. second-degree burns
2. a malignant melanoma
3. a breakdown of deep layers of the skin
4. a sunburn

1. substitution
2. translocation
3. missense
4. transversion

Explain why patients with Xeroderma pigmentosa are more prone to cancer than the rest of the population.

1. Xeroderma pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, thymine dimers are formed and they are not able to repair this defect. These dimers distort the structure of DNA and cause them to have a high risk of contracting skin cancer.
2. Xeroderma pigmentosa patients can employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the thymine dimers are formed and they are able to repair this defect. These dimers do not distort the structure of DNA and they have moderate risk of contracting skin cancer.
3. Xeroderma pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the adjacent adenine forms dimers and they are not able to repair this defect. These dimers distort the structure of DNA and they have high risk of contracting skin cancer.
4. Xeroderma pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the adjacent thymine cannot form thymine dimers and they are not able to repair this defect. The non-formation of dimers distorts the structure of DNA and they have high risk of contracting skin cancer.

1. The fragment exposed to UV light contains thymine dimers. Thymines lying adjacent to each other can form thymine dimers when exposed to UV light. They can be repaired by nucleotide excision.
2. The fragment exposed to UV light contains adenine dimers. Adenines lying adjacent to each other can form dimers when exposed to UV light. They can be repaired by nucleotide excision.
3. The fragment exposed to UV light contains thymine dimers. Thymines lying parallel to each other can form thymine dimers when exposed to UV light. They can be repaired by nucleotide excision.
4. The fragment exposed to UV light contains thymine dimers. Thymines lying adjacent to each other can form thymine dimers when exposed to UV light. They can be synthesized by nucleotide excision.

1. Substitution mutations may cause a different amino acid to be placed at a specific location, causing small changes in the protein. Frameshift mutations usually cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring.
2. Substitution mutations may cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. Frameshift mutations may cause a different amino acid to be placed at a specific location, causing small changes in a protein.
3. Substitution mutations may cause a different amino acid to be placed at a specific location, resulting in major changes to the protein and leading to radically different characteristics in the offspring. Frameshift mutations cause multiple amino acid differences in a protein, leading to small changes in the protein.
4. Substitution mutations result in a different amino acid being placed at a specific position in a protein, causing small changes. Silent mutations could result in new characteristics possessed by an offspring when a stop codon is substituted for an amino acid.