A-level-biology讲义6-Genetic-control

F Genetic control

Ⅰ Syllabus Contents

1 Structure and replication of DNA Structure: double helix

The basic unit is nucleotide. A nucleotide is made of a 5-carbon sugar(deoxyribose), a phophate and a nitrogen base. Four nitrogen bases in DNA are: A(adenine), C(cytosine), G(guanine) and T(thymine). A = T(2 hydrogen bonds) CG(3 hydrogen bonds)

Semi-conservative replication: the DNA double helix unzips, and each strand serves as a template for the formation of a new strand. The new molecules each consist of one old strand and one new strand. Direction: from 5' to 3' end.

2 Role of DNA in protein synthesis

Ⅱ Samples of A-LEVEL biology examination for analyzing

1 describe the structure of RNA and DNA and explain the importance of base pairing and the different hydrogen bonding between bases;

the structure of DNA and RNA Stand Unit Pentose sugar Phosphate group Nitrogen-containing base Base pairing DNA Double strands Nucleotide Deoxyribose Yes A,T,C,G A = T(2 hydrogen bonds) CG(3 hydrogen bonds) RNA Single strand Nucleotide Ribose Yes A,U,C,G A = U(2 hydrogen bonds) CG(3 hydrogen bonds)

Complementary pairing

Base complementary pairing is very important for DNA replication and transcription. It ensures the accuracy of genetic information transmission. The number of four kinds of different bases and their sequence in a DNA molecule represent some genetic information. During DNA semi-conserve replication or transcription, one old DNA strand act as a template, four kinds of bases are pairing by A pair with T in DNA replication or A pair with U in transcription and C pair with G. There are little error occurred during base pairing.

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Sample:

1.(9700_w11_qp_11,21) What makes the exact copying of DNA molecules possible? A base pairing

B hydrogen bonding between nucleotides

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C sugar-phosphate backbone D the double helix shape Key: A

Interpretion: Base complementary pairing is very important for DNA replication. It ensures the accuracy of genetic information transmission. The number of four kinds of different bases and their sequence in a DNA molecule represent some genetic information. During DNA semi-conserve replication, one old DNA strand act as a template, four kinds of bases are pairing by A pair with T in DNA replication and C pair with G. There are little error occurred during base pairing.

2 explain how DNA replicates semi-conservatively during interphase;

DNA molecules replicate during interphase. The hydrogen bonds between the bases break, allowing free nucleotides to fall into position opposite their complementary ones on each strand of the original DNA molecule. Adjacent nucleotides are then linked, through their phosphates and sugars, to form new strands. Two complete new molecules are thus formed from one old one, each new molecule containing one old strand and one new.

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Sample:

1.(9700_w11_qp_11,23) Bacteria were grown in a medium containing 15N. After several generations, all of the DNA contained 15N. Some of these bacteria were transferred to a medium containing the common isotope of nitrogen, 14N. The bacteria were allowed to divide once. The DNA of some of these bacteria was extracted and analysed. This DNA was all hybrid DNA containing equal amounts of 14N and 15N. The remaining bacteria were left in the medium with 14N and allowed to divide one more time. The DNA of some of these bacteria was extracted and analysed. What is the composition of this DNA? A 25 % hybrid DNA B 50 % hybrid DNA C 75 % hybrid DNA D 100 % hybrid DNA Key: B

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Interpretion:

In the first DNA replication, one old DNA strand contains 15N act as template and replicates a new DNA strand contain 14N, so does the other old one. As a result, one DNA molecule contain 15N is divided into two DNA molecules contain both 15N and 14N, namely hybrid DNA.

In the second DNA replication, one old DNA strand contains 15N act as template and replicates a new DNA strand contain 14N, formed a hybrid DNA. However, the other old DNA strand contains 14N and replicates a new DNA strand also contain 14N, formed a non-hybrid DNA.

In other words, only 50% DNA produced in the second DNA replication are hybrid DNA.

3 state that a gene is a sequence of nucleotides as part of a DNA molecule, which codes for a polypeptide and state that a mutation is a change in the sequence that may result in an altered polypeptide;

The sequence of bases on a DNA molecules codes for the sequence of amino acid in a protein or polypeptide. Each amino acid is coded for by three bases. A length of DNA coding for one complete protein or polypeptide is a gene.

Mutation can be defined as an unpredictable change in the base sequence in a DNA molecule(gene mutation) or in the structure or number of chromosomes (chromosome mutation). New alleles arise by gene mutation. Gene mutation include base substitution, deletions or additions.

Base additions or deletions usually have a very significant effect on the structure, and therefore the function, of the polypeptide that the allele codes for. Base additions or deletions always have large effects, because they alter every set of three bases that 'follows' them in the DNA molecule. They are said to cause frame shifts in the code. Often, the effects are so large that the protein that is made is totally useless. Or they may introduce a 'stop' triplet part way through a gene, so that a complete protein is never made at all.

Base substitutions, on the other hand, often have no effect at all because many amino acids have more than one triplet code, so even if one base is changed the same amino acid is still coded for. However, base substitutions can have very large effects. If, for example, the base sequence ATG mutated to ATT, this has produced a 'stop'

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triplet, so the synthesis of the protein would stop at this point. Sample:

1.(9700_w11_qp_11,22) Which molecule has its synthesis directly controlled by DNA? A amylase B cholesterol C glycogen D phospholipid Key: A

Interpretion: amylase is protein and protein synthesis is controlled by DNA.

4 describe the way in which the nucleotide sequence codes for the amino acid sequence in a polypeptide with reference to the nucleotide sequence for HbA (normal) and HbS (sickle cell) alleles of the gene for the β-haemoglobin polypeptide;

The haemoglobin molecule, which is made up of two α chains and two βchains, is nearly spherical.

The gene which codes for the amino acid sequence in the βchains is not the same in everyone. In most people, the βchains begin with the amino acid sequence:

Val-His-Leu-Thr-Pro-Glu-Glu-Lys

But in some people, the base sequence CTT is replaced by CAT, and the amino acid sequence becomes:

Val-His-Leu-Thr-Pro-Val-Glu-Lys

This small difference in the amino acid sequence makes little difference to the haemoglobin molecule when it is combined with oxygen. But when it is not combined with oxygen, the 'unusual' βchains make the haemoglobin molecule much less soluble. The molecules tend to stick to each other, forming long fibres inside the red blood cells. The red cells are pulled out of shape, into a half-moon or sickle shape. Sickle cell anaemia is a severe hereditary disease and may be fatal.

HbA (normal) and HbS (sickle cell)

(left is HbA, right is HbS)

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Sample:

1.(9700_s10_qp_11,23) What is the minimum number of base substitutions required to change the nucleotide sequence of the HbA (normal) allele to the HbS (sickle cell) allele? A 1 B 2 C 3 D 4 Key: A

Interpretion: the base sequence CTT is replaced by CAT, and the amino acid Glu becomes Val.

5 describe how the information on DNA is used during transcription and translation to construct polypeptides, including the role of messenger RNA (mRNA), transfer RNA (tRNA) and the ribosomes; Transcription:

During protein synthesis, a complementary copy of the base sequence on a gene is made, by building a molecule of mRNA against one DNA strand. The mRNA then moves to a ribosome in the cytoplasm.

Translation:

tRNA molecules with complementary triplets of bases temporarily pair with the base triplets on mRNA, bring appropriate amino acids. As two amino acids are held side by side, a peptide bond forms between them. The ribosome moves along the mRNA molecule, so that appropriate amino acids are gradually linked together, following the sequence laid down by the base sequence on the mRNA.

Types and roles of RNA

mRNA: copy the genetic information stored in the DNA. rRNA: components of ribosomes.

Role of ribosomes: site of protein or polypeptide synthesis.

tRNA: transfer the amino acids to the ribosome in the translation.

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Sample:

1.(9700_s11_qp_11,22) The following events occur during transcription. 1 Bonds break between complementary bases. 2 Bonds form between complementary bases. 3 Sugar-phosphate bonds form.

4 Free nucleotides pair with complementary nucleotides.

Before the mRNA leaves the nucleus, which events will have occurred twice? A 1 and 2 only B 1, 3 and 4 only C 2, 3 and 4 only D 1, 2, 3 and 4 Key: A

Interpretion: the first of bonds break between complementary bases occurs when DNA double-strand unwind. The second of bonds break between complementary bases occurs when mRNA separates from one DNA strand. The first of bonds form between complementary bases occurs when mRNA is produced in transcription. The second of bonds form between complementary bases occurs when DNA recovers to wind after mRNA

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separates from one DNA strand.

Ⅲ Practices

1.(9700_w11_qp_11,20) In a genetic engineering experiment a piece of double-stranded DNA containing 6000 nucleotides coding for a specific polypeptide is transcribed and translated. What is the total number of amino acids in this polypeptide? A 500 B 1000 C 2000 D 3000

2.(9700_s11_qp_11,21) The mechanism of action of four drugs that inhibit DNA replication is stated below.

● Aphidicholine inhibits DNA polymerase.

● Cytarabine is converted into a molecule that can substitute for a DNA nucleotide and also inhibits DNA repair mechanisms.

● Epirubicin inhibits an enzyme involved in the unwinding of DNA and separation of strands. ● Hydroxycarbamide inhibits an enzyme involved in the production of deoxyribonucleotides. Which row correctly matches a drug to an explanation of the mechanism of action? explanation of mechanism of action decreased pool of DNA strands not DNA damaged exposed DNA available available as during replication template strands nucleotides inhibits templates for and cell death unable to be copied chain elongation transcription occurs A aphidicholine epirubicin cytarabine hydroxycarbamide B epirubicin cytarabine hydroxycarbamide aphidicholine C hydroxycarbamide aphidicholine epirubicin cytarabine D hydroxycarbamide epirubicin cytarabine aphidicholine

3.(9700_s11_qp_11,23) Which type of sugar and types of bonds are found in a DNA molecule? type of sugar types of bonds A non-reducing hydrogen and ionic B non-reducing hydrogen and peptide C Reducing covalent and hydrogen D reducing hydrogen and peptide 4.(9700_w10_qp_11,20) DNA was extracted from the salivary glands of a fruit fly and a human cheek cell. In which way did the DNA molecules differ? A in the ratio of adenine to thymine B in the sequence of the nucleotides C in the type of pentose sugar D in the types of nucleotide

5.(9700_w10_qp_11,21) Which statement describes the semi-conservative replication of DNA? A Parental DNA is broken down into nucleotides and reassembled with new nucleotides. B Parental DNA is split into triplets and new triplets are added.

C Parental DNA is split into two strands, each of which is replicated.

D Parental DNA remains intact and a new daughter DNA copy is built from new nucleotides.

6.(9700_w10_qp_11,22) The table shows the percentages of nitrogenous bases in four samples of nucleic

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acids.

Which base is adenine? sample percentage of nitrogenous bases A B C D uracil 1 19 31 30 19 Nil 2 27 23 24 26 Nil 3 25 25 Nil 25 25 4 17 32 33 18 Nil

7.(9700_w10_qp_11,23) The table shows the role of four different proteins involved in DNA replication.

single-strand DNA protein helicase topoisomerase binding protein polymerase binds to unwinds the breaks and synthesises separated role parental DNA rejoins strand DNA strands to double helix the DNA strands of DNA stabilise them Which shows the function of these proteins?

8.(9700_s10_qp_11,21) What would be the result of analysing part of a DNA molecule? A hexose sugars and phosphates in equal proportion, and an equal number of cytosine and guanine bases

B nucleotides and phosphates in equal proportion, and an equal number of adenine and cytosine bases

C pentose sugars and phosphates in equal proportion, and an equal number of adenine and thymine bases

D twice as many phosphates as pentose sugars, and an equal number of adenine and guanine

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Bases

9.(9700_s10_qp_11,22) DNA is said to replicate in a semi-conservative way.

Results of Meselson and Stahl’s experiments gave overwhelming support to this theory. They used E. coli which has a generation time of 50 minutes.

Here are the steps in their experiment but they are in the wrong order. P All bacteria contain 15N DNA.

Q All bacteria contain hybrid DNA (15N DNA and 14N DNA). R Bacteria contain either all 14N DNA or hybrid DNA. S Bacteria grown in a 15N medium for many generations.

T Bacteria transferred to a 14N medium and sampled every 50 minutes.

Which sequence of letters shows the correct order of the steps in the experiment? A P → Q → R → S → T B P → S → T → R → Q C S → P → T → Q → R D S → R → Q → P → T

10.(9700_s10_qp_11,24) In a DNA molecule, the base sequence AGT codes for the amino acid serine. What is the base sequence of the anti-codon on the tRNA to which serine becomes attached? A AGU B GAU C TCA D UCA

11.(9700_s10_qp_21,5 ) (a) Name the stage during the mitotic cell cycle when replication of DNA occurs. ..................................................................................................................................... [1] (b) Fig. 5.1 shows details of DNA replication.

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(i) Name the bonds shown by the dashed lines on Fig. 5.1.

.............................................................................................................................. [1] (ii) Name the nitrogenous bases, M and O.

M .............................................................................................................................. O .......................................................................................................................... [1] (c) Explain why DNA replication is described as semi-conservative.

......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ..................................................................................................................................... [2] (d) The enzyme that catalyses the replication of DNA checks for errors in the process and corrects them. This makes sure that the cells produced in mitosis are genetically identical.

Explain why checking for errors and correcting them is necessary.

......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... .........................................................................................................................................

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..................................................................................................................................... [2] [Total: 7]

Key and interpretion

1.key: B

Interpretion: each DNA strand has 60002＝3000nucleotides. The total number of amino acids is 30003＝1000 because only one DNA strand is translated into the polypeptide. 2.Key: D Interpretion:

①Hydroxycarbamide inhibits an enzyme involved in the production of deoxyribonucleotides. As a result, there are less deoxyribonucleotides could be used for DNA chain elongation.

②Epirubicin inhibits an enzyme involved in the unwinding of DNA and separation of strands. DNA cannot unwind and transcription cannot occur.

③Cytarabine is converted into a molecule that can substitute for a DNA nucleotide and also inhibits DNA repair mechanisms. Therefore DNA replication will be disturbed and cell will die.

④Aphidicholine inhibits DNA polymerase. Exposed DNA template strands are unable to be copied

without DNA polymerase because it can help link the sugar and innermost phosphate groups of next-door nucleotides together. 3.Key: C

Interpretion: Deoxyribose is reducing sugar. Reducing sugars also contain glucose, fructose, galactose, lactose and maltose.

4.Key: B

Interpretion: different species' genetic information is different because their DNA molecules have different sequences of the nucleotides and the numbers of four nitrogen-containing bases. The ratios of adenine to thymine are both 1 for A pairing with T and their numbers are equal. The type of pentose sugar in all kinds of DNA molecules is all deoxyribose. The types of nucleotides in all kinds of DNA molecules are all four kinds, including Adenine deoxyribose nucleotide, thymine deoxyribose nucleotide, cytosine deoxyribose nucleotide and guanine deoxyribose nucleotide.

5.Key: C

Interpretion: Semi-conservative replication: the DNA double helix unzips, and each strand serves as a template

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for the formation of a new strand. The new molecules each consist of one old strand and one new strand. 6.Key: B

Interpretion: Sample 3 has uracil and has nil C, so it is a mRNA strand and C is thymine. In DNA double-strand molecules, the number of adenine is equal to that of thymine. Therefore we can infer that B is adenine because the number of B is nearly equal to that of C in sample 1,2 and 4. 7.Key: D Interpretion:

①helicase: unwind the double helix into two strands.

②topoisomerase: cut and rejoin the helix to prevent tangling.

③single-strand binding protein: help unwinding DNA strands not to recoil together.

④DNA polymerase: catalyze the elongation of the new DNA strands. Add the nucleotides alongside the naked strand of DNA. 8.Key: C

Interpretion: the unit of DNA molecule is nucleotide. A nucleotide is made of a 5-carbon sugar(deoxyribose), a phophate and a nitrogen base. Four nitrogen bases in DNA are: A(adenine), C(cytosine), G(guanine) and T(thymine). The number of A is equal to T and that of C is equal to G. 9.Key: C 10.Key: A

Interpretion: AGT in DNA→UCA in mRNA→AGU in tRNA. 11.Key:

(a) Interphase

(b) (i) hydrogen bond

(ii)M is adenine ; O is cytosine.

(c)Semi-conservative replication: the DNA double helix unzips, and each strand serves as a template for the formation of a new strand. The new molecules each consist of one old strand and one new strand.

(d) there may be errors in the process of DNA replication and lead to mutation. Mutation can be defined as an unpredictable change in the base sequence in a DNA molecule(gene mutation) or in the structure or number of chromosomes (chromosome mutation). New alleles arise by gene mutation. Gene mutation include base substitution, deletions or additions.

Mutation may lead to altered proteins produced and impair the function of proteins. For example, there may be altered antigen produced, which will make one own cells are rejected by immune system and lead to autoimmunity disease. Another example is that sickle cell anaemia, which is caused by the base sequence CTT replaced by CAT. Although only one thymine is replaced by adenine, the whole 3D structure of haemoglobin is changed and sickle cell anaemia is very severe disease which may be fatal. In conclusion, checking for errors occurred during DNA replication and correcting them is very necessary.

Mark scheme:

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