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2.1 The sequence of amino acids in a protein molecule is coded for by DNA and RNA - NSC Life Sciences - Question 2 - 2017 - Paper 2

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2.1 The sequence of amino acids in a protein molecule is coded for by DNA and RNA. The table below shows some mRNA codons and the corresponding amino acids. mRNA CO... show full transcript

Worked Solution & Example Answer:2.1 The sequence of amino acids in a protein molecule is coded for by DNA and RNA - NSC Life Sciences - Question 2 - 2017 - Paper 2

Step 1

2.1.1 According to the table, how many codons code for phenylalanine?

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Answer

According to the table, there are two codons that code for phenylalanine: UUC and UUU.

Step 2

2.1.2 What is the anticodon for glutamic acid?

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Answer

The anticodon for glutamic acid, which is coded by the mRNA codon GAG, is CUC.

Step 3

2.1.3 (a) DNA base triplet for the LAST codon on this section of mRNA

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Answer

The LAST codon on this section of mRNA is ACC. The corresponding DNA base triplet would be TGG.

Step 4

2.1.3 (b) FIRST amino acid coded for by this section of mRNA

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Answer

The FIRST amino acid coded for by this section of mRNA, GAC, is aspartate.

Step 5

2.1.4 (a) Describe the mutation that occurred.

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Answer

The mutation that occurred involved a change in the fourth codon from AGC to AGU, indicating a substitution of C with U.

Step 6

2.1.4 (b) Explain the effect that the mutation described in QUESTION 2.1.4(a) will have on the resulting protein.

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Answer

Since both AGC and AGU code for the same amino acid, serine, there will be no effect on the resulting protein.

Step 7

2.1.5 Name and describe the process occurring in the nucleus which results in the formation of an mRNA molecule.

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Answer

The process is called transcription. During transcription, the DNA double helix unwinds, and the DNA strands serve as a template for synthesizing mRNA, which is complementary to the DNA sequence. This process is controlled by specific enzymes.

Step 8

2.2.1 Define a population.

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A population is a group of organisms of the same species that can interbreed and occupy a defined area at a certain time.

Step 9

2.2.2 Other than mutations, give THREE causes of variation in a population.

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Three causes of variation in a population are:

  1. Crossing over during meiosis.
  2. Random arrangement of chromosomes.
  3. Random fertilization.

Step 10

2.2.3 Explain why there were eventually more squirrels with the mutation on one side of the river.

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Answer

The squirrels with the advantageous mutation survived better in their environment, leading to an increase in their population over time due to natural selection.

Step 11

2.2.4 Explain what effect the process above has on the biodiversity in this ecosystem.

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Answer

As new species arise, the biodiversity in the ecosystem increases. With two separate species, there is greater variety of genetic traits and interactions within the environment.

Step 12

2.2.5 Describe what can be done to confirm that the squirrels belong to two different biological species.

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To confirm the two squirrels belong to two different species, they can be allowed to interbreed and checked for the production of fertile offspring. Alternatively, DNA tests can be conducted to compare both species.

Step 13

2.3.1 (a) Give the Phenotype of Senzo.

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Answer

Senzo's phenotype is colour-blind male.

Step 14

2.3.1 (b) Give the Genotype of Thuli.

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Thuli's genotype is X^X.

Step 15

2.3.2 Describe how Linda inherited colour-blindness.

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Answer

Linda inherited one recessive allele (X^d) from her father and one normal allele (X^X) from her mother, resulting in her being a carrier.

Step 16

2.3.3 Explain why there are generally more males than females with colour-blindness in a population.

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Answer

Males are more likely to be colour-blind because they have one X chromosome. If that X chromosome carries the recessive allele (X^d), they will express colour-blindness. In contrast, females need two copies of the recessive allele to be colour-blind.

Step 17

2.3.4 Use a genetic cross to show the possible genotypes and phenotypes of their children.

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Using the normal male genotype (X^Y) and the carrier female genotype (X^X^d):

P1: Normal Female (X^X^d) x Normal Male (X^Y) Gametes: X^d, X^ x X, Y

Children genotypes will be: X^X (normal female), X^Y (normal male), X^dY (colour blind male).

Hence, possible phenotypes: Normal females, Normal males, Colour-blind males.

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