16.2. The roles of genes in determining the phenotype
A subsection of Biology, 9700, through 16. Inheritance
Listing 10 of 116 questions
Coat colour in cats is determined by a sex-linked gene with two alleles coding for black and orange. When black cats are mated with orange cats: • the female offspring are always tortoiseshell (black and orange patches) • the male offspring are always the same colour as their mother. Explain what is meant by a sex-linked gene. sex-linked gene Using the symbols B for the allele for black coat and O for the allele for orange coat, complete the genetic diagram below. parental phenotypes tortoiseshell female parental genotypes gametes offspring genotypes offspring phenotypes black male Explain why a male cat cannot have a tortoiseshell coat.
9700_s13_qp_41
THEORY
2013
Paper 4, Variant 1
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Resistance to the poison warfarin is now extremely common in rats. Warfarin inhibits an enzyme in the liver, vitamin K epoxide reductase, that is necessary for the recycling of vitamin K. This vitamin is involved in the production of substances required for blood clotting. • Rats susceptible to warfarin die of internal bleeding. • Rats that are homozygous for resistance to warfarin do not suffer from internal bleeding when their diet provides more than 70 μg of vitamin K per kg body mass per day. • Heterozygous rats are resistant to warfarin when their diet provides about 10 μg of vitamin K per kg body mass per day. Using appropriate symbols, complete the genetic diagram to show how two resistant rats can produce warfarin-susceptible offspring. key to symbols parental phenotypes resistant male parental genotypes gametes offspring genotypes offspring phenotypes resistant female Rats that are homozygous for warfarin resistance have a low survival rate in the wild. Suggest why this is so. Warfarin can be safely given to humans who are at risk of unwanted blood clots. The clotting time of the blood is measured regularly and the warfarin dose is varied accordingly. Suggest, giving a reason, the type of inhibition warfarin has on the enzyme vitamin K epoxide reductase. type of inhibition reason The allele for warfarin resistance may have originated by a single base substitution and resulted in a modified vitamin K epoxide reductase. Explain how a single base substitution may affect the phenotype of an organism.
9700_s13_qp_42
THEORY
2013
Paper 4, Variant 2
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Coat colour in cats is determined by a sex-linked gene with two alleles coding for black and orange. When black cats are mated with orange cats: • the female offspring are always tortoiseshell (black and orange patches) • the male offspring are always the same colour as their mother. Explain what is meant by a sex-linked gene. sex-linked gene Using the symbols B for the allele for black coat and O for the allele for orange coat, complete the genetic diagram below. parental phenotypes tortoiseshell female parental genotypes gametes offspring genotypes offspring phenotypes black male Explain why a male cat cannot have a tortoiseshell coat.
9700_s13_qp_43
THEORY
2013
Paper 4, Variant 3
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Deer mice, Peromyscus maniculatus, are small rodents that live in North America. Like all mammals, their blood contains haemoglobin which combines with oxygen in the lungs, and unloads its oxygen in respiring tissues. Deer mice show variation in their genotypes for the genes that code for the α-polypeptide chain of haemoglobin. In most populations of deer mice, the majority of individuals have the genotype A1A1, while a smaller number have the genotype A0A0. In mice with the genotype A1A1, the amino acid at position 64 in the α-polypeptide chain is aspartic acid. In mice with the genotype A0A0, the amino acid at this position is glycine. Suggest how the change from aspartic acid to glycine in the α-polypeptide chain could have been brought about. The genotypes of deer mice from three different populations, each living at a different altitude, were analysed. shows the relative proportions of deer mice with aspartic acid (white areas) and glycine (black areas) at position 64 in the α-polypeptide of their haemoglobin. altitude 4347 m 1158 m 620 m Describe the effect of altitude on the frequency of the haemoglobin alleles in these populations of deer mice. The partial pressure of oxygen is relatively low at high altitudes. Haemoglobin containing glycine at position 64 in the α-polypeptide chain has a higher affinity for oxygen than haemoglobin with aspartic acid at this position. Suggest how natural selection could account for the difference in allele frequency in deer mice living at high altitudes and low altitudes.
9700_s14_qp_41
THEORY
2014
Paper 4, Variant 1
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Myostatin is a protein that is produced in mammalian skeletal muscle cells. It circulates in the blood and acts on muscle tissue to slow down further differentiation and growth. In thoroughbred racehorses, a mutation involving the substitution of a single nucleotide has been identified in the MSTN gene which codes for myostatin. At the site of this mutation, the DNA nucleotide has either a cytosine (C) base or a thymine (T) base, giving race horses three possible genotypes for this mutation: CC, CT or TT. At two years of age, racehorses with the MSTN CC genotype have greater muscle mass than those with the TT genotype. Suggest an explanation for this difference. Racehorses that had won races of different distances were tested to determine their MSTN genotype. The results are shown in . genotype CC percentage of winners with each genotype 1.0 1.2 1.4 1.6 1.8 race distance / km 2.6 2.0 2.2 2.4 CT TT With reference to , describe the effect of the MSTN genotype on the ability of racehorses to win races of different lengths. Modern thoroughbred racehorses are the result of many years of artificial selection. Explain: what is meant by artificial selection how genetic tests for the MSTN genotype can help in the selective breeding of racehorses.
9700_s14_qp_42
THEORY
2014
Paper 4, Variant 2
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Deer mice, Peromyscus maniculatus, are small rodents that live in North America. Like all mammals, their blood contains haemoglobin which combines with oxygen in the lungs, and unloads its oxygen in respiring tissues. Deer mice show variation in their genotypes for the genes that code for the α-polypeptide chain of haemoglobin. In most populations of deer mice, the majority of individuals have the genotype A1A1, while a smaller number have the genotype A0A0. In mice with the genotype A1A1, the amino acid at position 64 in the α-polypeptide chain is aspartic acid. In mice with the genotype A0A0, the amino acid at this position is glycine. Suggest how the change from aspartic acid to glycine in the α-polypeptide chain could have been brought about. The genotypes of deer mice from three different populations, each living at a different altitude, were analysed. shows the relative proportions of deer mice with aspartic acid (white areas) and glycine (black areas) at position 64 in the α-polypeptide of their haemoglobin. altitude 4347 m 1158 m 620 m Describe the effect of altitude on the frequency of the haemoglobin alleles in these populations of deer mice. The partial pressure of oxygen is relatively low at high altitudes. Haemoglobin containing glycine at position 64 in the α-polypeptide chain has a higher affinity for oxygen than haemoglobin with aspartic acid at this position. Suggest how natural selection could account for the difference in allele frequency in deer mice living at high altitudes and low altitudes.
9700_s14_qp_43
THEORY
2014
Paper 4, Variant 3
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The fruit fly, Drosophila melanogaster, has eyes, a striped abdomen and wings longer than its abdomen. This is called a ‘wild-type’ fly. Mutation has resulted in many variations of these features. Table 6.1 shows diagrams of a wild-type fly and three other flies, each of which shows one recessive mutation. Table 6.1 H\HV DEGRPHQ ZLQJGHVFULSWLRQ present striped long present black long absent striped long present striped short Using appropriate symbols, complete the genetic diagram below. symbols parental phenotypes with eyes black abdomen X no eyes striped abdomen parental genotypes gametes offspring genotypes offspring phenotypes with eyes black abdomen no eyes black abdomen with eyes striped abdomen no eyes striped abdomen State how you would carry out a test cross. A cross was carried out between a fly heterozygous for striped abdomen and long wings and a fly with a black abdomen and short wings. The results are shown below in Table 6.2. Table 6.2 offspring number striped abdomen long wing black abdomen long wing striped abdomen short wing black abdomen short wing total A chi-squared test (χ2) was carried out on these data. Complete Table 6.3 and calculate the value of χ2. Table 6.3 observed number (O) expected number (E) O − E (O − E)2 (O − E)2 E χ2 = Σ (O − E)2 E Σ = sum of χ2 Table 6.4 shows χ2 values. Table 6.4 degrees of freedom probability 0.50 0.20 0.10 0.05 0.02 0.01 0.001 2.37 4.64 6.25 7.82 9.84 11.34 16.27 Using Table 6.4, explain what conclusions can be made about the results of the χ2 test.
9700_s16_qp_41
THEORY
2016
Paper 4, Variant 1
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The fruit fly, Drosophila melanogaster, has eyes, a striped abdomen and wings longer than its abdomen. This is called a ‘wild-type’ fly. Mutation has resulted in many variations of these features. Table 6.1 shows diagrams of a wild-type fly and three other flies, each of which shows one recessive mutation. Table 6.1 H\HV DEGRPHQ ZLQJGHVFULSWLRQ present striped long present black long absent striped long present striped short Using appropriate symbols, complete the genetic diagram below. symbols parental phenotypes with eyes black abdomen X no eyes striped abdomen parental genotypes gametes offspring genotypes offspring phenotypes with eyes black abdomen no eyes black abdomen with eyes striped abdomen no eyes striped abdomen State how you would carry out a test cross. A cross was carried out between a fly heterozygous for striped abdomen and long wings and a fly with a black abdomen and short wings. The results are shown below in Table 6.2. Table 6.2 offspring number striped abdomen long wing black abdomen long wing striped abdomen short wing black abdomen short wing total A chi-squared test (χ2) was carried out on these data. Complete Table 6.3 and calculate the value of χ2. Table 6.3 observed number (O) expected number (E) O − E (O − E)2 (O − E)2 E χ2 = Σ (O − E)2 E Σ = sum of χ2 Table 6.4 shows χ2 values. Table 6.4 degrees of freedom probability 0.50 0.20 0.10 0.05 0.02 0.01 0.001 2.37 4.64 6.25 7.82 9.84 11.34 16.27 Using Table 6.4, explain what conclusions can be made about the results of the χ2 test.
9700_s16_qp_43
THEORY
2016
Paper 4, Variant 3
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Questions Discovered
116