16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
The Labrador retriever is a modern breed of dog that can have yellow, black or brown fur and pale, black or brown noses. The inheritance of fur and nose colour is the result of the interaction between genes at two different loci, the B locus and the E locus. shows a Labrador retriever. Table 1.1 shows how gene interaction results in different phenotypes. Table 1.1 alleles at B locus alleles at E locus phenotype B_ ee yellow fur black nose bb ee yellow fur pale nose B_ E_ black fur black nose bb E_ brown fur brown nose A male Labrador retriever, heterozygous at the B locus and homozygous recessive at the E locus, was mated with a female Labrador retriever heterozygous at both loci. Explain the terms locus and homozygous. locus homozygous Use a genetic diagram to show the possible genotypes and phenotypes of the offspring from the mating between the two Labrador retrievers. parental phenotypes parental genotypes gametes offspring genotypes and phenotypes
9700_w16_qp_43
THEORY
2016
Paper 4, Variant 3
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In the fruit fly, Drosophila melanogaster, two different genes control body colour and eye colour. • G/g are alleles of the body colour gene. • G results in grey body, g results in black body. • R/r are alleles of the eye colour gene. • R results in red eyes, r results in brown eyes. Each gene is autosomal. A dihybrid cross was carried out using a fly with a grey body and red eyes crossed with a fly with a black body and brown eyes. Both parents were homozygous for both genes. The offspring from the F1 generation were crossed to obtain the F2 offspring. A statistical test showed that the results of the cross were significantly different from those expected. State the name of the statistical test used and state the expected phenotypic ratio for the F2 generation. statistical test expected ratio A test cross can be carried out in order to identify flies from an F2 generation that are heterozygous for both genes. Draw a genetic diagram to show how a test cross between a heterozygous grey‑bodied, red‑eyed F2 fly and a fly with a black body and brown eyes can produce four different offspring phenotypes. Use the symbols G/g and R/r. The results of the test cross in are shown in Table 2.1. These results are significantly different from the expected results. Table 2.1 phenotypes of offspring of test cross number of individuals grey body, red eyes grey body, brown eyes black body, red eyes black body, brown eyes Describe how these results are different from the expected results and explain why they are different.
9700_w18_qp_41
THEORY
2018
Paper 4, Variant 1
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In the fruit fly, Drosophila melanogaster, two different genes control body colour and eye colour. • G/g are alleles of the body colour gene. • G results in grey body, g results in black body. • R/r are alleles of the eye colour gene. • R results in red eyes, r results in brown eyes. Each gene is autosomal. A dihybrid cross was carried out using a fly with a grey body and red eyes crossed with a fly with a black body and brown eyes. Both parents were homozygous for both genes. The offspring from the F1 generation were crossed to obtain the F2 offspring. A statistical test showed that the results of the cross were significantly different from those expected. State the name of the statistical test used and state the expected phenotypic ratio for the F2 generation. statistical test expected ratio A test cross can be carried out in order to identify flies from an F2 generation that are heterozygous for both genes. Draw a genetic diagram to show how a test cross between a heterozygous grey‑bodied, red‑eyed F2 fly and a fly with a black body and brown eyes can produce four different offspring phenotypes. Use the symbols G/g and R/r. The results of the test cross in are shown in Table 2.1. These results are significantly different from the expected results. Table 2.1 phenotypes of offspring of test cross number of individuals grey body, red eyes grey body, brown eyes black body, red eyes black body, brown eyes Describe how these results are different from the expected results and explain why they are different.
9700_w18_qp_43
THEORY
2018
Paper 4, Variant 3
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shows a cat displaying the recessive phenotype for two unlinked genes. • H/h controls hair length. The allele for short hair is dominant to the allele for long hair. • R/r controls coat pattern. The allele for ‘full colour’, with pigment on all parts of the body, is dominant to the allele for ‘pointed’, where the pigment is restricted to the ears, face, paws and tail. Draw a genetic diagram to predict the offspring genotypes and phenotypes when the cat in is crossed with a cat that is heterozygous for the hair length gene and heterozygous for the coat pattern gene. parent phenotypes: parent genotypes: gametes: F1 genotypes: F1 phenotypes: Scientists isolated two sections of DNA thought to correspond with the allele for full colour (R) and the allele for pointed . Sequencing these two DNA sections showed that the DNA sequence for full colour, R, had a restriction site for the restriction enzyme HpaII. This restriction site did not occur in the DNA sequence r because of a single nucleotide substitution. The scientists then carried out an analysis of three generations of cats. Each cat was assessed for three features: • coat pattern, full colour or pointed • the presence or absence of the HpaII restriction site • the pair of alleles present at a variable marker locus thought to lie close to the R/r locus. The marker locus has seven different alleles designated as 1, 2, 3, 4, 5, 6 and 7. shows the relationships of these cats and the results of the assessment. – – – – – – – + – + – + – – – – HpaII restriction site absent HpaII restriction site present are alleles at the marker locus full colour male pointed male full colour female pointed female Key – + 1–7 – – + + – – Identify evidence from to support these statements: the pointed phenotype is due to a recessive allele the R/r gene is located on an autosome the marker locus and R/r are closely linked.
9700_w20_qp_41
THEORY
2020
Paper 4, Variant 1
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shows a cat displaying the recessive phenotype for two unlinked genes. • H/h controls hair length. The allele for short hair is dominant to the allele for long hair. • R/r controls coat pattern. The allele for ‘full colour’, with pigment on all parts of the body, is dominant to the allele for ‘pointed’, where the pigment is restricted to the ears, face, paws and tail. Draw a genetic diagram to predict the offspring genotypes and phenotypes when the cat in is crossed with a cat that is heterozygous for the hair length gene and heterozygous for the coat pattern gene. parent phenotypes: parent genotypes: gametes: F1 genotypes: F1 phenotypes: Scientists isolated two sections of DNA thought to correspond with the allele for full colour (R) and the allele for pointed . Sequencing these two DNA sections showed that the DNA sequence for full colour, R, had a restriction site for the restriction enzyme HpaII. This restriction site did not occur in the DNA sequence r because of a single nucleotide substitution. The scientists then carried out an analysis of three generations of cats. Each cat was assessed for three features: • coat pattern, full colour or pointed • the presence or absence of the HpaII restriction site • the pair of alleles present at a variable marker locus thought to lie close to the R/r locus. The marker locus has seven different alleles designated as 1, 2, 3, 4, 5, 6 and 7. shows the relationships of these cats and the results of the assessment. – – – – – – – + – + – + – – – – HpaII restriction site absent HpaII restriction site present are alleles at the marker locus full colour male pointed male full colour female pointed female Key – + 1–7 – – + + – – Identify evidence from to support these statements: the pointed phenotype is due to a recessive allele the R/r gene is located on an autosome the marker locus and R/r are closely linked.
9700_w20_qp_43
THEORY
2020
Paper 4, Variant 3
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The Labrador is a variety of domestic dog. Labradors have fur that can be brown, black or yellow. In Labradors, TYRP1 is one gene that codes for fur colour. This gene has two alleles, B and b. • The dominant allele, B, codes for the enzyme tyrosinase that functions in the pathway to produce melanin, leading to black fur. • The production of melanin in Labradors is very similar to the production of melanin in humans. • The recessive allele, b, codes for an enzyme that results in the production of a brown form of melanin, leading to brown fur. Outline how melanin may be produced in Labradors to produce black fur. Another gene, MC1R, interacts with TYRP1. MC1R has two alleles, E and e. • The dominant allele, E, allows the alleles of TYRP1 to be expressed. • The recessive allele, e, prevents the alleles of TYRP1 from being expressed. • When no form of melanin is produced the Labrador will have yellow fur. Construct a genetic diagram to show the ratio of possible offspring from a cross between a black male Labrador, heterozygous for both genes, and a yellow female Labrador, heterozygous for TYRP1. parental black × yellow phenotype parental genotype gametes offspring genotypes offspring phenotypes ratio State the term used to describe a protein that is involved in the control of gene expression in eukaryotes.
9700_w21_qp_42
THEORY
2021
Paper 4, Variant 2
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In some varieties of domestic cats the gene for fur colour is located on the X chromosome. This gene has two alleles. One allele codes for black fur and the other allele codes for ginger fur. The two alleles are codominant so a heterozygous cat will have fur with patches of black and ginger colours. A cat with fur of two colours is known as a tortoiseshell. Using appropriate symbols, construct a genetic diagram to show the results of a cross between a female tortoiseshell cat and a male ginger cat. symbols parent phenotypes tortoiseshell female ginger male parent genotypes gametes offspring genotypes offspring phenotypes In humans the TYR gene is involved in the production of a dark pigment, melanin, in some cells. Describe how the expression of the TYR gene leads to the production of melanin.
9700_w22_qp_41
THEORY
2022
Paper 4, Variant 1
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The determination of sex in domestic turkeys is different from that in humans. The sex chromosomes in turkeys are named Z and W. Male turkeys are ZZ and female turkeys are ZW. The gene for feather colour is located only on the Z chromosome. • The dominant allele codes for bronze feathers. • The recessive allele codes for brown feathers. Define the terms dominant and recessive. dominant recessive Using suitable symbols, construct a genetic diagram to show the results of a cross between a heterozygous bronze male turkey and a brown female turkey. symbols parent phenotypes bronze male brown female parent genotypes gametes offspring genotypes offspring phenotypes Explain how you would carry out a test cross to determine the genotype of a bronze male turkey.
9700_w22_qp_42
THEORY
2022
Paper 4, Variant 2
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Questions Discovered
199