Directly related questions
- 22N.1A.SL.TZ0.13: What is a difference between two alleles of a gene? A. Their positions on homologous...
- 22N.1A.SL.TZ0.13: What is a difference between two alleles of a gene? A. Their positions on homologous...
-
22N.1A.SL.TZ0.15:
A couple have four children whose blood groups are A, B and AB. What is the likely combination of the parents’ genotypes?
A. IAi and IBi
B. IAi and IBIB
C. IAIB and ii
D. IAIA and IBIB
-
22N.1A.SL.TZ0.15:
A couple have four children whose blood groups are A, B and AB. What is the likely combination of the parents’ genotypes?
A. IAi and IBi
B. IAi and IBIB
C. IAIB and ii
D. IAIA and IBIB
- 22N.1A.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
- 22N.1A.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
-
22N.2.SL.TZ0.4a:
Draw the symbol for individual X on the diagram.
-
22N.2.SL.TZ0.4a:
Draw the symbol for individual X on the diagram.
-
22N.2.SL.TZ0.a:
Draw the symbol for individual X on the diagram.
-
22N.2.SL.TZ0.4b:
Calculate the probability of male Y having an allele for the disorder.
-
22N.2.SL.TZ0.4b:
Calculate the probability of male Y having an allele for the disorder.
-
22N.2.SL.TZ0.b:
Calculate the probability of male Y having an allele for the disorder.
- 22N.2.SL.TZ0.4c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
- 22N.2.SL.TZ0.4c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
- 22N.2.SL.TZ0.c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
- SPM.1A.HL.TZ0.31: Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X chromosome....
- SPM.1A.HL.TZ0.31: Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X chromosome....
- SPM.1A.HL.TZ0.31: Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X chromosome....
- SPM.1A.HL.TZ0.31: Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X chromosome....
-
SPM.2.HL.TZ0.7b:
Explain whether these results fit the predicted Mendelian ratios for this cross.
-
SPM.2.HL.TZ0.7b:
Explain whether these results fit the predicted Mendelian ratios for this cross.
-
SPM.2.HL.TZ0.7b:
Explain whether these results fit the predicted Mendelian ratios for this cross.
-
SPM.2.HL.TZ0.b:
Explain whether these results fit the predicted Mendelian ratios for this cross.
- SPM.2.HL.TZ0.7a: Identify the recombinant phenotypes amongst the offspring.
- SPM.2.HL.TZ0.7a: Identify the recombinant phenotypes amongst the offspring.
- SPM.2.HL.TZ0.7a: Identify the recombinant phenotypes amongst the offspring.
- SPM.2.HL.TZ0.a: Identify the recombinant phenotypes amongst the offspring.
-
SPM.2.SL.TZ0.7b:
Outline one example of a human disease caused by an autosomal recessive allele.
-
SPM.2.SL.TZ0.7b:
Outline one example of a human disease caused by an autosomal recessive allele.
-
SPM.2.SL.TZ0.7b:
Outline one example of a human disease caused by an autosomal recessive allele.
-
SPM.2.SL.TZ0.b:
Outline one example of a human disease caused by an autosomal recessive allele.
Sub sections and their related questions
D3.2.1. Production of haploid gametes in parents and their fusion to form a diploid zygote as the means of inheritance
- 21M.1A.SL.TZ1.18: Which genotype would be normally found in a gamete? A. Rr B. RS C. rStt D. TUt
- 22M.1A.SL.TZ1.14: Chromosome numbers vary between species. Which statement refers to humans? A. An egg cell has 22...
- 23M.2.SL.TZ2.4cii: Sex is determined in the same way in pangolins as in humans. State how many autosomes there are...
-
23M.1A.SL.TZ1.10:
Scientists sequenced the genes in each chromosome of chimpanzees (Pan troglodytes) and humans (Homo sapiens). The graph shows the mean divergence between the genes of these species by chromosome.
[Source: Material from: Mikkelsen, T.S. et al, Initial sequence of the chimpanzee genome and
comparison with the human genome, 2005 Nature, reproduced with permission of SNCSC.]
What can be deduced from this data?
A. Autosomes are more similar than Y chromosomes.B. There is the same number of chromosomes in humans and chimpanzees.
C. Humans are more closely related to chimpanzees than to other species.
D. Smaller chromosomes are more similar than larger chromosomes.
-
23M.1A.SL.TZ1.14:
Scientists sequenced the genes in each chromosome of chimpanzees (Pan troglodytes) and humans (Homo sapiens). The graph shows the mean divergence between the genes of these species by chromosome.
[Source: Material from: Mikkelsen, T.S. et al, Initial sequence of the chimpanzee genome and
comparison with the human genome, 2005 Nature, reproduced with permission of SNCSC.]
What can be deduced from this data?
A. Autosomes are more similar than Y chromosomes.B. There is the same number of chromosomes in humans and chimpanzees.
C. Humans are more closely related to chimpanzees than to other species.
D. Smaller chromosomes are more similar than larger chromosomes.
- 21M.1A.SL.TZ1.18: Which genotype would be normally found in a gamete? A. Rr B. RS C. rStt D. TUt
- 22M.1A.SL.TZ1.14: Chromosome numbers vary between species. Which statement refers to humans? A. An egg cell has 22...
- 23M.2.SL.TZ2.4cii: Sex is determined in the same way in pangolins as in humans. State how many autosomes there are...
- 23M.2.SL.TZ2.ii: Sex is determined in the same way in pangolins as in humans. State how many autosomes there are...
-
23M.1A.SL.TZ1.10:
Scientists sequenced the genes in each chromosome of chimpanzees (Pan troglodytes) and humans (Homo sapiens). The graph shows the mean divergence between the genes of these species by chromosome.
[Source: Material from: Mikkelsen, T.S. et al, Initial sequence of the chimpanzee genome and
comparison with the human genome, 2005 Nature, reproduced with permission of SNCSC.]
What can be deduced from this data?
A. Autosomes are more similar than Y chromosomes.B. There is the same number of chromosomes in humans and chimpanzees.
C. Humans are more closely related to chimpanzees than to other species.
D. Smaller chromosomes are more similar than larger chromosomes.
-
23M.1A.SL.TZ1.14:
Scientists sequenced the genes in each chromosome of chimpanzees (Pan troglodytes) and humans (Homo sapiens). The graph shows the mean divergence between the genes of these species by chromosome.
[Source: Material from: Mikkelsen, T.S. et al, Initial sequence of the chimpanzee genome and
comparison with the human genome, 2005 Nature, reproduced with permission of SNCSC.]
What can be deduced from this data?
A. Autosomes are more similar than Y chromosomes.B. There is the same number of chromosomes in humans and chimpanzees.
C. Humans are more closely related to chimpanzees than to other species.
D. Smaller chromosomes are more similar than larger chromosomes.
D3.2.2. Methods for conducting genetic crosses in flowering plants
- SPM.1A.HL.TZ0.31: Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X chromosome....
-
23M.2.SL.TZ2.2b:
Using a Punnett grid, explain the results of this cross.
-
23M.2.SL.TZ2.2c:
Suggest a reason for the F2 generation not corresponding exactly to the expected ratio of purple-flowered and white-flowered plants.
-
23M.1A.SL.TZ2.13:
The pedigree chart shows the inheritance of three generations of Duane syndrome, a condition caused by a dominant allele that affects alignment of the eyes.
[Source: Yang, M.-M., Ho, M. et al., 2013. Pedigree of a Chinese family with Duane retraction syndrome. [diagram online]
Available at: https://www.researchgate.net/figure/Pedigree-of-a-Chinese-family-with-Duane-retraction-syndrome-
Squares-men-circles_fig1_236921765 [Accessed 5 October 2021]. Public domain.]If individuals II:1 and II:2 had a third child, what is the probability that the child would have Duane syndrome?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
-
23M.1A.SL.TZ2.16:
The pedigree chart shows the inheritance of three generations of Duane syndrome, a condition caused by a dominant allele that affects alignment of the eyes.
[Source: Yang, M.-M., Ho, M. et al., 2013. Pedigree of a Chinese family with Duane retraction syndrome. [diagram online]
Available at: https://www.researchgate.net/figure/Pedigree-of-a-Chinese-family-with-Duane-retraction-syndrome-
Squares-men-circles_fig1_236921765 [Accessed 5 October 2021]. Public domain.]If individuals II:1 and II:2 had a third child, what is the probability that the child would have Duane syndrome?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
- SPM.1A.HL.TZ0.31: Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X chromosome....
- SPM.1A.HL.TZ0.31: Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X chromosome....
- SPM.1A.HL.TZ0.31: Female grasshoppers have XX sex chromosomes and males have XO, signifying a single X chromosome....
-
23M.2.SL.TZ2.2b:
Using a Punnett grid, explain the results of this cross.
-
23M.2.SL.TZ2.2c:
Suggest a reason for the F2 generation not corresponding exactly to the expected ratio of purple-flowered and white-flowered plants.
-
23M.2.SL.TZ2.b:
Using a Punnett grid, explain the results of this cross.
-
23M.2.SL.TZ2.c:
Suggest a reason for the F2 generation not corresponding exactly to the expected ratio of purple-flowered and white-flowered plants.
-
23M.1A.SL.TZ2.13:
The pedigree chart shows the inheritance of three generations of Duane syndrome, a condition caused by a dominant allele that affects alignment of the eyes.
[Source: Yang, M.-M., Ho, M. et al., 2013. Pedigree of a Chinese family with Duane retraction syndrome. [diagram online]
Available at: https://www.researchgate.net/figure/Pedigree-of-a-Chinese-family-with-Duane-retraction-syndrome-
Squares-men-circles_fig1_236921765 [Accessed 5 October 2021]. Public domain.]If individuals II:1 and II:2 had a third child, what is the probability that the child would have Duane syndrome?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
-
23M.1A.SL.TZ2.16:
The pedigree chart shows the inheritance of three generations of Duane syndrome, a condition caused by a dominant allele that affects alignment of the eyes.
[Source: Yang, M.-M., Ho, M. et al., 2013. Pedigree of a Chinese family with Duane retraction syndrome. [diagram online]
Available at: https://www.researchgate.net/figure/Pedigree-of-a-Chinese-family-with-Duane-retraction-syndrome-
Squares-men-circles_fig1_236921765 [Accessed 5 October 2021]. Public domain.]If individuals II:1 and II:2 had a third child, what is the probability that the child would have Duane syndrome?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
D3.2.3. Genotype as the combination of alleles inherited by an organism
-
21M.2.SL.TZ2.5a:
Autosomal genes are located in chromosomes that are not sex chromosomes. The inheritance of autosomal genes is affected by whether the genes are linked or unlinked. Explain the two types of inheritance, using the example of parents that are heterozygous for two genes A and B.
- 22M.1A.SL.TZ2.13: Which statement defines alleles? A. They are the different forms of a gene that have the same...
- 19N.1A.SL.TZ0.10: Which statements are characteristics of alleles? I. Alleles differ significantly in number of...
- 22N.1A.SL.TZ0.13: What is a difference between two alleles of a gene? A. Their positions on homologous...
-
21M.2.SL.TZ2.5a:
Autosomal genes are located in chromosomes that are not sex chromosomes. The inheritance of autosomal genes is affected by whether the genes are linked or unlinked. Explain the two types of inheritance, using the example of parents that are heterozygous for two genes A and B.
-
21M.2.SL.TZ2.a:
Autosomal genes are located in chromosomes that are not sex chromosomes. The inheritance of autosomal genes is affected by whether the genes are linked or unlinked. Explain the two types of inheritance, using the example of parents that are heterozygous for two genes A and B.
- 22M.1A.SL.TZ2.13: Which statement defines alleles? A. They are the different forms of a gene that have the same...
- 19N.1A.SL.TZ0.10: Which statements are characteristics of alleles? I. Alleles differ significantly in number of...
- 22N.1A.SL.TZ0.13: What is a difference between two alleles of a gene? A. Their positions on homologous...
D3.2.4. Phenotype as the observable traits of an organism resulting from genotype and environmental factors
-
21M.2.SL.TZ2.5c:
Congenital lactase deficiency is a type of lactose intolerance that occurs in infants. It is inherited in an autosomal recessive pattern. Calculate the chance of congenital lactose intolerance in a child whose parents are both carriers for the disorder, showing fully how you reached your answer.
-
21M.2.SL.TZ2.5c:
Congenital lactase deficiency is a type of lactose intolerance that occurs in infants. It is inherited in an autosomal recessive pattern. Calculate the chance of congenital lactose intolerance in a child whose parents are both carriers for the disorder, showing fully how you reached your answer.
-
21M.2.SL.TZ2.c:
Congenital lactase deficiency is a type of lactose intolerance that occurs in infants. It is inherited in an autosomal recessive pattern. Calculate the chance of congenital lactose intolerance in a child whose parents are both carriers for the disorder, showing fully how you reached your answer.
D3.2.5. Effects of dominant and recessive alleles on phenotype
- 22M.1A.SL.TZ1.16: Huntington’s disease is an autosomal dominant genetic disease. What are the chances of two...
- 21N.1A.SL.TZ1.12: Huntington’s disease can develop in middle age and leads to death of brain cells. It is carried...
-
19M.1A.SL.TZ1.15:
Creeper in chickens is a condition in which the chickens are born with very short legs. The creeper allele (C) is dominant over the normal allele (c). Embryos which are homozygous for the dominant allele fail to develop into viable chickens and die before they hatch. What phenotypic ratio would you expect in the live offspring of a cross between two creeper chickens?
A. All creeper
B. 1 creeper; 2 normal
C. 2 creeper; 1 normal
D. 3 creeper; 1 normal
- 19N.1A.SL.TZ0.10: Which statements are characteristics of alleles? I. Alleles differ significantly in number of...
- 20N.1A.SL.TZ0.14: Tall heterozygous pea plants were crossed and the resulting seeds grown. Out of 360 plants, 270...
- 23M.2.SL.TZ1.5a: Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.5b:
Deduce reasons for the difference between the results of Cross 2 and Cross 3.
- 23M.2.SL.TZ1.5c: There were 79 progeny in Cross 4. Predict the expected results by completing the table.
-
23M.2.SL.TZ1.5d:
Identify, using one recognition feature visible in the photo, the phylum in which D. plantarius is classified.
- 23M.2.SL.TZ1.5a: Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.5b:
Deduce reasons for the difference between the results of Cross 2 and Cross 3.
- 23M.2.SL.TZ1.5c: There were 79 progeny in Cross 4. Predict the expected results by completing the table.
- 23M.2.SL.TZ1.5d: The proportion of banded and unbanded individuals in a population of D. plantarius can change....
-
23M.1A.SL.TZ2.13:
The pedigree chart shows the inheritance of three generations of Duane syndrome, a condition caused by a dominant allele that affects alignment of the eyes.
[Source: Yang, M.-M., Ho, M. et al., 2013. Pedigree of a Chinese family with Duane retraction syndrome. [diagram online]
Available at: https://www.researchgate.net/figure/Pedigree-of-a-Chinese-family-with-Duane-retraction-syndrome-
Squares-men-circles_fig1_236921765 [Accessed 5 October 2021]. Public domain.]If individuals II:1 and II:2 had a third child, what is the probability that the child would have Duane syndrome?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
-
23M.1A.SL.TZ2.16:
The pedigree chart shows the inheritance of three generations of Duane syndrome, a condition caused by a dominant allele that affects alignment of the eyes.
[Source: Yang, M.-M., Ho, M. et al., 2013. Pedigree of a Chinese family with Duane retraction syndrome. [diagram online]
Available at: https://www.researchgate.net/figure/Pedigree-of-a-Chinese-family-with-Duane-retraction-syndrome-
Squares-men-circles_fig1_236921765 [Accessed 5 October 2021]. Public domain.]If individuals II:1 and II:2 had a third child, what is the probability that the child would have Duane syndrome?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
- 22M.1A.SL.TZ1.16: Huntington’s disease is an autosomal dominant genetic disease. What are the chances of two...
- 21N.1A.SL.TZ1.12: Huntington’s disease can develop in middle age and leads to death of brain cells. It is carried...
-
19M.1A.SL.TZ1.15:
Creeper in chickens is a condition in which the chickens are born with very short legs. The creeper allele (C) is dominant over the normal allele (c). Embryos which are homozygous for the dominant allele fail to develop into viable chickens and die before they hatch. What phenotypic ratio would you expect in the live offspring of a cross between two creeper chickens?
A. All creeper
B. 1 creeper; 2 normal
C. 2 creeper; 1 normal
D. 3 creeper; 1 normal
- 19N.1A.SL.TZ0.10: Which statements are characteristics of alleles? I. Alleles differ significantly in number of...
- 20N.1A.SL.TZ0.14: Tall heterozygous pea plants were crossed and the resulting seeds grown. Out of 360 plants, 270...
- 23M.2.SL.TZ1.5a: Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.5b:
Deduce reasons for the difference between the results of Cross 2 and Cross 3.
- 23M.2.SL.TZ1.5c: There were 79 progeny in Cross 4. Predict the expected results by completing the table.
-
23M.2.SL.TZ1.5d:
Identify, using one recognition feature visible in the photo, the phylum in which D. plantarius is classified.
- 23M.2.SL.TZ1.a: Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.b:
Deduce reasons for the difference between the results of Cross 2 and Cross 3.
- 23M.2.SL.TZ1.c: There were 79 progeny in Cross 4. Predict the expected results by completing the table.
-
23M.2.SL.TZ1.d:
Identify, using one recognition feature visible in the photo, the phylum in which D. plantarius is classified.
- 23M.2.SL.TZ1.5a: Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.5b:
Deduce reasons for the difference between the results of Cross 2 and Cross 3.
- 23M.2.SL.TZ1.5c: There were 79 progeny in Cross 4. Predict the expected results by completing the table.
- 23M.2.SL.TZ1.5d: The proportion of banded and unbanded individuals in a population of D. plantarius can change....
- 23M.2.SL.TZ1.a: Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.b:
Deduce reasons for the difference between the results of Cross 2 and Cross 3.
- 23M.2.SL.TZ1.c: There were 79 progeny in Cross 4. Predict the expected results by completing the table.
- 23M.2.SL.TZ1.d: The proportion of banded and unbanded individuals in a population of D. plantarius can change....
-
23M.1A.SL.TZ2.13:
The pedigree chart shows the inheritance of three generations of Duane syndrome, a condition caused by a dominant allele that affects alignment of the eyes.
[Source: Yang, M.-M., Ho, M. et al., 2013. Pedigree of a Chinese family with Duane retraction syndrome. [diagram online]
Available at: https://www.researchgate.net/figure/Pedigree-of-a-Chinese-family-with-Duane-retraction-syndrome-
Squares-men-circles_fig1_236921765 [Accessed 5 October 2021]. Public domain.]If individuals II:1 and II:2 had a third child, what is the probability that the child would have Duane syndrome?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
-
23M.1A.SL.TZ2.16:
The pedigree chart shows the inheritance of three generations of Duane syndrome, a condition caused by a dominant allele that affects alignment of the eyes.
[Source: Yang, M.-M., Ho, M. et al., 2013. Pedigree of a Chinese family with Duane retraction syndrome. [diagram online]
Available at: https://www.researchgate.net/figure/Pedigree-of-a-Chinese-family-with-Duane-retraction-syndrome-
Squares-men-circles_fig1_236921765 [Accessed 5 October 2021]. Public domain.]If individuals II:1 and II:2 had a third child, what is the probability that the child would have Duane syndrome?
A. 25 %
B. 50 %
C. 75 %
D. 100 %
D3.2.6. Phenotypic plasticity as the capacity to develop traits suited to the environment experienced by an organism, by varying patterns of gene expression
NoneD3.2.7. Phenylketonuria as an example of a human disease due to a recessive allele
-
SPM.2.SL.TZ0.7b:
Outline one example of a human disease caused by an autosomal recessive allele.
-
SPM.2.SL.TZ0.7b:
Outline one example of a human disease caused by an autosomal recessive allele.
-
SPM.2.SL.TZ0.7b:
Outline one example of a human disease caused by an autosomal recessive allele.
-
SPM.2.SL.TZ0.b:
Outline one example of a human disease caused by an autosomal recessive allele.
D3.2.8. Single-nucleotide polymorphisms and multiple alleles in gene pools
- 19N.1A.SL.TZ0.10: Which statements are characteristics of alleles? I. Alleles differ significantly in number of...
- 19N.1A.SL.TZ0.10: Which statements are characteristics of alleles? I. Alleles differ significantly in number of...
D3.2.9. ABO blood groups as an example of multiple alleles
- 19N.2.SL.TZ0.6b: Outline how two parents could have a child with any of the four ABO blood groups.
- 19N.2.SL.TZ0.7b: Outline how two parents could have a child with any of the four ABO blood groups.
- 19M.2.SL.TZ2.7b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
-
21M.2.SL.TZ1.7a:
Outline the inheritance of blood types in the ABO blood system in humans.
- 20N.1A.SL.TZ0.15: What are all the possible phenotypes of children born to a mother with blood group AB and a...
- 21M.1A.SL.TZ2.15: A woman with blood type A has three children with a man who has blood type AB. The first child...
- 19N.2.SL.TZ0.6b: Outline how two parents could have a child with any of the four ABO blood groups.
- 19N.2.SL.TZ0.b: Outline how two parents could have a child with any of the four ABO blood groups.
- 19N.2.SL.TZ0.7b: Outline how two parents could have a child with any of the four ABO blood groups.
- 19N.2.SL.TZ0.b: Outline how two parents could have a child with any of the four ABO blood groups.
- 19M.2.SL.TZ2.7b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
- 19M.2.SL.TZ2.b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
-
21M.2.SL.TZ1.7a:
Outline the inheritance of blood types in the ABO blood system in humans.
-
21M.2.SL.TZ1.a:
Outline the inheritance of blood types in the ABO blood system in humans.
- 20N.1A.SL.TZ0.15: What are all the possible phenotypes of children born to a mother with blood group AB and a...
- 21M.1A.SL.TZ2.15: A woman with blood type A has three children with a man who has blood type AB. The first child...
D3.2.10. Incomplete dominance and codominance
- 20N.1A.SL.TZ0.35: Andalusian fowl have varied colours and types of feathers. The allele for black feathers is...
-
21M.2.SL.TZ2.5a:
Autosomal genes are located in chromosomes that are not sex chromosomes. The inheritance of autosomal genes is affected by whether the genes are linked or unlinked. Explain the two types of inheritance, using the example of parents that are heterozygous for two genes A and B.
- 19N.1A.SL.TZ0.15: A variety of Pelargonium has yellow leaves. When plants of this variety are crossed, the...
- 19M.1A.SL.TZ2.15: The children in a family have blood groups A, B and O. What are the genotypes of their...
- 19M.2.SL.TZ2.7b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
- 20N.2.SL.TZ0.5a: Identify the genotype of the male wire-haired dog.
- 20N.2.SL.TZ0.5b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
-
22N.1A.SL.TZ0.15:
A couple have four children whose blood groups are A, B and AB. What is the likely combination of the parents’ genotypes?
A. IAi and IBi
B. IAi and IBIB
C. IAIB and ii
D. IAIA and IBIB
- 20N.1A.SL.TZ0.35: Andalusian fowl have varied colours and types of feathers. The allele for black feathers is...
-
21M.2.SL.TZ2.5a:
Autosomal genes are located in chromosomes that are not sex chromosomes. The inheritance of autosomal genes is affected by whether the genes are linked or unlinked. Explain the two types of inheritance, using the example of parents that are heterozygous for two genes A and B.
-
21M.2.SL.TZ2.a:
Autosomal genes are located in chromosomes that are not sex chromosomes. The inheritance of autosomal genes is affected by whether the genes are linked or unlinked. Explain the two types of inheritance, using the example of parents that are heterozygous for two genes A and B.
- 19N.1A.SL.TZ0.15: A variety of Pelargonium has yellow leaves. When plants of this variety are crossed, the...
- 19M.1A.SL.TZ2.15: The children in a family have blood groups A, B and O. What are the genotypes of their...
- 19M.2.SL.TZ2.7b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
- 19M.2.SL.TZ2.b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
- 20N.2.SL.TZ0.5a: Identify the genotype of the male wire-haired dog.
- 20N.2.SL.TZ0.5b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
- 20N.2.SL.TZ0.a: Identify the genotype of the male wire-haired dog.
- 20N.2.SL.TZ0.b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
-
22N.1A.SL.TZ0.15:
A couple have four children whose blood groups are A, B and AB. What is the likely combination of the parents’ genotypes?
A. IAi and IBi
B. IAi and IBIB
C. IAIB and ii
D. IAIA and IBIB
D3.2.11. Sex determination in humans and inheritance of genes on sex chromosomes
-
19M.1A.SL.TZ1.16:
Which pedigree chart is consistent with the inheritance of red-green colour blindness?
[Source: © International Baccalaureate Organization 2019]
-
19N.1A.SL.TZ0.16:
A pair of alleles controls the secretion of antigens corresponding to blood group in saliva. Examine the pedigree chart.
[Source: © International Baccalaureate Organization 2019]
Based on this pedigree chart, which best describes the allele conferring antigen secretion in saliva?
A. Dominant
B. Recessive
C. Sex-linked
D. Co-dominant
-
22M.2.SL.TZ2.6b:
A family has a history of hemophilia that is caused by a sex-linked recessive allele. A woman from this family is a carrier and marries a man who does not have the allele. Showing your working, determine the probability of their children having the disease.
- 19M.2.SL.TZ2.7b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
-
19M.1A.SL.TZ1.13:
The system of sex determination in chimpanzees (Pan troglodytes) is the same as in other mammals. A chimpanzee has 48 chromosomes in the nuclei of its body cells.
What can be deduced from this information?
A. The sex of the chimpanzee
B. The number of genes in each chromosome
C. Whether non-disjunction has occurred
D. The number of autosomes in a diploid cell
-
23M.2.SL.TZ1.6c:
Explain how the presence of a Y chromosome in the cells of a human embryo causes it to develop as a male.
-
19M.1A.SL.TZ1.16:
Which pedigree chart is consistent with the inheritance of red-green colour blindness?
[Source: © International Baccalaureate Organization 2019]
-
19N.1A.SL.TZ0.16:
A pair of alleles controls the secretion of antigens corresponding to blood group in saliva. Examine the pedigree chart.
[Source: © International Baccalaureate Organization 2019]
Based on this pedigree chart, which best describes the allele conferring antigen secretion in saliva?
A. Dominant
B. Recessive
C. Sex-linked
D. Co-dominant
-
22M.2.SL.TZ2.6b:
A family has a history of hemophilia that is caused by a sex-linked recessive allele. A woman from this family is a carrier and marries a man who does not have the allele. Showing your working, determine the probability of their children having the disease.
-
22M.2.SL.TZ2.b:
A family has a history of hemophilia that is caused by a sex-linked recessive allele. A woman from this family is a carrier and marries a man who does not have the allele. Showing your working, determine the probability of their children having the disease.
- 19M.2.SL.TZ2.7b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
- 19M.2.SL.TZ2.b: Mendel found the same pattern of inheritance in all the crosses that he performed. Outline, with...
-
19M.1A.SL.TZ1.13:
The system of sex determination in chimpanzees (Pan troglodytes) is the same as in other mammals. A chimpanzee has 48 chromosomes in the nuclei of its body cells.
What can be deduced from this information?
A. The sex of the chimpanzee
B. The number of genes in each chromosome
C. Whether non-disjunction has occurred
D. The number of autosomes in a diploid cell
-
23M.2.SL.TZ1.6c:
Explain how the presence of a Y chromosome in the cells of a human embryo causes it to develop as a male.
-
23M.2.SL.TZ1.c:
Explain how the presence of a Y chromosome in the cells of a human embryo causes it to develop as a male.
D3.2.12. Haemophilia as an example of a sex-linked genetic disorder
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
-
21N.2.SL.TZ0.2a.i:
Draw a Punnett square to show all the possible genotypes of Queen Victoria’s children.
- 21N.2.SL.TZ0.2a.ii: Deduce the genotype of Queen Victoria’s daughter Alice.
-
22M.2.SL.TZ2.6b:
A family has a history of hemophilia that is caused by a sex-linked recessive allele. A woman from this family is a carrier and marries a man who does not have the allele. Showing your working, determine the probability of their children having the disease.
-
19N.1A.HL.TZ0.12:
The pedigree chart shows the inheritance of hemophilia in some of the descendants of Queen Victoria. Which letter points to a family member certain to be heterozygous?
[Source: © International Baccalaureate Organization 2019]
- 22M.1A.SL.TZ2.15: The pedigree chart shows the inheritance of hemochromatosis, a genetic disease which causes an...
- 22N.2.SL.TZ0.4c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
-
21N.2.SL.TZ0.2a.i:
Draw a Punnett square to show all the possible genotypes of Queen Victoria’s children.
- 21N.2.SL.TZ0.2a.ii: Deduce the genotype of Queen Victoria’s daughter Alice.
-
21N.2.SL.TZ0.a.i:
Draw a Punnett square to show all the possible genotypes of Queen Victoria’s children.
- 21N.2.SL.TZ0.a.ii: Deduce the genotype of Queen Victoria’s daughter Alice.
-
22M.2.SL.TZ2.6b:
A family has a history of hemophilia that is caused by a sex-linked recessive allele. A woman from this family is a carrier and marries a man who does not have the allele. Showing your working, determine the probability of their children having the disease.
-
22M.2.SL.TZ2.b:
A family has a history of hemophilia that is caused by a sex-linked recessive allele. A woman from this family is a carrier and marries a man who does not have the allele. Showing your working, determine the probability of their children having the disease.
-
19N.1A.HL.TZ0.12:
The pedigree chart shows the inheritance of hemophilia in some of the descendants of Queen Victoria. Which letter points to a family member certain to be heterozygous?
[Source: © International Baccalaureate Organization 2019]
- 22M.1A.SL.TZ2.15: The pedigree chart shows the inheritance of hemochromatosis, a genetic disease which causes an...
- 22N.2.SL.TZ0.4c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
- 22N.2.SL.TZ0.c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
D3.2.13. Pedigree charts to deduce patterns of inheritance of genetic disorders
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 21N.1A.SL.TZ0.15: In humans, male pattern baldness is caused by a recessive sex-linked gene found only on the X...
-
19M.1A.SL.TZ2.14:
The pedigree chart shows a family affected by cystic fibrosis.
[Source: © International Baccalaureate Organization 2019]
What is the genotype of the affected boy’s father?
A. AA only
B. Either AA or Aa
C. Aa only
D. aa only
-
19N.1A.HL.TZ0.12:
The pedigree chart shows the inheritance of hemophilia in some of the descendants of Queen Victoria. Which letter points to a family member certain to be heterozygous?
[Source: © International Baccalaureate Organization 2019]
- 20N.1A.SL.TZ0.13: In the pedigree chart, individuals affected by a genetic disease are shown as shaded symbols....
-
22N.2.SL.TZ0.4a:
Draw the symbol for individual X on the diagram.
-
22N.2.SL.TZ0.4b:
Calculate the probability of male Y having an allele for the disorder.
- 22N.2.SL.TZ0.4c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.4a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.4b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 22M.2.SL.TZ1.a: State the probability that Edward had hemophilia.
-
22M.2.SL.TZ1.b:
Explain the reasons for none of the females in the pedigree chart having hemophilia.
- 21N.1A.SL.TZ0.15: In humans, male pattern baldness is caused by a recessive sex-linked gene found only on the X...
-
19M.1A.SL.TZ2.14:
The pedigree chart shows a family affected by cystic fibrosis.
[Source: © International Baccalaureate Organization 2019]
What is the genotype of the affected boy’s father?
A. AA only
B. Either AA or Aa
C. Aa only
D. aa only
-
19N.1A.HL.TZ0.12:
The pedigree chart shows the inheritance of hemophilia in some of the descendants of Queen Victoria. Which letter points to a family member certain to be heterozygous?
[Source: © International Baccalaureate Organization 2019]
- 20N.1A.SL.TZ0.13: In the pedigree chart, individuals affected by a genetic disease are shown as shaded symbols....
-
22N.2.SL.TZ0.4a:
Draw the symbol for individual X on the diagram.
-
22N.2.SL.TZ0.4b:
Calculate the probability of male Y having an allele for the disorder.
- 22N.2.SL.TZ0.4c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
-
22N.2.SL.TZ0.a:
Draw the symbol for individual X on the diagram.
-
22N.2.SL.TZ0.b:
Calculate the probability of male Y having an allele for the disorder.
- 22N.2.SL.TZ0.c: Explain how the information in the box labelled B indicates that the gene is not sex-linked.
D3.2.14. Continuous variation due to polygenic inheritance and/or environmental factors
-
22M.2.SL.TZ1.8c:
Explain the reasons for variation in human height.
- 21N.2.SL.TZ0.6c: Distinguish between continuous and discrete variation, using examples.
-
23M.1A.SL.TZ1.34:
A total of 271 164 people were tested for blood groups in Mexico. The pie chart summarizes the ABO blood group distribution.
[Source: Canizalez-Román, A. et al., 2018. Blood Groups Distribution and Gene Diversity
of the ABO and Rh (D) Loci in the Mexican Population.
BioMed Research International, (Article ID 1925619).
http://dx.doi.org/10.1155/2018/1925619. Public domain.]What can be concluded from the ABO blood group distribution in Mexico?
A. Allele frequencies are not the same for all blood group alleles.
B. The majority of the Mexican population shows a co-dominant phenotype.
C. Most of the Mexican population can receive blood from all blood groups.
D. Only 2 % of the Mexican population have a heterozygous genotype for blood groups.
-
22M.2.SL.TZ1.8c:
Explain the reasons for variation in human height.
-
22M.2.SL.TZ1.c:
Explain the reasons for variation in human height.
- 21N.2.SL.TZ0.6c: Distinguish between continuous and discrete variation, using examples.
- 21N.2.SL.TZ0.c: Distinguish between continuous and discrete variation, using examples.
-
23M.1A.SL.TZ1.34:
A total of 271 164 people were tested for blood groups in Mexico. The pie chart summarizes the ABO blood group distribution.
[Source: Canizalez-Román, A. et al., 2018. Blood Groups Distribution and Gene Diversity
of the ABO and Rh (D) Loci in the Mexican Population.
BioMed Research International, (Article ID 1925619).
http://dx.doi.org/10.1155/2018/1925619. Public domain.]What can be concluded from the ABO blood group distribution in Mexico?
A. Allele frequencies are not the same for all blood group alleles.
B. The majority of the Mexican population shows a co-dominant phenotype.
C. Most of the Mexican population can receive blood from all blood groups.
D. Only 2 % of the Mexican population have a heterozygous genotype for blood groups.
D3.2.15. Box-and-whisker plots to represent data for a continuous variable such as student height
NoneD3.2.16. Segregation and independent assortment of unlinked genes in meiosis
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 19M.2.HL.TZ1.2d.i: A fly that is homozygous dominant for both body colour and wing size mates with a fly that is...
- 19M.2.HL.TZ1.2d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 19M.2.HL.TZ1.2d.i: A fly that is homozygous dominant for both body colour and wing size mates with a fly that is...
- 19M.2.HL.TZ1.2d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
- 19M.2.HL.TZ1.d.i: A fly that is homozygous dominant for both body colour and wing size mates with a fly that is...
- 19M.2.HL.TZ1.d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
D3.2.17. Punnett grids for predicting genotypic and phenotypic ratios in dihybrid crosses involving pairs of unlinked autosomal genes
- 19N.2.HL.TZ0.2a: State the type of inheritance shown.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
-
19M.1A.HL.TZ2.20:
William Bateson and Reginald Punnett used the sweet pea (Lathyrus odoratus) in genetics studies in the early 20th century. Pure-breeding plants that produced purple flowers and long pollen grains were crossed with pure-breeding plants that produced red flowers and round pollen grains. The resulting offspring all produced purple flowers and long pollen grains. Two of the F1 generation plants were crossed. The table shows the ratio of phenotypes in the F2 generation.
What is an explanation for these experimental results?
A. Purple flowers and long pollen grains are dominant and the alleles have assorted independently.
B. The genes for flower colour and pollen shape are linked and all plants producing long pollen grains are recombinants.
C. The genes for flower colour and pollen shape are linked and all plants producing red flowers are recombinants.
D. Plants producing purple flowers and round pollen grains arose through crossing over.
- 19M.2.HL.TZ1.2d.i: A fly that is homozygous dominant for both body colour and wing size mates with a fly that is...
-
21M.1A.HL.TZ1.36:
In fruit flies (Drosophila melanogaster), grey bodies (b+) are dominant to black bodies (b) and normal wings (vg+) are dominant to vestigial wings (vg). Homozygous vestigial winged, black bodied flies were crossed with individuals that were heterozygous for both traits. 2300 individuals were counted and the phenotypes observed were recorded as shown.
965 normal wings, grey bodies
944 vestigial wings, black bodies
206 vestigial wings, grey bodies
185 normal wings, black bodiesWhich statement is valid?
A. The predicted phenotypic ratio was 9:3:3: 1.
B. There is independent assortment of wings but not body colour.
C. The expected number of vestigial winged, grey bodied flies was 575.
D. The traits are on different chromosomes.
- 21N.1A.HL.TZ1.35: A dihybrid cross was carried out between two plants to determine whether the genes for seed shape...
- 22N.1A.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
-
SPM.2.HL.TZ0.7b:
Explain whether these results fit the predicted Mendelian ratios for this cross.
-
SPM.2.HL.TZ0.7b:
Explain whether these results fit the predicted Mendelian ratios for this cross.
- 19N.2.HL.TZ0.2a: State the type of inheritance shown.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.a: State the type of inheritance shown.
- 19N.2.HL.TZ0.b: Identify the recombinants.
-
19M.1A.HL.TZ2.20:
William Bateson and Reginald Punnett used the sweet pea (Lathyrus odoratus) in genetics studies in the early 20th century. Pure-breeding plants that produced purple flowers and long pollen grains were crossed with pure-breeding plants that produced red flowers and round pollen grains. The resulting offspring all produced purple flowers and long pollen grains. Two of the F1 generation plants were crossed. The table shows the ratio of phenotypes in the F2 generation.
What is an explanation for these experimental results?
A. Purple flowers and long pollen grains are dominant and the alleles have assorted independently.
B. The genes for flower colour and pollen shape are linked and all plants producing long pollen grains are recombinants.
C. The genes for flower colour and pollen shape are linked and all plants producing red flowers are recombinants.
D. Plants producing purple flowers and round pollen grains arose through crossing over.
- 19M.2.HL.TZ1.2d.i: A fly that is homozygous dominant for both body colour and wing size mates with a fly that is...
- 19M.2.HL.TZ1.d.i: A fly that is homozygous dominant for both body colour and wing size mates with a fly that is...
-
21M.1A.HL.TZ1.36:
In fruit flies (Drosophila melanogaster), grey bodies (b+) are dominant to black bodies (b) and normal wings (vg+) are dominant to vestigial wings (vg). Homozygous vestigial winged, black bodied flies were crossed with individuals that were heterozygous for both traits. 2300 individuals were counted and the phenotypes observed were recorded as shown.
965 normal wings, grey bodies
944 vestigial wings, black bodies
206 vestigial wings, grey bodies
185 normal wings, black bodiesWhich statement is valid?
A. The predicted phenotypic ratio was 9:3:3: 1.
B. There is independent assortment of wings but not body colour.
C. The expected number of vestigial winged, grey bodied flies was 575.
D. The traits are on different chromosomes.
- 21N.1A.HL.TZ1.35: A dihybrid cross was carried out between two plants to determine whether the genes for seed shape...
- 22N.1A.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
-
SPM.2.HL.TZ0.7b:
Explain whether these results fit the predicted Mendelian ratios for this cross.
-
SPM.2.HL.TZ0.b:
Explain whether these results fit the predicted Mendelian ratios for this cross.
D3.2.18. Loci of human genes and their polypeptide products
- 19N.2.HL.TZ0.2a: State the type of inheritance shown.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.2a: State the type of inheritance shown.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.a: State the type of inheritance shown.
- 19N.2.HL.TZ0.b: Identify the recombinants.
D3.2.19. Autosomal gene linkage
- 19N.2.HL.TZ0.2a: State the type of inheritance shown.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
-
21N.2.HL.TZ0.2c:
Explain gene linkage and its effects on inheritance.
- 23M.1A.SL.TZ2.32: In the fruit fly Drosophila, the alleles for red eyes and brown bodies are dominant to the...
- 19N.2.HL.TZ0.2a: State the type of inheritance shown.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.a: State the type of inheritance shown.
- 19N.2.HL.TZ0.b: Identify the recombinants.
-
21N.2.HL.TZ0.2c:
Explain gene linkage and its effects on inheritance.
-
21N.2.HL.TZ0.c:
Explain gene linkage and its effects on inheritance.
- 23M.1A.SL.TZ2.32: In the fruit fly Drosophila, the alleles for red eyes and brown bodies are dominant to the...
D3.2.20. Recombinants in crosses involving two linked or unlinked genes
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 19N.2.HL.TZ0.2a: State the type of inheritance shown.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- SPM.2.HL.TZ0.7a: Identify the recombinant phenotypes amongst the offspring.
- 23M.1A.SL.TZ2.32: In the fruit fly Drosophila, the alleles for red eyes and brown bodies are dominant to the...
- SPM.2.HL.TZ0.7a: Identify the recombinant phenotypes amongst the offspring.
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1A.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 19N.2.HL.TZ0.2a: State the type of inheritance shown.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.a: State the type of inheritance shown.
- 19N.2.HL.TZ0.b: Identify the recombinants.
- SPM.2.HL.TZ0.7a: Identify the recombinant phenotypes amongst the offspring.
- SPM.2.HL.TZ0.a: Identify the recombinant phenotypes amongst the offspring.
- 23M.1A.SL.TZ2.32: In the fruit fly Drosophila, the alleles for red eyes and brown bodies are dominant to the...
D3.2.21. Use of a chi-squared test on data from dihybrid crosses
- 19N.2.HL.TZ0.2c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
- 22M.1A.SL.TZ2.17: A group of students used quadrat sampling to gather data on the presence of two plant species in...
- 19N.2.HL.TZ0.2c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
- 19N.2.HL.TZ0.c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
- 22M.1A.SL.TZ2.17: A group of students used quadrat sampling to gather data on the presence of two plant species in...