Syllabus sections » |
Topic 10: Genetics and evolution
Description
Overview of the essential ideas for this topic.
10.1: Meiosis leads to independent assortment of chromosomes and unique composition of alleles in daughter cells.
10.2: Genes may be linked or unlinked and are inherited accordingly.
10.3: Gene pools change over time.
Directly related questions
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17N.2.HL.TZ0.05a:
Describe the process of crossing over.
- 20N.1.HL.TZ0.35: Andalusian fowl have varied colours and types of feathers. The allele for black feathers is...
- 20N.1.HL.TZ0.35: Andalusian fowl have varied colours and types of feathers. The allele for black feathers is...
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20N.2.HL.TZ0.8b:
Describe the changes that occur in gene pools during speciation.
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20N.2.HL.TZ0.8b:
Describe the changes that occur in gene pools during speciation.
-
20N.2.HL.TZ0.b:
Describe the changes that occur in gene pools during speciation.
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17N.2.HL.TZ0.05a:
Describe the process of crossing over.
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17N.2.HL.TZ0.a:
Describe the process of crossing over.
- 17N.2.HL.TZ0.05b: Explain the reason for linked genes not following the pattern of inheritance discovered by Mendel.
- 17N.2.HL.TZ0.05b: Explain the reason for linked genes not following the pattern of inheritance discovered by Mendel.
- 21M.1.HL.TZ1.35: Which statement is valid regarding chromatids? A. Sister chromatids separate during meiosis...
- 17N.1.HL.TZ0.35: How do the concepts of gradualism and punctuated equilibrium differ? A. The timing of...
- 17N.2.HL.TZ0.b: Explain the reason for linked genes not following the pattern of inheritance discovered by Mendel.
- 21M.1.HL.TZ2.36: Which cell is a polyploid zygote produced by fusion of one haploid and one diploid gamete?
- 17N.1.HL.TZ0.35: How do the concepts of gradualism and punctuated equilibrium differ? A. The timing of...
- 21M.1.HL.TZ2.36: Which cell is a polyploid zygote produced by fusion of one haploid and one diploid gamete?
- 17N.1.HL.TZ0.36: In a plant, dark leaves are dominant to pale leaves and yellow seeds are dominant to white...
- 17N.1.HL.TZ0.36: In a plant, dark leaves are dominant to pale leaves and yellow seeds are dominant to white...
- 21M.1.HL.TZ1.35: Which statement is valid regarding chromatids? A. Sister chromatids separate during meiosis...
- 21M.1.HL.TZ2.35: What process occurs in both mitosis and meiosis? A. Formation of chiasmata B. Reduction...
- 21N.2.HL.TZ0.2b.ii: State the stage of meiosis where chiasmata formation may occur.
- 21M.1.HL.TZ2.35: What process occurs in both mitosis and meiosis? A. Formation of chiasmata B. Reduction...
- 21N.2.HL.TZ0.6c: Distinguish between continuous and discrete variation, using examples.
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21M.2.HL.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.
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21M.2.HL.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.
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21M.2.HL.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.
- 21N.1.HL.TZ1.35: A dihybrid cross was carried out between two plants to determine whether the genes for seed shape...
- 21N.2.HL.TZ0.2b.ii: State the stage of meiosis where chiasmata formation may occur.
- 21N.1.HL.TZ1.35: A dihybrid cross was carried out between two plants to determine whether the genes for seed shape...
- 21N.2.HL.TZ0.6c: Distinguish between continuous and discrete variation, using examples.
- 21N.2.HL.TZ0.c: Distinguish between continuous and discrete variation, using examples.
- 21N.2.HL.TZ0.b.ii: State the stage of meiosis where chiasmata formation may occur.
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22M.2.HL.TZ2.3d:
When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.
- 22M.2.HL.TZ2.3a: State the alternative hypothesis for this study.
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22M.2.HL.TZ2.3d:
When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.
- 22M.2.HL.TZ2.3a: State the alternative hypothesis for this study.
- 22M.2.HL.TZ2.a: State the alternative hypothesis for this study.
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22M.2.HL.TZ1.8c:
Explain the reasons for variation in human height.
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22M.2.HL.TZ2.d:
When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.
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22M.2.HL.TZ2.3b:
To calculate chi-squared, expected values must first be calculated. Assuming that there is no association between the two species, calculate the expected number of quadrats in which both species would be present, showing your working.
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22M.2.HL.TZ2.3b:
To calculate chi-squared, expected values must first be calculated. Assuming that there is no association between the two species, calculate the expected number of quadrats in which both species would be present, showing your working.
-
22M.2.HL.TZ2.b:
To calculate chi-squared, expected values must first be calculated. Assuming that there is no association between the two species, calculate the expected number of quadrats in which both species would be present, showing your working.
-
22M.2.HL.TZ2.3c:
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
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22M.2.HL.TZ1.8c:
Explain the reasons for variation in human height.
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22M.2.HL.TZ1.c:
Explain the reasons for variation in human height.
-
22M.2.HL.TZ2.3c:
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
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22M.2.HL.TZ2.7c:
Explain how speciation occurs, including the different processes of isolation and selection.
-
22M.2.HL.TZ2.c:
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
- 22M.1.HL.TZ1.35: What is polyploidy? A. Having an extra set of chromosomes B. Having an extra sex chromosome C....
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22M.2.HL.TZ2.7c:
Explain how speciation occurs, including the different processes of isolation and selection.
-
22M.2.HL.TZ2.c:
Explain how speciation occurs, including the different processes of isolation and selection.
- 22M.1.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1.HL.TZ1.35: What is polyploidy? A. Having an extra set of chromosomes B. Having an extra sex chromosome C....
- 22M.1.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
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22M.1.HL.TZ2.35:
When a cell divides by meiosis, chiasmata can be observed. Which are features of chiasmata?
I. They are points of attachment between chromatids of non-homologous chromosomes.
II. They occur during meiosis I.
III. They increase stability of bivalents.
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
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22M.1.HL.TZ2.36:
Three-spined stickleback fish (Gasterosteus aculeatus) vary in the number of armour plates. The graph shows the frequency of individuals with low, partial or complete plating in a three-spined stickleback population living in Kennedy Lake, Vancouver Island, Canada.
[Source: Reprinted from Current Biology, 24, Marchinko, K.B., Matthews, B., Arnegard, M.E., Rogers, S.M. and Schluter, D., Maintenance of a Genetic Polymorphism with Disruptive Natural Selection in Stickleback.
2014. pp.1289–1292 with permission from Elsevier.]Which type of natural selection could result in this pattern of variation in the population?
A. Disruptive
B. Directional
C. Stabilizing
D. Convergent
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22M.1.HL.TZ2.35:
When a cell divides by meiosis, chiasmata can be observed. Which are features of chiasmata?
I. They are points of attachment between chromatids of non-homologous chromosomes.
II. They occur during meiosis I.
III. They increase stability of bivalents.
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
- 18N.1.HL.TZ0.36: The image shows variation in height of adult humans. What can explain the variation? A. One...
-
22M.1.HL.TZ2.36:
Three-spined stickleback fish (Gasterosteus aculeatus) vary in the number of armour plates. The graph shows the frequency of individuals with low, partial or complete plating in a three-spined stickleback population living in Kennedy Lake, Vancouver Island, Canada.
[Source: Reprinted from Current Biology, 24, Marchinko, K.B., Matthews, B., Arnegard, M.E., Rogers, S.M. and Schluter, D., Maintenance of a Genetic Polymorphism with Disruptive Natural Selection in Stickleback.
2014. pp.1289–1292 with permission from Elsevier.]Which type of natural selection could result in this pattern of variation in the population?
A. Disruptive
B. Directional
C. Stabilizing
D. Convergent
- 18N.1.HL.TZ0.36: The image shows variation in height of adult humans. What can explain the variation? A. One...
- 18M.1.HL.TZ2.21: Which example shows disruptive selection? A. Giraffe necks have become longer over time. B....
- 18N.2.HL.TZ0.8c: Research suggests that many living plant species are polyploid. Explain how polyploidy occurs...
- 18N.2.HL.TZ0.8c: Research suggests that many living plant species are polyploid. Explain how polyploidy occurs...
- 18N.2.HL.TZ0.c: Research suggests that many living plant species are polyploid. Explain how polyploidy occurs...
- 18M.1.HL.TZ2.21: Which example shows disruptive selection? A. Giraffe necks have become longer over time. B....
- 18M.1.HL.TZ1.10: What is always passed to the next generation as a result of sexual reproduction? A. Homologous...
- 18M.1.HL.TZ1.35: What forms when two different chromatids of the same homologous pair cross over? A. Daughter...
- 18M.1.HL.TZ1.10: What is always passed to the next generation as a result of sexual reproduction? A. Homologous...
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18M.1.HL.TZ1.36:
What could account for this distribution of height in a population?
[Source: Graph adapted from Six Minutes http://sixminutes.dlugan.com/good-public-speaker-average/]
A. Gene linkage
B. Dominant alleles
C. Independent assortment
D. Multiple genes
- 18M.1.HL.TZ2.17: Which process could cause non-disjunction if it occurred during meiosis? A. Sister chromatids do...
- 18M.1.HL.TZ1.35: What forms when two different chromatids of the same homologous pair cross over? A. Daughter...
- 18M.1.HL.TZ2.17: Which process could cause non-disjunction if it occurred during meiosis? A. Sister chromatids do...
- 18M.2.HL.TZ2.3a.i: The micrograph shows a plant cell of Lilium grandiflorum during meiosis. Identify, giving...
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18M.1.HL.TZ1.36:
What could account for this distribution of height in a population?
[Source: Graph adapted from Six Minutes http://sixminutes.dlugan.com/good-public-speaker-average/]
A. Gene linkage
B. Dominant alleles
C. Independent assortment
D. Multiple genes
- 19M.1.HL.TZ1.35: Natural selection can operate in different ways. What is the effect of disruptive selection? A....
- 18M.2.HL.TZ2.3a.ii: Outline the law of independent assortment.
- 18M.2.HL.TZ2.3a.ii: Outline the law of independent assortment.
- 18M.2.HL.TZ2.a.ii: Outline the law of independent assortment.
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19M.1.HL.TZ2.19:
The cheetah (Acinonyx jubatus) is a large cat found in Africa. It has been discovered that organs could be transferred between any two individuals without rejection of the organ.
What is the probable reason for this?
A. Cheetahs have poor reproductive success.
B. Cheetahs have high heterozygosity.
C. Cheetahs have a large gene pool.
D. Cheetahs have a small gene pool.
- 18M.2.HL.TZ2.3a.i: The micrograph shows a plant cell of Lilium grandiflorum during meiosis. Identify, giving...
- 18M.2.HL.TZ2.a.i: The micrograph shows a plant cell of Lilium grandiflorum during meiosis. Identify, giving...
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19M.1.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.1.HL.TZ1.35: Natural selection can operate in different ways. What is the effect of disruptive selection? A....
-
19M.1.HL.TZ2.19:
The cheetah (Acinonyx jubatus) is a large cat found in Africa. It has been discovered that organs could be transferred between any two individuals without rejection of the organ.
What is the probable reason for this?
A. Cheetahs have poor reproductive success.
B. Cheetahs have high heterozygosity.
C. Cheetahs have a large gene pool.
D. Cheetahs have a small gene pool.
- 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...
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19M.1.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.TZ1.2d.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...
- 19N.2.HL.TZ0.2c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
- 19M.2.HL.TZ1.d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.1.HL.TZ0.36: A hunter tends to kill the bigger individuals of a population for their meat or for large...
- 19N.1.HL.TZ0.36: A hunter tends to kill the bigger individuals of a population for their meat or for large...
- 19N.2.HL.TZ0.2c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.b: Identify the recombinants.
- 19N.2.HL.TZ0.c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
-
22N.2.HL.TZ0.8b:
Describe polyploidy and how it can lead to speciation.
- 19N.1.HL.TZ0.35: Which event happens in meiosis II but not in meiosis I? A. Spindle microtubules attach to...
- 19N.1.HL.TZ0.35: Which event happens in meiosis II but not in meiosis I? A. Spindle microtubules attach to...
- 22N.1.HL.TZ0.36: Two kinds of wolf spider rub specialized body parts together in order to produce distinct sounds...
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22N.2.HL.TZ0.8b:
Describe polyploidy and how it can lead to speciation.
- 22N.1.HL.TZ0.36: Two kinds of wolf spider rub specialized body parts together in order to produce distinct sounds...
- 22N.1.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
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22N.2.HL.TZ0.b:
Describe polyploidy and how it can lead to speciation.
- 19M.2.HL.TZ1.7b: Explain how natural selection can lead to speciation.
- 22N.1.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
- 20N.2.HL.TZ0.5a: Identify the genotype of the male wire-haired dog.
- 20N.2.HL.TZ0.5a: Identify the genotype of the male wire-haired dog.
- 20N.2.HL.TZ0.a: Identify the genotype of the male wire-haired dog.
- 20N.2.HL.TZ0.5b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
- 19M.2.HL.TZ1.7b: Explain how natural selection can lead to speciation.
- 20N.2.HL.TZ0.5b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
- 20N.2.HL.TZ0.b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
- 20N.1.HL.TZ0.36: Many commercially produced bananas are triploid instead of diploid. The nucleus of a triploid...
- 20N.1.HL.TZ0.36: Many commercially produced bananas are triploid instead of diploid. The nucleus of a triploid...
- 19M.2.HL.TZ1.b: Explain how natural selection can lead to speciation.
-
22N.2.HL.TZ0.2b:
Explain the reason that Morgan’s results did not agree with expected Mendelian ratios in a dihybrid cross.
-
21N.1.HL.TZ1.36:
The graph shows the relationship between mass at birth and the percentage that die shortly after birth for Scottish Blackface lambs.
[Source: Dwyer, C.M., Conington, J., Corbiere, F., Holmoy, I.H., Muri, K., Nowak, R., Rooke, J., Vipond, J. and Gautier,
J.-M., 2016. Invited review: Improving neonatal survival in small ruminants: science into practice. Animal, 10(3),
pp. 449–459.]What type of selection for the lambs is shown in the graph?
A. Disruptive selection, as there is a drop in mortality at intermediate birth masses
B. Stabilizing selection, as lambs with low or high birth mass are less likely to survive
C. Directional selection, as lambs with a high birth mass have high mortality
D. There is no evidence in the graph of selection, as survival frequency is not shown
-
21N.1.HL.TZ1.36:
The graph shows the relationship between mass at birth and the percentage that die shortly after birth for Scottish Blackface lambs.
[Source: Dwyer, C.M., Conington, J., Corbiere, F., Holmoy, I.H., Muri, K., Nowak, R., Rooke, J., Vipond, J. and Gautier,
J.-M., 2016. Invited review: Improving neonatal survival in small ruminants: science into practice. Animal, 10(3),
pp. 449–459.]What type of selection for the lambs is shown in the graph?
A. Disruptive selection, as there is a drop in mortality at intermediate birth masses
B. Stabilizing selection, as lambs with low or high birth mass are less likely to survive
C. Directional selection, as lambs with a high birth mass have high mortality
D. There is no evidence in the graph of selection, as survival frequency is not shown
-
21M.1.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.
-
21M.1.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.
-
22N.2.HL.TZ0.2b:
Explain the reason that Morgan’s results did not agree with expected Mendelian ratios in a dihybrid cross.
-
22N.2.HL.TZ0.b:
Explain the reason that Morgan’s results did not agree with expected Mendelian ratios in a dihybrid cross.
-
23M.1.HL.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.
- 21N.2.HL.TZ0.2b.i: Draw the same chromosomes to show their structure at the same stage of meiosis if there had been...
- 21N.2.HL.TZ0.2b.i: Draw the same chromosomes to show their structure at the same stage of meiosis if there had been...
- 21N.2.HL.TZ0.b.i: Draw the same chromosomes to show their structure at the same stage of meiosis if there had been...
-
21N.2.HL.TZ0.2c:
Explain gene linkage and its effects on inheritance.
-
23M.1.HL.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.
-
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.1.HL.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.1.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.
-
23M.1.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.
-
23M.1.HL.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.1.HL.TZ1.11:
The micrograph shows a cross section of a lily (Lilium longiflorum) anther.
[Source: Sinhyu / iStock.]
What can be found in X?
A. Haploid nuclei produced by meiosis
B. Diploid nuclei produced by fertilization
C. Haploid nuclei produced by mitosis
D. Diploid nuclei produced by pollination
-
23M.1.HL.TZ1.11:
The micrograph shows a cross section of a lily (Lilium longiflorum) anther.
[Source: Sinhyu / iStock.]
What can be found in X?
A. Haploid nuclei produced by meiosis
B. Diploid nuclei produced by fertilization
C. Haploid nuclei produced by mitosis
D. Diploid nuclei produced by pollination
-
23M.2.HL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.2.HL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.2.HL.TZ1.a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.a:
Explain the conclusion that can be drawn from Cross 1.
Sub sections and their related questions
10.1 Meiosis
- 17N.2.HL.TZ0.05b: Explain the reason for linked genes not following the pattern of inheritance discovered by Mendel.
- 18M.1.HL.TZ1.10: What is always passed to the next generation as a result of sexual reproduction? A. Homologous...
- 18M.1.HL.TZ1.35: What forms when two different chromatids of the same homologous pair cross over? A. Daughter...
- 18M.2.HL.TZ2.3a.i: The micrograph shows a plant cell of Lilium grandiflorum during meiosis. Identify, giving...
- 18M.2.HL.TZ2.3a.ii: Outline the law of independent assortment.
- 18M.1.HL.TZ2.17: Which process could cause non-disjunction if it occurred during meiosis? A. Sister chromatids do...
- 19N.1.HL.TZ0.35: Which event happens in meiosis II but not in meiosis I? A. Spindle microtubules attach to...
- 20N.1.HL.TZ0.35: Andalusian fowl have varied colours and types of feathers. The allele for black feathers is...
- 21M.1.HL.TZ1.35: Which statement is valid regarding chromatids? A. Sister chromatids separate during meiosis...
- 21M.1.HL.TZ2.35: What process occurs in both mitosis and meiosis? A. Formation of chiasmata B. Reduction...
- 21N.2.HL.TZ0.2b.i: Draw the same chromosomes to show their structure at the same stage of meiosis if there had been...
- 21N.2.HL.TZ0.2b.ii: State the stage of meiosis where chiasmata formation may occur.
-
22M.1.HL.TZ2.35:
When a cell divides by meiosis, chiasmata can be observed. Which are features of chiasmata?
I. They are points of attachment between chromatids of non-homologous chromosomes.
II. They occur during meiosis I.
III. They increase stability of bivalents.
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
-
23M.1.HL.TZ1.11:
The micrograph shows a cross section of a lily (Lilium longiflorum) anther.
[Source: Sinhyu / iStock.]
What can be found in X?
A. Haploid nuclei produced by meiosis
B. Diploid nuclei produced by fertilization
C. Haploid nuclei produced by mitosis
D. Diploid nuclei produced by pollination
- 19N.1.HL.TZ0.35: Which event happens in meiosis II but not in meiosis I? A. Spindle microtubules attach to...
- 20N.1.HL.TZ0.35: Andalusian fowl have varied colours and types of feathers. The allele for black feathers is...
- 21M.1.HL.TZ1.35: Which statement is valid regarding chromatids? A. Sister chromatids separate during meiosis...
- 21M.1.HL.TZ2.35: What process occurs in both mitosis and meiosis? A. Formation of chiasmata B. Reduction...
- 21N.2.HL.TZ0.2b.i: Draw the same chromosomes to show their structure at the same stage of meiosis if there had been...
- 21N.2.HL.TZ0.2b.ii: State the stage of meiosis where chiasmata formation may occur.
- 21N.2.HL.TZ0.b.i: Draw the same chromosomes to show their structure at the same stage of meiosis if there had been...
- 21N.2.HL.TZ0.b.ii: State the stage of meiosis where chiasmata formation may occur.
-
22M.1.HL.TZ2.35:
When a cell divides by meiosis, chiasmata can be observed. Which are features of chiasmata?
I. They are points of attachment between chromatids of non-homologous chromosomes.
II. They occur during meiosis I.
III. They increase stability of bivalents.
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
- 17N.2.HL.TZ0.05b: Explain the reason for linked genes not following the pattern of inheritance discovered by Mendel.
- 17N.2.HL.TZ0.b: Explain the reason for linked genes not following the pattern of inheritance discovered by Mendel.
- 18M.1.HL.TZ1.10: What is always passed to the next generation as a result of sexual reproduction? A. Homologous...
-
23M.1.HL.TZ1.11:
The micrograph shows a cross section of a lily (Lilium longiflorum) anther.
[Source: Sinhyu / iStock.]
What can be found in X?
A. Haploid nuclei produced by meiosis
B. Diploid nuclei produced by fertilization
C. Haploid nuclei produced by mitosis
D. Diploid nuclei produced by pollination
- 18M.1.HL.TZ1.35: What forms when two different chromatids of the same homologous pair cross over? A. Daughter...
- 18M.2.HL.TZ2.3a.i: The micrograph shows a plant cell of Lilium grandiflorum during meiosis. Identify, giving...
- 18M.2.HL.TZ2.3a.ii: Outline the law of independent assortment.
- 18M.2.HL.TZ2.a.i: The micrograph shows a plant cell of Lilium grandiflorum during meiosis. Identify, giving...
- 18M.2.HL.TZ2.a.ii: Outline the law of independent assortment.
- 18M.1.HL.TZ2.17: Which process could cause non-disjunction if it occurred during meiosis? A. Sister chromatids do...
10.2 Inheritance
- 17N.1.HL.TZ0.36: In a plant, dark leaves are dominant to pale leaves and yellow seeds are dominant to white...
- 18N.1.HL.TZ0.36: The image shows variation in height of adult humans. What can explain the variation? A. One...
-
19M.1.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.TZ1.2d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.2c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
- 20N.2.HL.TZ0.5a: Identify the genotype of the male wire-haired dog.
- 20N.2.HL.TZ0.5b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
- 20N.1.HL.TZ0.36: Many commercially produced bananas are triploid instead of diploid. The nucleus of a triploid...
-
21M.1.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.
-
21M.2.HL.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.
- 21N.1.HL.TZ1.35: A dihybrid cross was carried out between two plants to determine whether the genes for seed shape...
-
21N.2.HL.TZ0.2c:
Explain gene linkage and its effects on inheritance.
- 21N.2.HL.TZ0.6c: Distinguish between continuous and discrete variation, using examples.
- 22M.1.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
-
22M.2.HL.TZ1.8c:
Explain the reasons for variation in human height.
- 22M.2.HL.TZ2.3a: State the alternative hypothesis for this study.
-
22M.2.HL.TZ2.3b:
To calculate chi-squared, expected values must first be calculated. Assuming that there is no association between the two species, calculate the expected number of quadrats in which both species would be present, showing your working.
-
22M.2.HL.TZ2.3c:
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
-
22M.2.HL.TZ2.3d:
When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.
-
22N.2.HL.TZ0.2b:
Explain the reason that Morgan’s results did not agree with expected Mendelian ratios in a dihybrid cross.
- 22N.1.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
-
23M.2.HL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.1.HL.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.1.HL.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.
-
23M.2.SL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.1.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.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.2c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
- 19N.2.HL.TZ0.b: Identify the recombinants.
- 19N.2.HL.TZ0.c: The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio...
- 20N.2.HL.TZ0.5a: Identify the genotype of the male wire-haired dog.
- 20N.2.HL.TZ0.5b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
- 20N.2.HL.TZ0.a: Identify the genotype of the male wire-haired dog.
- 20N.2.HL.TZ0.b: Using a Punnett square, determine how a smooth-haired puppy could be produced in the offspring.
- 20N.1.HL.TZ0.36: Many commercially produced bananas are triploid instead of diploid. The nucleus of a triploid...
-
21M.1.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.
-
21M.2.HL.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.HL.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.
- 21N.1.HL.TZ1.35: A dihybrid cross was carried out between two plants to determine whether the genes for seed shape...
-
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.
- 21N.2.HL.TZ0.6c: Distinguish between continuous and discrete variation, using examples.
- 21N.2.HL.TZ0.c: Distinguish between continuous and discrete variation, using examples.
- 22M.1.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
-
22M.2.HL.TZ1.8c:
Explain the reasons for variation in human height.
-
22M.2.HL.TZ1.c:
Explain the reasons for variation in human height.
- 22M.2.HL.TZ2.3a: State the alternative hypothesis for this study.
-
22M.2.HL.TZ2.3b:
To calculate chi-squared, expected values must first be calculated. Assuming that there is no association between the two species, calculate the expected number of quadrats in which both species would be present, showing your working.
-
22M.2.HL.TZ2.3c:
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
-
22M.2.HL.TZ2.3d:
When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.
- 22M.2.HL.TZ2.a: State the alternative hypothesis for this study.
-
22M.2.HL.TZ2.b:
To calculate chi-squared, expected values must first be calculated. Assuming that there is no association between the two species, calculate the expected number of quadrats in which both species would be present, showing your working.
-
22M.2.HL.TZ2.c:
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
-
22M.2.HL.TZ2.d:
When the data in the table were used to calculate chi-squared, the calculated value was 0.056. The critical value is 3.84. Explain the conclusion that can be drawn from the calculated and critical values for chi-squared.
-
22N.2.HL.TZ0.2b:
Explain the reason that Morgan’s results did not agree with expected Mendelian ratios in a dihybrid cross.
-
22N.2.HL.TZ0.b:
Explain the reason that Morgan’s results did not agree with expected Mendelian ratios in a dihybrid cross.
- 22N.1.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
- 17N.1.HL.TZ0.36: In a plant, dark leaves are dominant to pale leaves and yellow seeds are dominant to white...
-
23M.2.HL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.2.HL.TZ1.a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.1.HL.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.1.HL.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.
-
23M.2.SL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.2.SL.TZ1.a:
Explain the conclusion that can be drawn from Cross 1.
-
23M.1.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.
- 18N.1.HL.TZ0.36: The image shows variation in height of adult humans. What can explain the variation? A. One...
-
19M.1.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.TZ1.2d.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...
10.3 Gene pools and speciation
- 17N.1.HL.TZ0.35: How do the concepts of gradualism and punctuated equilibrium differ? A. The timing of...
-
18M.1.HL.TZ1.36:
What could account for this distribution of height in a population?
[Source: Graph adapted from Six Minutes http://sixminutes.dlugan.com/good-public-speaker-average/]
A. Gene linkage
B. Dominant alleles
C. Independent assortment
D. Multiple genes
- 18M.1.HL.TZ2.21: Which example shows disruptive selection? A. Giraffe necks have become longer over time. B....
- 18N.2.HL.TZ0.8c: Research suggests that many living plant species are polyploid. Explain how polyploidy occurs...
- 19M.1.HL.TZ1.35: Natural selection can operate in different ways. What is the effect of disruptive selection? A....
-
19M.1.HL.TZ2.19:
The cheetah (Acinonyx jubatus) is a large cat found in Africa. It has been discovered that organs could be transferred between any two individuals without rejection of the organ.
What is the probable reason for this?
A. Cheetahs have poor reproductive success.
B. Cheetahs have high heterozygosity.
C. Cheetahs have a large gene pool.
D. Cheetahs have a small gene pool.
- 19M.2.HL.TZ1.7b: Explain how natural selection can lead to speciation.
- 19N.1.HL.TZ0.36: A hunter tends to kill the bigger individuals of a population for their meat or for large...
-
20N.2.HL.TZ0.8b:
Describe the changes that occur in gene pools during speciation.
- 21M.1.HL.TZ2.36: Which cell is a polyploid zygote produced by fusion of one haploid and one diploid gamete?
-
21N.1.HL.TZ1.36:
The graph shows the relationship between mass at birth and the percentage that die shortly after birth for Scottish Blackface lambs.
[Source: Dwyer, C.M., Conington, J., Corbiere, F., Holmoy, I.H., Muri, K., Nowak, R., Rooke, J., Vipond, J. and Gautier,
J.-M., 2016. Invited review: Improving neonatal survival in small ruminants: science into practice. Animal, 10(3),
pp. 449–459.]What type of selection for the lambs is shown in the graph?
A. Disruptive selection, as there is a drop in mortality at intermediate birth masses
B. Stabilizing selection, as lambs with low or high birth mass are less likely to survive
C. Directional selection, as lambs with a high birth mass have high mortality
D. There is no evidence in the graph of selection, as survival frequency is not shown
- 22M.1.HL.TZ1.35: What is polyploidy? A. Having an extra set of chromosomes B. Having an extra sex chromosome C....
-
22M.1.HL.TZ2.36:
Three-spined stickleback fish (Gasterosteus aculeatus) vary in the number of armour plates. The graph shows the frequency of individuals with low, partial or complete plating in a three-spined stickleback population living in Kennedy Lake, Vancouver Island, Canada.
[Source: Reprinted from Current Biology, 24, Marchinko, K.B., Matthews, B., Arnegard, M.E., Rogers, S.M. and Schluter, D., Maintenance of a Genetic Polymorphism with Disruptive Natural Selection in Stickleback.
2014. pp.1289–1292 with permission from Elsevier.]Which type of natural selection could result in this pattern of variation in the population?
A. Disruptive
B. Directional
C. Stabilizing
D. Convergent
-
22M.2.HL.TZ2.7c:
Explain how speciation occurs, including the different processes of isolation and selection.
-
22N.2.HL.TZ0.8b:
Describe polyploidy and how it can lead to speciation.
- 22N.1.HL.TZ0.36: Two kinds of wolf spider rub specialized body parts together in order to produce distinct sounds...
- 19N.1.HL.TZ0.36: A hunter tends to kill the bigger individuals of a population for their meat or for large...
-
20N.2.HL.TZ0.8b:
Describe the changes that occur in gene pools during speciation.
-
20N.2.HL.TZ0.b:
Describe the changes that occur in gene pools during speciation.
- 21M.1.HL.TZ2.36: Which cell is a polyploid zygote produced by fusion of one haploid and one diploid gamete?
-
21N.1.HL.TZ1.36:
The graph shows the relationship between mass at birth and the percentage that die shortly after birth for Scottish Blackface lambs.
[Source: Dwyer, C.M., Conington, J., Corbiere, F., Holmoy, I.H., Muri, K., Nowak, R., Rooke, J., Vipond, J. and Gautier,
J.-M., 2016. Invited review: Improving neonatal survival in small ruminants: science into practice. Animal, 10(3),
pp. 449–459.]What type of selection for the lambs is shown in the graph?
A. Disruptive selection, as there is a drop in mortality at intermediate birth masses
B. Stabilizing selection, as lambs with low or high birth mass are less likely to survive
C. Directional selection, as lambs with a high birth mass have high mortality
D. There is no evidence in the graph of selection, as survival frequency is not shown
- 22M.1.HL.TZ1.35: What is polyploidy? A. Having an extra set of chromosomes B. Having an extra sex chromosome C....
-
22M.1.HL.TZ2.36:
Three-spined stickleback fish (Gasterosteus aculeatus) vary in the number of armour plates. The graph shows the frequency of individuals with low, partial or complete plating in a three-spined stickleback population living in Kennedy Lake, Vancouver Island, Canada.
[Source: Reprinted from Current Biology, 24, Marchinko, K.B., Matthews, B., Arnegard, M.E., Rogers, S.M. and Schluter, D., Maintenance of a Genetic Polymorphism with Disruptive Natural Selection in Stickleback.
2014. pp.1289–1292 with permission from Elsevier.]Which type of natural selection could result in this pattern of variation in the population?
A. Disruptive
B. Directional
C. Stabilizing
D. Convergent
-
22M.2.HL.TZ2.7c:
Explain how speciation occurs, including the different processes of isolation and selection.
-
22M.2.HL.TZ2.c:
Explain how speciation occurs, including the different processes of isolation and selection.
-
22N.2.HL.TZ0.8b:
Describe polyploidy and how it can lead to speciation.
-
22N.2.HL.TZ0.b:
Describe polyploidy and how it can lead to speciation.
- 22N.1.HL.TZ0.36: Two kinds of wolf spider rub specialized body parts together in order to produce distinct sounds...
- 17N.1.HL.TZ0.35: How do the concepts of gradualism and punctuated equilibrium differ? A. The timing of...
-
18M.1.HL.TZ1.36:
What could account for this distribution of height in a population?
[Source: Graph adapted from Six Minutes http://sixminutes.dlugan.com/good-public-speaker-average/]
A. Gene linkage
B. Dominant alleles
C. Independent assortment
D. Multiple genes
- 18M.1.HL.TZ2.21: Which example shows disruptive selection? A. Giraffe necks have become longer over time. B....
- 18N.2.HL.TZ0.8c: Research suggests that many living plant species are polyploid. Explain how polyploidy occurs...
- 18N.2.HL.TZ0.c: Research suggests that many living plant species are polyploid. Explain how polyploidy occurs...
- 19M.1.HL.TZ1.35: Natural selection can operate in different ways. What is the effect of disruptive selection? A....
-
19M.1.HL.TZ2.19:
The cheetah (Acinonyx jubatus) is a large cat found in Africa. It has been discovered that organs could be transferred between any two individuals without rejection of the organ.
What is the probable reason for this?
A. Cheetahs have poor reproductive success.
B. Cheetahs have high heterozygosity.
C. Cheetahs have a large gene pool.
D. Cheetahs have a small gene pool.
- 19M.2.HL.TZ1.7b: Explain how natural selection can lead to speciation.
- 19M.2.HL.TZ1.b: Explain how natural selection can lead to speciation.