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10.2 Inheritance
Description
| Nature of science: Looking for patterns, trends and discrepancies—Mendel used observations of the natural world to find and explain patterns and trends. Since then, scientists have looked for discrepancies and asked questions based on further observations to show exceptions to the rules. For example, Morgan discovered non-Mendelian ratios in his experiments with Drosophila. (3.1) |
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Understandings:
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Theory of knowledge:
Biology Topic 3.4 Inheritance Aims:
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Directly related questions
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22M.2.HL.TZ1.8c:
Explain the reasons for variation in human height.
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22M.2.HL.TZ1.8c:
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.
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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.TZ2.3c:
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.c:
State the number of degrees of freedom for this test to determine the critical value of chi-squared.
- 22M.1.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 22M.1.HL.TZ1.36: An individual is heterozygous for two linked genes . To investigate the frequency of crossing...
- 18N.1.HL.TZ0.36: The image shows variation in height of adult humans. What can explain the variation? A. One...
- 18N.1.HL.TZ0.36: The image shows variation in height of adult humans. What can explain the variation? A. One...
- 19M.2.HL.TZ1.2d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
- 19M.2.HL.TZ1.2d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
- 19M.2.HL.TZ1.d.ii: The offspring, which were all heterozygous for grey body and normal wings, were crossed with...
- 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...
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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.
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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.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...
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.2b: Identify the recombinants.
- 19N.2.HL.TZ0.b: 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.TZ0.2c: 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...
- 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.
- 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...
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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.
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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.
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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.1.HL.TZ1.35: A dihybrid cross was carried out between two plants to determine whether the genes for seed shape...
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21N.2.HL.TZ0.2c:
Explain gene linkage and its effects on inheritance.
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21N.2.HL.TZ0.2c:
Explain gene linkage and its effects on inheritance.
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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.TZ0.6c: Distinguish between continuous and discrete variation, using examples.
- 21N.2.HL.TZ0.c: Distinguish between continuous and discrete variation, using examples.
- 22M.2.HL.TZ2.3a: State the alternative hypothesis for this study.
- 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.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.
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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.
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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.
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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.
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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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22N.2.HL.TZ0.2b:
Explain the reason that Morgan’s results did not agree with expected Mendelian ratios in a dihybrid cross.
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22N.2.HL.TZ0.2b:
Explain the reason that Morgan’s results did not agree with expected Mendelian ratios in a dihybrid cross.
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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...
- 22N.1.HL.TZ0.35: Black, short-haired guinea pigs, heterozygous for both characteristics, were crossed. They...
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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.
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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.
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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.
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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.
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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.
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23M.2.HL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
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23M.2.SL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
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23M.2.HL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
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23M.2.HL.TZ1.a:
Explain the conclusion that can be drawn from Cross 1.
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23M.2.SL.TZ1.5a:
Explain the conclusion that can be drawn from Cross 1.
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23M.2.SL.TZ1.a:
Explain the conclusion that can be drawn from Cross 1.
