BSCI110B Chapter 15 Concept Check

1. What is the physical basis of Mendel’s laws? The physical basis for the law of segregation is the separation of homologs in . The physical basis for the law of independent assortment is the alternative arrangements of homologous chromosome pairs in .

During anaphase I of meiosis, homologous chromosomes separate and move to opposite poles of the cell. This is the physical basis for Mendel's law of segregation, which states that alleles for a trait separate during gamete formation. On the other hand, during metaphase I of meiosis, homologous chromosome pairs align randomly at the cell's equator. This random arrangement is the physical basis for Mendel's law of independent assortment, which states that alleles for different traits segregate independently of one another during gamete formation.

Explanation

During anaphase I of meiosis, homologous chromosomes separate and move to opposite poles of the cell. This is the physical basis for Mendel's law of segregation, which states that alleles for a trait separate during gamete formation. On the other hand, during metaphase I of meiosis, homologous chromosome pairs align randomly at the cell's equator. This random arrangement is the physical basis for Mendel's law of independent assortment, which states that alleles for different traits segregate independently of one another during gamete formation.

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2. About 5% of individuals with Down syndrome have a chromosomal translocation in which a third cop of chromosome 21 is attached to chromosome 14. If this translocation occurred in a parent’s gonad, how could it lead to Down syndrome in a child? In meiosis, a combined 14-21 chromosome will behave as one chromosome. If a gamete receives the combined 14-21 chromosome and a normal copy of chromosome 21, trisomy 21 will result when this gamete combines with a normal gamete during .

During fertilization, when the gamete with the combined 14-21 chromosome and a normal copy of chromosome 21 combines with a normal gamete, it results in trisomy 21. Trisomy 21 means that there are three copies of chromosome 21 instead of the usual two, which is characteristic of Down syndrome.

Explanation

During fertilization, when the gamete with the combined 14-21 chromosome and a normal copy of chromosome 21 combines with a normal gamete, it results in trisomy 21. Trisomy 21 means that there are three copies of chromosome 21 instead of the usual two, which is characteristic of Down syndrome.

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3. Mitochondrial genes are critical to the energy metabolism of cells, but mitochondrial disorders caused by mutations in these genes are generally not lethal. Why not? The situation is similar to that for chloroplasts. Each cell contains numerous mitochondria, and in affected individuals, most cells contain a variable mixture of normal and mutant mitochondria. The mitochondria carry out enough cellular respiration for survival.

Mitochondrial genes are critical for energy metabolism in cells. However, mitochondrial disorders caused by mutations in these genes are generally not lethal because each cell contains numerous mitochondria. In affected individuals, most cells contain a mixture of normal and mutant mitochondria. The normal mitochondria are able to carry out enough cellular respiration for survival, compensating for the dysfunctional mutant mitochondria. This allows the cells to still produce enough energy to sustain basic cellular functions and prevent lethality.

Explanation

Mitochondrial genes are critical for energy metabolism in cells. However, mitochondrial disorders caused by mutations in these genes are generally not lethal because each cell contains numerous mitochondria. In affected individuals, most cells contain a mixture of normal and mutant mitochondria. The normal mitochondria are able to carry out enough cellular respiration for survival, compensating for the dysfunctional mutant mitochondria. This allows the cells to still produce enough energy to sustain basic cellular functions and prevent lethality.

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4. Which one of Mendel’s laws relates to the inheritance of alleles for a single character? Which law relates to the inheritance of alleles for two characters in a dihybrid cross? Mendel derived the from following a single character, which states that the two alleles for a heritable character segregate during gamete formation and end up in different gametes. Using a dihybrid cross Mendel developed the , which states that each pair of alleles segregates independently of other pairs of alleles during gamete formation.

Mendel derived the law of segregation from following a single character, which states that the two alleles for a heritable character segregate during gamete formation and end up in different gametes. Using a dihybrid cross Mendel developed the law of independent assortment, which states that each pair of alleles segregates independently of other pairs of alleles during gamete formation.

Explanation

Mendel derived the law of segregation from following a single character, which states that the two alleles for a heritable character segregate during gamete formation and end up in different gametes. Using a dihybrid cross Mendel developed the law of independent assortment, which states that each pair of alleles segregates independently of other pairs of alleles during gamete formation.

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5. Propose a possible reason that the first naturally occurring mutant fruit fly Morgan saw involved a gene on a sex chromosome. To show the mutant phenotype, a male needs to possess mutant allele. If this gene had been on a pair of , two mutant alleles would have had to be present for an individual to show the mutant phenotype, a much less probable situation.

The reason the first naturally occurring mutant fruit fly that Morgan saw involved a gene on a sex chromosome is that for the mutant phenotype to be expressed, a male fly needs to possess the mutant allele. If the gene had been located on a pair of autosomes, two mutant alleles would have had to be present for an individual to show the mutant phenotype. This would be a much less probable situation compared to having the gene on a sex chromosome.

Explanation

The reason the first naturally occurring mutant fruit fly that Morgan saw involved a gene on a sex chromosome is that for the mutant phenotype to be expressed, a male fly needs to possess the mutant allele. If the gene had been located on a pair of autosomes, two mutant alleles would have had to be present for an individual to show the mutant phenotype. This would be a much less probable situation compared to having the gene on a sex chromosome.

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6. A white-eyed female Drosophila is mated with a red-eyed (wild-type) male, the reciprocal cross of that is shown. What phenotypes and genotypes do you predict for the offspring? Because the gene for this eye-color character is located on the X chromosome, all female offspring will be red-eyed ; all male offspring will inherit a Y chromosome from the father and be white-eyed .

In Drosophila, eye color is determined by a gene located on the X chromosome. The white-eyed female Drosophila has two copies of the white-eye gene (XW+XW), while the red-eyed male has one copy of the wild-type gene (XW) and one copy of the white-eye gene (XWY). During the reciprocal cross, all female offspring will inherit one X chromosome from the white-eyed female (XW) and one X chromosome from the red-eyed male (XW), resulting in them having two copies of the wild-type gene (XW+XW) and being red-eyed. On the other hand, all male offspring will inherit a Y chromosome from the father (Y) and an X chromosome from the mother (XW), resulting in them having one copy of the white-eye gene (XWY) and being white-eyed.

Explanation

In Drosophila, eye color is determined by a gene located on the X chromosome. The white-eyed female Drosophila has two copies of the white-eye gene (XW+XW), while the red-eyed male has one copy of the wild-type gene (XW) and one copy of the white-eye gene (XWY). During the reciprocal cross, all female offspring will inherit one X chromosome from the white-eyed female (XW) and one X chromosome from the red-eyed male (XW), resulting in them having two copies of the wild-type gene (XW+XW) and being red-eyed. On the other hand, all male offspring will inherit a Y chromosome from the father (Y) and an X chromosome from the mother (XW), resulting in them having one copy of the white-eye gene (XWY) and being white-eyed.

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7. Neither Tom nor Rhoda has Duchenne muscular dystrophy, but their firstborn son does have it. What is the probability that a second child of this couple will have the disease? What is the probability if the second child is a boy? A girl? There is a ¼ chance that the firstborn son does have it, a chance that the child will inherit a Y chromosome from the father and be male multiplied by a chance that he will inherit the X carrying the disease allele from his mother. If the child is a boy, there is a chance he will have the disease; a female would have zero chance (but chance of being a carrier).

The probability that a second child of this couple will have the disease is 1/4, since there is a 1/4 chance that the firstborn son has it. The probability if the second child is a boy is 1/2, as there is a 1/2 chance that the child will inherit a Y chromosome from the father. The probability if the second child is a girl is also 1/2, as there is a 1/2 chance that the child will inherit an X chromosome carrying the disease allele from the mother.

Explanation

The probability that a second child of this couple will have the disease is 1/4, since there is a 1/4 chance that the firstborn son has it. The probability if the second child is a boy is 1/2, as there is a 1/2 chance that the child will inherit a Y chromosome from the father. The probability if the second child is a girl is also 1/2, as there is a 1/2 chance that the child will inherit an X chromosome carrying the disease allele from the mother.

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8. During early embryonic development of female carriers for color blindness, the normal allele is inactivated by chance in about half the cells. Why, then, aren’t 50% of female carriers color-blind? The cells in the eye responsible for color vision must come from multiple cells in the early embryo. The descendants of of those cells express the allele for normal color vision and the allele for color blindness. Having the number of mature eye cells expressing the normal allele must be sufficient for normal color vision.

During early embryonic development, the inactivation of the normal allele by chance only occurs in about half of the cells. However, the cells responsible for color vision in the eye must come from multiple cells in the early embryo. The descendants of these cells will express both the normal allele and the allele for color blindness. Therefore, having only half of the cells expressing the normal allele is sufficient for normal color vision, as long as the number of mature eye cells expressing the normal allele is enough.

Explanation

During early embryonic development, the inactivation of the normal allele by chance only occurs in about half of the cells. However, the cells responsible for color vision in the eye must come from multiple cells in the early embryo. The descendants of these cells will express both the normal allele and the allele for color blindness. Therefore, having only half of the cells expressing the normal allele is sufficient for normal color vision, as long as the number of mature eye cells expressing the normal allele is enough.

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9. When two genes are located on the same chromosome, what is the physical basis for the production of recombinant offspring in a testcross between a dihybrid parent and a double-mutant (recessive) parent? Crossing over during in the parent produces some gametes with recombinant genotypes for the two genes. Offspring with a recombinant phenotype arise from fertilization of the recombinant gametes by homozygous recessive gametes from the double-mutant parent.

During meiosis I, crossing over occurs between homologous chromosomes, leading to the exchange of genetic material. This can result in the production of gametes with recombinant genotypes for the two genes located on the same chromosome. In a testcross between a dihybrid parent and a double-mutant (recessive) parent, the dihybrid parent is heterozygous for both genes. When fertilization occurs between the recombinant gametes produced during meiosis I and the homozygous recessive gametes from the double-mutant parent, offspring with a recombinant phenotype can be produced.

Explanation

During meiosis I, crossing over occurs between homologous chromosomes, leading to the exchange of genetic material. This can result in the production of gametes with recombinant genotypes for the two genes located on the same chromosome. In a testcross between a dihybrid parent and a double-mutant (recessive) parent, the dihybrid parent is heterozygous for both genes. When fertilization occurs between the recombinant gametes produced during meiosis I and the homozygous recessive gametes from the double-mutant parent, offspring with a recombinant phenotype can be produced.

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10. Genes A, B, and C are located on the same chromosome. Testcrosses show that the recombination frequency between A and B is 28% and between A and C is 12%. Can you determine the linear order of these genes? Explain. . To determine which possibility is correct, you need to know the recombination frequency between and .

Based on the given information, it is not possible to determine the linear order of genes A, B, and C. The recombination frequency between A and B is 28%, indicating that they are relatively far apart on the chromosome. The recombination frequency between A and C is 12%, suggesting that they are closer together. However, the information provided does not give any indication of the distance or order between B and C. Therefore, without knowing the recombination frequency between B and C, we cannot determine their linear order.

Explanation

Based on the given information, it is not possible to determine the linear order of genes A, B, and C. The recombination frequency between A and B is 28%, indicating that they are relatively far apart on the chromosome. The recombination frequency between A and C is 12%, suggesting that they are closer together. However, the information provided does not give any indication of the distance or order between B and C. Therefore, without knowing the recombination frequency between B and C, we cannot determine their linear order.

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11. More common than completely polyploidy animals are mosaic polyploids, animals that are diploid except for patches of polyploidy cells. How might a mosaic tetraploid, an animal with some cells containing four sets of chromosomes arise? At some point during development, one of the embryo’s cells may have to carry out mitosis after duplicating its chromosomes. Subsequent normal cell cycles would produce genetic copies of this tetraploid cell.

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Explanation

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12. The ABO blood type locus has been mapped on chromosome 9. A father who has type AB blood and a mother who has type O blood have a child with trisomy 9 and type A blood. Using this information, can you tell in which parent the nondisjunction occurred? . The child can be either I^AI^A or I^Aii. A sperm of genotype I^AI^A could result from nondisjunction in the father during , while an egg with the genotype ii could result from nondisjunction in the mother during either or .

The correct answer is "No; meiosis II; meiosis I; meiosis II". The child can have either genotype IAIA or IAii. If the father had a nondisjunction event during meiosis II, it would result in a sperm with genotype IAIA. If the mother had a nondisjunction event during meiosis I, it would result in an egg with genotype ii. Therefore, the nondisjunction event likely occurred in the father during meiosis II.

Explanation

The correct answer is "No; meiosis II; meiosis I; meiosis II". The child can have either genotype IAIA or IAii. If the father had a nondisjunction event during meiosis II, it would result in a sperm with genotype IAIA. If the mother had a nondisjunction event during meiosis I, it would result in an egg with genotype ii. Therefore, the nondisjunction event likely occurred in the father during meiosis II.

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13. Gene dosage, the number of active copies of a gene, is important to proper development. Identify and describe two processes that establish the proper dosage of certain genes. Inactivation of an chromosome in females and . Because of X inactivation, the effective dose of genes on the X chromosome is in males and females. As a result of genomic imprinting, allele of certain genes is phenotypically expressed.

During X inactivation, one of the X chromosomes in females is randomly inactivated, leading to an equal dosage of genes on the X chromosome in both males and females. This ensures that the proper dosage of genes is maintained. Genomic imprinting is another process that establishes gene dosage by causing the expression of only one allele of certain genes. This means that only one copy of the gene is active, resulting in the proper dosage of these genes.

Explanation

During X inactivation, one of the X chromosomes in females is randomly inactivated, leading to an equal dosage of genes on the X chromosome in both males and females. This ensures that the proper dosage of genes is maintained. Genomic imprinting is another process that establishes gene dosage by causing the expression of only one allele of certain genes. This means that only one copy of the gene is active, resulting in the proper dosage of these genes.

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14. Reciprocal crosses between two primrose varieties, A and B, produced the following results: A female X B male => offspring with all green (nonvariegated) leaves. B female X A male => offspring with spotted (variegated) leaves. Explain. The genes for leaf coloration are located in within the cytoplasm. Normally, only the parent transmits plastids genes to offspring. Since variegated offspring are produced only when the female parent is of the variety, we can conclude that variety B contains both the wild-type and mutant alleles of pigment genes, producing variegated leaves.

The genes for leaf coloration are located within the plastids, which are inherited maternally. Normally, only the maternal parent transmits plastid genes to offspring. In this case, variegated offspring are produced only when the female parent is of variety B. This suggests that variety B contains both the wild-type and mutant alleles of pigment genes, resulting in variegated leaves.

Explanation

The genes for leaf coloration are located within the plastids, which are inherited maternally. Normally, only the maternal parent transmits plastid genes to offspring. In this case, variegated offspring are produced only when the female parent is of variety B. This suggests that variety B contains both the wild-type and mutant alleles of pigment genes, resulting in variegated leaves.

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Bsci110b Chapter 15 Concept Check