Bsci110b Chapter 15 Concept Check

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.