Final answer:
In the G1 phase, a person with Down syndrome, who has trisomy 21, would have 39 allele copies (for the gene on chromosome 21) representing the potential for 13 different alleles.
Step-by-step explanation:
When considering a person with Down syndrome, who has trisomy 21 (three copies of chromosome 21), let's analyze each statement given:
- A) In the G1 phase of the cell cycle, before DNA replication, each chromosome is a single chromatid. Therefore, since individuals with Down syndrome have three copies of chromosome 21, they would have 39 allele copies (three copies of 13 alleles for the gene in question).
- B) In meiosis I anaphase, homologous chromosomes are segregating, so a person would have 23 individual chromosomes, assuming non-disjunction does not occur. However, an individual with Down syndrome is trisomic for chromosome 21, and nondisjunction is precisely the error that led to the condition. Still, since the question assumes a normal meiosis I anaphase, we can infer that there would be at least two, potentially three (depending on the disjunction of the extra chromosome 21), copies of each of the 13 different alleles for the gene.
- C) As explained in option A, in the G1 phase, a person with Down syndrome would have 39 allele copies if we consider an extra copy of all 13 alleles on chromosome 21 due to trisomy 21.
- D) In meiosis I anaphase, if nondisjunction occurs, we could have irregular numbers of alleles. However, in a normal scenario, each homologous chromosome is separated into different cells, which would result in having 26 allele copies after the separation of the homologs.
Based on this, the correct statement appears to be C) In the G1 phase, a person with Down syndrome has 39 allele copies; potential for 13 different alleles.