Suppose there are two genes on two different chromosomes, one gene called G and the other called D. An individual has the genotype GgDd. Which of the following drawings correctly shows cells in this individual after DNA replication but before cell division of the first meiosis? Assume no recombination/crossing-over occurs between the chromosomes. a) c) G-Ggg D|| Data Gg G-g D-d D-d b) d) GgGg D-a Da GG gg D-D d-d

Meiosis is reduction division. In this process, 4 different haploid daughter cells are produced from one diploid parent cell. There is Meiosis I and Meiosis II. Meiosis I can again divided in Prophage I Metaphase I Anaphase I Telophase I And Meiosis II also divided in Prophage II Metaphase II Anaphase II Telophase IIMeiosis I- this stage is also called reduction division. Because in this step 2 different haploid(x) daughter cells are produced from one diploid (2x) parent cell. That means half number of chromosomes are present in daughter cell than mother cell. Here the genotype of the individual is GgDd. So at the 1st stage, he/she contains 4 chromosomes with single chromatid. After replication, he/she contains 4 chromosomes with 2 chromatids. Hence the option A is the correct answer. Option B is not the correct answer because here G, g alleles are present on same chromosome, and D, d alleles are present on same chromosome. Replication is semiconservative so after replication same copy of gene will produce. So after replication of G chromosome- GG chromosome should produce (same for others). Option C is not the correct answer because here G and D genes should present on different chromosomes. But here both genes are present on same chromosome. Option D is not the correct answer because here G and D genes should present on different chromosomes. But here both genes are present on same chromosome.