Sister chromatids are attached along their lengths by protein complexes called cohesins. In mitosis, this attachment lasts until the end of metaphase, when enzymes cleave the cohesins, freeing the sister chromatids to move to opposite poles of the cell. In meiosis, sister chromatid cohesion is released in two steps, one at the start of anaphase I and one at anaphase II. In metaphase I, homologs are held together by cohesion between sister chromatid arms in regions beyond points of crossing over, where stretches of sister chromatids now belong to different chromosomes. The combination of crossing over and sister chromatid cohesion along the arms results in the formation of a chiasma. Chiasmata hold homologs together as the spindle forms for the first meiotic division. At the onset of anaphase I, the release of cohesion along sister chromatid arms allows homologs to separate. At anaphase II, the release of sister chromatid cohesion at the centromeres allows the sister chromatids to separate. Thus, sister chromatid cohesion and crossing over, acting together, play an essential role in the lining up of chromosomes by homologous pairs at metaphase I. • As seen in a karyotype, normal human somatic cells are diploid. They have 46 chromosomes made up of two sets of 23—one set from each parent. In human diploid cells, there are 22 homologous pairs of autosomes, each with a maternal and a paternal homolog. The 23rd pair, the sex chromosomes, determines whether the person is female (XX) or male (XY).
• At sexual maturity in the human life cycle, ovaries and testes (the gonads) produce haploid gametes by meiosis, each gamete containing a single set of 23 chromosomes (n 23). During fertilization, an egg and sperm unite, forming a diploid (2n 46) single-celled zygote, which develops into a multicellular organism by mitosis.
• Sexual life cycles differ in the timing of meiosis relative to fertilization and in the point(s) of the cycle at which a multicellular organism is produced by mitosis.
start by listing all genotypes we could get that fulfill this condition: ppyyRr, ppYyrr, Ppyyrr, PPyyrr, and ppyyrr. (Because the condition is at least two recessive traits, it includes the last genotype, which shows all three recessive traits.) Next, we calculate the probability for each of these genotypes resulting from our PpYyRr x Ppyyrr cross by multiplying together the individual probabilities for the allele pairs, just as we did in our dihybrid example. Note that in a cross involving heterozygous and homozygous allele pairs (for example, Yy yy), the probability of heterozygous offspring is 1 ⁄2 and the probability of homozygous offspring is 1 ⁄2. Finally, we use the addition rule to add the probabilities for all the different genotypes that fulfill the condition of at least two recessive traits Kenneth R. Miller, Levine
Kenneth R. Miller, Levine
Christy C. Hayhoe, Doug Hayhoe, Jeff Major, Maurice DiGiuseppe