Inheritance: Question 7

Worked solution

Part (a): Independent assortment at metaphase I and anaphase I

At metaphase I, the four pairs of homologous chromosomes each line up at the equator of the cell. The orientation of each pair, which homologue faces which pole, is random and is set independently for every pair; the orientation of one pair has no influence on the orientation of any other pair. At anaphase I, the two homologues of each pair are then pulled apart to opposite poles of the cell. Because each pair was orientated independently, the particular mixture of maternal and paternal chromosomes that ends up in each daughter cell is essentially random, and this randomness is what independent assortment contributes to genetic variation.

Part (b): Number of chromosome combinations from independent assortment

With 4 pairs of homologous chromosomes, each pair can end up in a gamete in one of 2 possible ways (as the maternal homologue or the paternal homologue), and this choice is made independently for each of the 4 pairs.

The number of possible combinations is therefore: $2^4 = 2 \times 2 \times 2 \times 2 = 16$

So, through independent assortment alone (with no crossing over), a single individual of this species can produce 16 genetically different combinations of whole chromosomes in its gametes.

Part (c): Why crossing over increases this number further

The calculation in part (b) treats each chromosome as a single, unchanging unit that is either “maternal” or “paternal.” Crossing over breaks this assumption: during prophase I, non-sister chromatids of each homologous pair exchange corresponding segments of DNA at chiasmata, before independent assortment separates the pairs. This creates recombinant chromatids that carry a mixture of maternal and paternal alleles along their length, rather than being purely maternal or purely paternal.

Because each of the 16 combinations calculated in part (b) can now be made up of chromosomes with many different possible internal allele combinations (rather than just two, purely maternal or purely paternal, versions of each chromosome), the true number of genetically distinct gametes the worm can produce is far greater than 16.

Part (d): Random fertilisation as a further source of variation

Independent assortment and crossing over both act during meiosis to make the gametes of a single individual genetically diverse. Random fertilisation acts on top of this: which of the very many possible gametes from one parent happens to fuse with which of the very many possible gametes from the other parent is a matter of chance.

Since the two parents’ gametes vary independently of one another, combining them at random multiplies the two parents’ independent sources of variation together, rather than simply adding them. This means the number of genetically different zygotes (and therefore offspring) that could potentially be produced by two parents is far greater than the number of genetically different gametes produced by either parent on its own.

Final answers

• (a) Each homologous pair orientates randomly and independently at metaphase I, then separates at anaphase I, randomising which maternal/paternal chromosomes travel together into each gamete.
• (b) Number of chromosome combinations $= 2^4 = 16$.
• (c) Crossing over creates recombinant chromosomes with new internal allele combinations, so the true number of distinct gametes is far greater than the 16 whole-chromosome combinations alone.
• (d) Random fertilisation combines the independently varying gametes of two parents by chance, multiplying their sources of variation and further increasing offspring diversity.