Written for this site from the class slides, with textbook page references for each statement. The chapter is on pages 747–759 of the course textbook. Image credits are on slide 14.
In meiosis I, homologous chromosomes separate, so each gamete gets one allele of each gene (segregation). Each pair of homologues lines up at random in metaphase I, so genes on different chromosomes are inherited independently (independent assortment). A parent RrYy makes RY, Ry, rY and ry gametes in equal numbers.
Crossing two double heterozygotes for unlinked genes gives a 9:3:3:1 phenotype ratio. A test cross of a double heterozygote with a double homozygous recessive gives 1:1:1:1.
Databases give each human gene’s locus and product. HBB (beta globin) and HBD (delta globin) lie next to each other on chromosome 11; HBA1 (alpha globin) is on chromosome 16; CFTR is on chromosome 7.
Genes on the same chromosome are linked and tend to be inherited together, so they do not assort independently. They are shown with alleles beside vertical lines, one line per homologous chromosome. The closer the loci, the stronger the linkage.
A recombinant has a combination of alleles not found in either parent. With linked genes, recombinants come only from crossing over between the loci, so a test cross gives two large parental classes and two small recombinant classes.
The null hypothesis is that observed numbers do not differ significantly from those expected (for example 9:3:3:1). Calculate χ2 = Σ(O − E)2/E, with degrees of freedom = classes − 1. If χ2 is greater than the critical value at p = 0.05 (7.81 for 3 degrees of freedom), reject the null hypothesis; the genes may be linked.
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Print this guide and complete it as you work through the slides and videos. It asks for your own explanations, so close the notes while you try each section.
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A printable worksheet covering all six statements, from quick recall to data questions and a longer answer. It has 40 marks and takes about 40 minutes, or you can do one section at a time.
Your teacher has the mark scheme.
You will need: Two coins per student, Punnett grid template, chi-squared table
Safety: No practical hazards.
Carry this out with your teacher's supervision and follow your school's lab rules.
Morgan's fruit fly results did not fit independent assortment. In 1913 his student Sturtevant used this to make the first genetic map.
The chi-squared cut-off is a convention. Consider how an agreed threshold shapes what counts as a real result.
Summarised in our own words. Use each code to find the exact wording in your textbook or the IB subject guide.
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