Genetics
- Monohybrid cross (Aa × Aa)
- genotypes 1 : 2 : 1, phenotypes 3 : 1
- Dihybrid cross
- phenotypes 9 : 3 : 3 : 1
- Hardy–Weinberg
- p + q = 1, p² + 2pq + q² = 1
- Independent genes
- multiply the probabilities
Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.
Tall (A) is dominant over short (a), and round seeds (B) over wrinkled (b). The genes are on different chromosomes. For the cross AaBb × Aabb, what is the probability P that an offspring is tall, round seeds?
Two AaBb parents are crossed. Gene A gives round seeds (dominant) or wrinkled seeds; gene B gives tall plants (dominant) or short plants. Click the one cell for an offspring that is shows wrinkled seeds and is homozygous for tall plants.
Where does RNA polymerase bind to start transcription of a gene?
Which checkpoint ensures all chromosomes are attached to the spindle before anaphase begins?
Answer: P = 0.375
Mendelian genetics: each parent passes on one allele of each gene, chosen at random. A Punnett square lists every combination of gametes, each equally likely. A dominant allele (capital letter) shows whenever it is present; a recessive one only in the homozygous form. Genes on different chromosomes are inherited independently, so their probabilities multiply.
Answer: aB × aB
Punnett square: one parent's gametes run along the top, the other's down the side. Each cell joins the gamete above it with the gamete beside it, and every cell is equally likely. A recessive trait only shows when both alleles are recessive; for an X-linked trait, sons get their only X from the mother.
A homozygous genotype needs the same gamete from both parents, so only one cell in the 16 fits.
Answer: D · At the promoter
RNA polymerase, helped by transcription factors in eukaryotes, binds the promoter and begins making RNA from the template strand.
Answer: A · The spindle assembly (metaphase) checkpoint
The spindle checkpoint delays anaphase until every kinetochore is attached, preventing unequal chromosome distribution.
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