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 short, wrinkled seeds?
Two AaBb parents are crossed. Gene A gives yellow seeds (dominant) or green seeds; gene B gives black fur (dominant) or white fur. Click the one cell for an offspring that is shows green seeds and is homozygous for black fur.
During which phase of mitosis do sister chromatids separate?
Tumour suppressor genes such as p53 normally:
Answer: P = 0.0625
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: B · Anaphase
In anaphase cohesin is cleaved and chromatids are pulled to opposite poles by spindle microtubules.
Answer: A · Stop the cell cycle or trigger apoptosis when DNA is damaged
p53 arrests the cycle for repair or triggers apoptosis; mutations that inactivate it let damaged cells keep dividing.
ExamProblems · examproblems.com/courses/biology