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, wrinkled seeds?
Two AaBb parents are crossed. Gene A gives black fur (dominant) or white fur; gene B gives tall plants (dominant) or short plants. Click the one cell for an offspring that is shows both white fur and short plants.
Cyclin-dependent kinases (CDKs) that drive the cell cycle become active when they bind to:
A non-competitive inhibitor binds an enzyme away from the active site. How does it change the kinetics?
Answer: P = 0.1875
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: C · Cyclins
CDKs are only active when bound to cyclins, whose levels rise and fall through the cycle. Active cyclin–CDK complexes phosphorylate targets that push the cell forward.
Answer: A · Vmax falls and Km stays the same
A pure non-competitive inhibitor reduces the number of working enzyme molecules, lowering Vmax. Substrate binding is unaffected, so Km stays the same.
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