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
Biology: Genetics and Cell Biology: cram pack
Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.
1Formulas and methods
2Practice problems
1.GeneticsMedium
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?
AaBb × AaBb
P =
2.GeneticsHard
In a population of 1000 in Hardy–Weinberg equilibrium, 1% show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 1000q² = 0.01
N =
3.Cell biologyMedium
What is the net ATP yield of glycolysis per molecule of glucose?
- A1 ATP
- B4 ATP
- C2 ATP
- D36 ATP
4.Cell biologyHard
How does a cell mark a damaged or short-lived protein for destruction by the proteasome?
- ABy adding mannose-6-phosphate
- BBy adding a signal sequence for the ER
- CBy attaching a chain of ubiquitin molecules
- DBy removing all its phosphate groups
3Solutions
1.Genetics
Answer: P = 0.5625
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.
Each gene on its own
P(tall) = 0.75
P(round seeds) = 0.75
Multiply
P = 0.75·0.75 = 0.5625
Answer
P = 0.5625
2.Genetics
Answer: N = 810
Hardy–Weinberg: in a large, randomly mating population with no selection, the allele frequencies p (dominant) and q (recessive), p + q = 1, give genotype frequencies p², 2pq and q².
Allele frequencies
q = √0.01 = 0.1
p = 0.9
Homozygous dominant
p² = 0.81
N = 1000·0.81 = 810
Answer
N = 810
3.Cell biology
Answer: C · 2 ATP
Why this is the answer
Glycolysis uses 2 ATP and produces 4, a net gain of 2 ATP, plus 2 NADH and 2 pyruvate per glucose.
Answer
C · 2 ATP
4.Cell biology
Answer: C · By attaching a chain of ubiquitin molecules
Why this is the answer
Enzymes attach a polyubiquitin chain to the target protein. The proteasome recognises this chain, unfolds the protein and breaks it into short peptides.
Answer
C · By attaching a chain of ubiquitin molecules
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