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 short, round seeds?
AaBb × aabb
P =
2.GeneticsHard
A recessive condition affects 1 in 400 of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.0025
f =
3.Cell biologyMedium
Which enzyme makes the second messenger cyclic AMP when activated by a Gs protein?
- APhospholipase C
- BPhosphodiesterase
- CProtein kinase A
- DAdenylyl cyclase
4.Cell biologyHard
Some antibiotics, such as tetracyclines, block bacterial protein synthesis with little effect on human cytoplasmic ribosomes. Why?
- AHuman ribosomes are inside the nucleus
- BBacteria use DNA instead of mRNA as the template for translation on ribosomes
- CHuman cells do not make proteins
- DBacterial 70S ribosomes differ in structure from eukaryotic 80S ribosomes
3Solutions
1.Genetics
Answer: P = 0.25
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(short) = 0.5
P(round seeds) = 0.5
Multiply
P = 0.5·0.5 = 0.25
Answer
P = 0.25
2.Genetics
Answer: f = 0.095
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 = √0.0025 = 0.05
p = 1 − 0.05 = 0.95
Carriers
2pq = 2·0.95·0.05 = 0.095
Answer
f = 0.095
3.Cell biology
Answer: D · Adenylyl cyclase
Why this is the answer
Activated Gs stimulates adenylyl cyclase to convert ATP into cAMP. cAMP activates protein kinase A, and phosphodiesterase breaks cAMP down.
Answer
D · Adenylyl cyclase
4.Cell biology
Answer: D · Bacterial 70S ribosomes differ in structure from eukaryotic 80S ribosomes
Why this is the answer
Bacterial ribosomes (70S: 30S + 50S) differ from cytoplasmic eukaryotic ribosomes (80S: 40S + 60S), so drugs can target them selectively.
Answer
D · Bacterial 70S ribosomes differ in structure from eukaryotic 80S ribosomes
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