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
red-green colour blindness is X-linked recessive (allele b). A carrier mother, XBXb, has children with an unaffected father, XBY. The mother's gametes run down the side. Click the cell for an unaffected son.
♀ XBXb♂ XBY
- 1Xb × Y
- 2Xb × XB
- 3XB × Y
- 4XB × XB
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?
- AProtein kinase A
- BAdenylyl cyclase
- CPhospholipase C
- DPhosphodiesterase
4.Cell biologyHard
In E. coli, under which conditions is the lac operon most strongly transcribed?
- ABoth lactose and glucose present
- BBoth lactose and glucose absent
- CLactose present and glucose absent
- DLactose absent and glucose present
3Solutions
1.Genetics
Answer: XB × Y
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.
Fill the square
XbY, XbXB, XBY, XBXB
Find the cell
XB × Y = XBY
Answer
Cell XB × Y → XBY
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: B · 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
B · Adenylyl cyclase
4.Cell biology
Answer: C · Lactose present and glucose absent
Why this is the answer
Lactose (as allolactose) removes the repressor. Low glucose raises cAMP, which activates CAP to boost transcription. Both are needed for maximal expression.
Answer
C · Lactose present and glucose absent
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