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, wrinkled seeds?
AaBb × Aabb
P =
2.GeneticsHard
In a population of 500 in Hardy–Weinberg equilibrium, 1% show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 500q² = 0.01
N =
3.Cell biologyMedium
Which enzyme makes the second messenger cyclic AMP when activated by a Gs protein?
- APhosphodiesterase
- BPhospholipase C
- CAdenylyl cyclase
- DProtein kinase A
4.Cell biologyHard
In eukaryotic mRNA processing, splicing removes:
- AThe 5′ cap
- BThe poly-A tail
- CExons
- DIntrons
3Solutions
1.Genetics
Answer: P = 0.375
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(wrinkled seeds) = 0.5
Multiply
P = 0.75·0.5 = 0.375
Answer
P = 0.375
2.Genetics
Answer: N = 405
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 = 500·0.81 = 405
Answer
N = 405
3.Cell biology
Answer: C · 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
C · Adenylyl cyclase
4.Cell biology
Answer: D · Introns
Why this is the answer
The spliceosome removes introns and joins exons; the 5′ cap and poly-A tail are added for stability and export.
Answer
D · Introns
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