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
A recessive condition affects 1 in 10000 of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.0001
f =
3.Cell biologyMedium
Which enzyme makes the second messenger cyclic AMP when activated by a Gs protein?
- APhospholipase C
- BProtein kinase A
- CPhosphodiesterase
- DAdenylyl cyclase
4.Cell biologyHard
In Griffith's 1928 experiment, mice injected with heat-killed smooth (S) pneumococci plus live rough (R) bacteria died. What did this show?
- AR bacteria are virulent on their own
- BA substance from the dead S cells transformed R cells into a virulent form
- CMice make S bacteria from R bacteria
- DHeat does not kill S bacteria
3Solutions
1.Genetics
Answer: P = 0.125
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) = 0.5
P(wrinkled seeds) = 0.25
Multiply
P = 0.5·0.25 = 0.125
Answer
P = 0.125
2.Genetics
Answer: f = 0.0198
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 = √0.0001 = 0.01
p = 1 − 0.01 = 0.99
Carriers
2pq = 2·0.99·0.01 = 0.0198
Answer
f = 0.0198
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: B · A substance from the dead S cells transformed R cells into a virulent form
Why this is the answer
Something from the heat-killed S strain changed living R bacteria into the virulent S type. Griffith called this transformation; the "transforming principle" was later shown to be DNA.
Answer
B · A substance from the dead S cells transformed R cells into a virulent form
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