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, wrinkled seeds?
AaBb × aabb
P =
2.GeneticsHard
In a population of 500 in Hardy–Weinberg equilibrium, 1 in 1600 show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 500q² = 0.000625
N =
3.Cell biologyMedium
The contractile ring that pinches an animal cell in two during cytokinesis is made of:
- ACellulose fibres
- BMicrotubules and kinesin
- CKeratin filaments
- DActin and myosin
4.Cell biologyHard
A double-stranded DNA sample contains 30% adenine. What percentage is guanine?
- A40%
- B70%
- C20%
- D30%
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(wrinkled seeds) = 0.5
Multiply
P = 0.5·0.5 = 0.25
Answer
P = 0.25
2.Genetics
Answer: N = 475.3
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 = √0.000625 = 0.025
p = 0.975
Homozygous dominant
p² = 0.9506
N = 500·0.9506 = 475.3
Answer
N = 475.3
3.Cell biology
Answer: D · Actin and myosin
Why this is the answer
A ring of actin filaments and myosin II contracts under the membrane, forming the cleavage furrow that separates the two daughter cells.
Answer
D · Actin and myosin
4.Cell biology
Answer: C · 20%
Why this is the answer
A = T = 30%, so A + T = 60%. The remaining 40% is split equally between G and C, so G = 20%.
Answer
C · 20%
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