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, round seeds?
AaBb × aaBb
P =
2.GeneticsHard
In a population of 1000 in Hardy–Weinberg equilibrium, 1% show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 1000q² = 0.01
N =
3.Cell biologyMedium
Why does DNA replication need primase?
- APrimase proofreads the new DNA
- BPrimase joins the Okazaki fragments together once their RNA primers have been removed
- CDNA polymerase can only extend an existing strand, so a short RNA primer is needed
- DPrimase unwinds the DNA double helix
4.Cell biologyHard
Bacteria moved to a much colder environment change the fatty acids in their membranes. Which change keeps the membrane fluid?
- AFewer membrane proteins
- BMore unsaturated fatty acids
- CMore saturated fatty acids
- DLonger fatty acid tails
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) = 0.5
P(round seeds) = 0.75
Multiply
P = 0.5·0.75 = 0.375
Answer
P = 0.375
2.Genetics
Answer: N = 810
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 = 1000·0.81 = 810
Answer
N = 810
3.Cell biology
Answer: C · DNA polymerase can only extend an existing strand, so a short RNA primer is needed
Why this is the answer
Primase makes a short RNA primer with a free 3′ end. DNA polymerase extends it; the primer is later removed and replaced with DNA.
Answer
C · DNA polymerase can only extend an existing strand, so a short RNA primer is needed
4.Cell biology
Answer: B · More unsaturated fatty acids
Why this is the answer
Double bonds in unsaturated fatty acids create kinks that prevent tight packing, keeping the membrane fluid at low temperatures.
Answer
B · More unsaturated fatty acids
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