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 in 10000 show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 1000q² = 0.0001
N =
3.Cell biologyMedium
The Golgi apparatus mainly:
- AReplicates DNA
- BBreaks down old organelles and engulfed bacteria with enzymes
- CProduces ATP
- DModifies, sorts and packages proteins for transport
4.Cell biologyHard
Why can translation in bacteria begin before transcription of the mRNA is finished?
- ABacterial ribosomes can read the DNA template directly in the cytoplasm
- BBacterial mRNA is spliced faster
- CBacteria do not use mRNA
- DThere is no nuclear envelope separating DNA from ribosomes
3Solutions
1.Genetics
Answer: P = 0.5625
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(round seeds) = 0.75
Multiply
P = 0.75·0.75 = 0.5625
Answer
P = 0.5625
2.Genetics
Answer: N = 980.1
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 = 0.99
Homozygous dominant
p² = 0.9801
N = 1000·0.9801 = 980.1
Answer
N = 980.1
3.Cell biology
Answer: D · Modifies, sorts and packages proteins for transport
Why this is the answer
The Golgi modifies proteins (e.g. glycosylation) and sorts them into vesicles for secretion, membranes or lysosomes.
Answer
D · Modifies, sorts and packages proteins for transport
4.Cell biology
Answer: D · There is no nuclear envelope separating DNA from ribosomes
Why this is the answer
In prokaryotes DNA lies in the cytoplasm, so ribosomes can attach to mRNA while it is still being made. In eukaryotes mRNA must be processed and exported first.
Answer
D · There is no nuclear envelope separating DNA from ribosomes
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