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 1000 in Hardy–Weinberg equilibrium, 9% show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 1000q² = 0.09
N =
3.Cell biologyMedium
Which codon starts translation and which amino acid does it code for?
- AUGA, coding for methionine
- BAUG, coding for methionine
- CAUG, coding for glycine
- DUAA, coding for leucine
4.Cell biologyHard
In damaged tissue, cells swell, their membranes burst and the area becomes inflamed. Which type of cell death is this?
- ASenescence
- BNecrosis
- CMitosis
- DApoptosis
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(short) = 0.25
P(wrinkled seeds) = 0.5
Multiply
P = 0.25·0.5 = 0.125
Answer
P = 0.125
2.Genetics
Answer: N = 490
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 = √0.09 = 0.3
p = 0.7
Homozygous dominant
p² = 0.49
N = 1000·0.49 = 490
Answer
N = 490
3.Cell biology
Answer: B · AUG, coding for methionine
Why this is the answer
AUG is the usual start codon and codes for methionine. UAA, UAG and UGA are stop codons and code for no amino acid.
Answer
B · AUG, coding for methionine
4.Cell biology
Answer: B · Necrosis
Why this is the answer
Necrosis is uncontrolled death with swelling, membrane rupture and leakage of contents that trigger inflammation. Apoptosis is orderly shrinkage into apoptotic bodies that are removed quietly.
Answer
B · Necrosis
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