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, round seeds?
AaBb × Aabb
P =
2.GeneticsHard
In a population of 2000 in Hardy–Weinberg equilibrium, 1 in 10000 show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 2000q² = 0.0001
N =
3.Cell biologyMedium
Active transport differs from facilitated diffusion because it:
- ANeeds no proteins
- BUses energy to move substances against their concentration gradient
- COnly moves water
- DMoves substances down their gradient faster by using carrier proteins
4.Cell biologyHard
In E. coli, under which conditions is the lac operon most strongly transcribed?
- ABoth lactose and glucose absent
- BBoth lactose and glucose present
- CLactose present and glucose absent
- DLactose absent and glucose present
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(round seeds) = 0.5
Multiply
P = 0.25·0.5 = 0.125
Answer
P = 0.125
2.Genetics
Answer: N = 1960
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 = 2000·0.9801 = 1960
Answer
N = 1960
3.Cell biology
Answer: B · Uses energy to move substances against their concentration gradient
Why this is the answer
Active transport uses ATP (or another gradient) to move molecules against their gradient; facilitated diffusion is passive.
Answer
B · Uses energy to move substances against their concentration gradient
4.Cell biology
Answer: C · Lactose present and glucose absent
Why this is the answer
Lactose (as allolactose) removes the repressor. Low glucose raises cAMP, which activates CAP to boost transcription. Both are needed for maximal expression.
Answer
C · Lactose present and glucose absent
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