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 2000 in Hardy–Weinberg equilibrium, 1 in 400 show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 2000q² = 0.0025
N =
3.Cell biologyMedium
Glycolysis takes place in the:
- AInner mitochondrial membrane
- BMitochondrial matrix
- CCytoplasm
- DNucleus
4.Cell biologyHard
Why is DNA replication called semi-conservative?
- AThe new DNA is completely new
- BEach new double helix contains one original strand and one new strand
- COnly half of the DNA is copied
- DMutations are conserved in half of the daughter cells after every division
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 = 1805
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 = √0.0025 = 0.05
p = 0.95
Homozygous dominant
p² = 0.9025
N = 2000·0.9025 = 1805
Answer
N = 1805
3.Cell biology
Answer: C · Cytoplasm
Why this is the answer
Glycolysis splits glucose into two pyruvate in the cytoplasm, producing a net 2 ATP and 2 NADH.
Answer
C · Cytoplasm
4.Cell biology
Answer: B · Each new double helix contains one original strand and one new strand
Why this is the answer
Meselson and Stahl showed that after one replication the DNA had intermediate density, consistent with one old and one new strand.
Answer
B · Each new double helix contains one original strand and one new strand
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