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
Two AaBb parents are crossed. Gene A gives tall plants (dominant) or short plants; gene B gives black fur (dominant) or white fur. Click the one cell for an offspring that is homozygous for tall plants and shows white fur.
AaBb × AaBb
- 1ab × Ab
- 2ab × aB
- 3ab × ab
- 4ab × AB
- 5Ab × Ab
- 6Ab × aB
- 7Ab × ab
- 8Ab × AB
- 9aB × Ab
- 10aB × aB
- 11aB × ab
- 12aB × AB
- 13AB × Ab
- 14AB × aB
- 15AB × ab
- 16AB × AB
3.Cell biologyMedium
Why does inserting a single nucleotide into a coding sequence usually have a large effect?
- AIt changes the sugar in the DNA backbone
- BIt shifts the reading frame of all following codons
- CIt stops the gene from being transcribed into messenger RNA
- DIt always changes one amino acid only
4.Cell biologyHard
A mutation in the lac repressor stops it binding to the operator. What is the result?
- AThe lac genes are expressed all the time, even without lactose
- BThe lac genes can never be expressed
- CThe bacteria can no longer take up glucose
- DThe lac genes are expressed only when glucose is present in the medium
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: Ab × Ab
Punnett square: one parent's gametes run along the top, the other's down the side. Each cell joins the gamete above it with the gamete beside it, and every cell is equally likely. A recessive trait only shows when both alleles are recessive; for an X-linked trait, sons get their only X from the mother.
Genotype asked for
AAbb
Gametes
A homozygous genotype needs the same gamete from both parents, so only one cell in the 16 fits.
Ab × Ab = AAbb
Answer
Cell Ab × Ab → AAbb
3.Cell biology
Answer: B · It shifts the reading frame of all following codons
Why this is the answer
An insertion that is not a multiple of three causes a frameshift. Every codon downstream is read differently, often soon reaching a stop codon.
Answer
B · It shifts the reading frame of all following codons
4.Cell biology
Answer: A · The lac genes are expressed all the time, even without lactose
Why this is the answer
If the repressor cannot bind the operator, RNA polymerase is never blocked, so the operon is constitutively expressed.
Answer
A · The lac genes are expressed all the time, even without lactose
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