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, wrinkled seeds?
AaBb × aaBb
P =
2.GeneticsHard
In a population of 5000 in Hardy–Weinberg equilibrium, 1 in 1600 show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 5000q² = 0.000625
N =
3.Cell biologyMedium
A newly made protein has an N-terminal signal sequence. Where is it directed?
- AInto the mitochondrial matrix
- BDirectly to the cytosol for use there
- CInto the nucleus
- DInto the endoplasmic reticulum
4.Cell biologyHard
In Tay–Sachs disease, a ganglioside builds up in nerve cells because of a missing enzyme. In which organelle does the build-up occur?
- AMitochondrion
- BSmooth ER
- CPeroxisome
- DLysosome
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(tall) = 0.5
P(wrinkled seeds) = 0.25
Multiply
P = 0.5·0.25 = 0.125
Answer
P = 0.125
2.Genetics
Answer: N = 4753
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 = √0.000625 = 0.025
p = 0.975
Homozygous dominant
p² = 0.9506
N = 5000·0.9506 = 4753
Answer
N = 4753
3.Cell biology
Answer: D · Into the endoplasmic reticulum
Why this is the answer
The signal recognition particle (SRP) binds the signal sequence and brings the ribosome to the ER membrane, where the protein enters the ER lumen or membrane.
Answer
D · Into the endoplasmic reticulum
4.Cell biology
Answer: D · Lysosome
Why this is the answer
Tay–Sachs is a lysosomal storage disease. Lack of hexosaminidase A means GM2 ganglioside cannot be broken down and accumulates in neuronal lysosomes.
Answer
D · Lysosome
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