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
In a population of 500 in Hardy–Weinberg equilibrium, 4% show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 500q² = 0.04
N =
3.Cell biologyMedium
How does glucose enter most body cells, such as red blood cells?
- ABy primary active transport using ATP directly at the membrane
- BBy facilitated diffusion through GLUT carrier proteins
- CBy simple diffusion through the lipid bilayer
- DBy phagocytosis
4.Cell biologyHard
Why can translation in bacteria begin before transcription of the mRNA is finished?
- ABacteria do not use mRNA
- BThere is no nuclear envelope separating DNA from ribosomes
- CBacterial mRNA is spliced faster
- DBacterial ribosomes can read the DNA template directly in the cytoplasm
3Solutions
1.Genetics
Answer: P = 0.5625
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) = 0.75
P(round seeds) = 0.75
Multiply
P = 0.75·0.75 = 0.5625
Answer
P = 0.5625
2.Genetics
Answer: N = 320
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 = √0.04 = 0.2
p = 0.8
Homozygous dominant
p² = 0.64
N = 500·0.64 = 320
Answer
N = 320
3.Cell biology
Answer: B · By facilitated diffusion through GLUT carrier proteins
Why this is the answer
Glucose uses GLUT carrier proteins to move down its concentration gradient without using ATP. This is facilitated diffusion.
Answer
B · By facilitated diffusion through GLUT carrier proteins
4.Cell biology
Answer: B · There is no nuclear envelope separating DNA from ribosomes
Why this is the answer
In prokaryotes DNA lies in the cytoplasm, so ribosomes can attach to mRNA while it is still being made. In eukaryotes mRNA must be processed and exported first.
Answer
B · There is no nuclear envelope separating DNA from ribosomes
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