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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

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

2Practice problems

1.GeneticsMedium
♂♀XH​YXH​Xh​

haemophilia is X-linked recessive (allele h). A carrier mother, XHXh, has children with an unaffected father, XHY. The mother's gametes run down the side. Click the cell for an unaffected son.

♀ XHXh♂ XHY
  1. 1XH × XH
  2. 2XH × Y
  3. 3Xh × XH
  4. 4Xh × Y
2.GeneticsHard
AA?AaAaaap (A)q (a)p (A)q (a)q² = 0.0001N = 5000

In a population of 5000 in Hardy–Weinberg equilibrium, 1 in 10000 show a recessive trait. How many individuals N are expected to be homozygous dominant?

N = 5000q² = 0.0001
N =
3.Cell biologyMedium

Cyclin-dependent kinases (CDKs) that drive the cell cycle become active when they bind to:

  1. ACyclins
  2. BHistones
  3. CRibosomes
  4. DTubulins
4.Cell biologyHard

How can one human gene produce several different proteins?

  1. ABy reading the gene in all three frames at once
  2. BBy copying the gene into different chromosomes
  3. CBy alternative splicing of its pre-mRNA
  4. DBy using a different genetic code in each tissue

3Solutions

1.Genetics

Answer: XH × Y

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.

Fill the square

XHXH, XHY, XhXH, XhY

Find the cell

XH × Y = XHY

Answer

Cell XH × Y → XHY
2.Genetics

Answer: N = 4901

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 = 5000·0.9801 = 4901

Answer

N = 4901
3.Cell biology

Answer: A · Cyclins

Why this is the answer

CDKs are only active when bound to cyclins, whose levels rise and fall through the cycle. Active cyclin–CDK complexes phosphorylate targets that push the cell forward.

Answer

A · Cyclins
4.Cell biology

Answer: C · By alternative splicing of its pre-mRNA

Why this is the answer

Alternative splicing joins different combinations of exons from the same pre-mRNA, producing related proteins from one gene.

Answer

C · By alternative splicing of its pre-mRNA

ExamProblems · examproblems.com/courses/biology