Topic 2 & 3: Genetics Review

1. Topic 2 & 3: Genetics Review

2. Syllabus Statements 2.4.1 Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and phosphate. 2.4.2 State the name of the 4 bases in DNA 2.4.3 Outline how the DNA nucleotides are linked together by covalent bonds into a single strand. 2.4.4 Explain how a DNA double helix is formed using complimentary base pairing and hydrogen bonds. 2.4.5 Draw a simple diagram of the molecular structure of DNA. 2.5.1 State that DNA replication is semi-conservative. 2.5.2 Explain DNA replication in terms of unwinding of the double helix and separation of the strands by helicase, followed by formation of the new complementary strand by DNA polymerase. 2.5.3 Explain the significance of complementary base pairing in the conservation of the base sequence of DNA. 2.6.1 Compare the structure of RNA and DNA 2.6.2 Outline DNA transcription in terms of the formation of an RNA strand complementary to the DNA strand by RNA polymerase. 2.6.3 Describe the genetic code in terms of codons composed of triplets of bases. 2.6.4 Explain the process of translation, leading to peptide linkage formation. 2.6.5 Define the terms degenerate and universal as they relate to the genetic code. 2.6.6 Explain the relationship between one gene and one polypeptide.

3. Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and phosphate. Four DNA bases are Thyamine Adenine Cytosine Guanine Nucleotide

4. Outline the bonding in the DNA strand Covalent bond

5. Structural Details Antiparallel Strands Complementary Base Pairing

6. What is meant by semiconservative replication of DNA? Complementary base pairing ensures that resulting strands are equal to initial

7. How Does replication happen? • Helicase: which unwinds the DNA double helix and separates the strands by breaking the hydrogen bonds • DNA Polymerase which links up the nucleotides to form the new strand of DNA. This 2nd step, after the unwinding and separation of strands, involves having the single strands act as templates for the new strands. Free nucleotides are present in large numbers around the replication fork. The bases of these nucleotides form hydrogen bonds with the bases of the parent strand. DNA polymerase is the main enzyme involved. Daughter DNA molecules each rewind into a double helix.

8. List 3 ways RNA is different from DNA • RNA nucleotides contain the sugar ribose. Ribose has one more hydroxyl than deoxyribose. • Uracil, a pyrimidine, is unique to RNA and is similar to thymine (A, C, G, U). • RNA is single stranded.

9. Transcription

10. What is a codon? • A 3 base sequence in mRNA • Each codes for a particular amino acid • 64 possible codons • 20 amino acids • Stops & starts also

11. Define • Degenerate  • Amino acids are coded for by multiple different codon sequences. As many as 6 sequences in some cases for one amino acid • Universal  • DNA code is the same in all living things. The gene for a bacterial polypeptide will create the same polypeptide in any eukaryote

12. Translation Initiation

13. Translation Elongation

14. Translation Termination

15. So Here’s a DNA Strand ATTCGGCCACATTTC 1. Write out the complementary strand TAAGCCGGTGTAAAG • Write out the RNA transcript of the original strand UAAGCCGGUGUAAAG • Write out the first 3 tRNA anticodons AUU CGG CCA

16. Syllabus Statements • 3.1.1: State that eukaryotic chromosomes are made of DNA and protein • 3.1.2: State that in karyotyping chromosomes are arranged in pairs according to their structure • 3.1.3: Describe one application of karyotyping (cross reference with 3.2.5) • 3.1.4: Define gene, allele, genome • 3.1.5: Define gene mutation • 3.1.6: Explain the consequence of a base substitution mutation in relationto the process of transcription & translation, using the example of sickle cell anemia

17. Chromosomes = composed of DNA and histone proteins in eukaryotes

18. Definitions • Gene = • Allele = • Genome = • Gene mutation =

19. Human Karyotype • Can display an individual’s somatic-cell metaphase chromosomes • Karyotype when arranged in a standard sequence. • Display of chromosomes arranged by size, position of centromere and staining pattern. • Often made from lymphocytes or amniocytes.

20. Sickle Cell Disease

21. Sickle Cell Disease • Replacement of #6 amino acid, glutamic acid, with valine substitutes a non-polar amino acid for a polar amino acid. • #6 amino acid is on the OUTSIDE of the hemoglobin molecule. • Hydrophobic interaction set up by valine causes change in shape of hemoglobin which leads to sickling of the red blood cell.

22. Syllabus Statements II • 3.3.1: Define  genotype, phenotype, dominant allele, recessive allele, codominant alleles, locus, homozygous, heterozygous, carrier, test cross • 3.3.2: Construct a Punnett grid • 3.3.3: Construct a pedigree chart • 3.3.4: State that some genes have more than two alleles (multiple alleles) • 3.3.5: Describe ABO blood groups as an example of codominance and multiple alleles • 3.3.6: Outline how sex chromosomes determine gender by referring to the inheritance of X and Y chromosomes in humans

23. Some Definitions Dominant allele = Recessive allele = Codominant alleles = Locus =

24. 3.3.7: State that some genes are present on the X chromosome and absent from the shorter Y chromosome in humans. • 3.3.8: Define Sex linkage • 3.3.9: State two examples of sex linkage • 3.3.10: State that human females can be homozygous or heterozygous • 3.3.11: Explain that female carriers are heterozygous for X-linked recessive alleles • 3.3.12: Calculate and predict the genotypic and phenotypic ratios of offspring of monohybrid crosses involving any of the above patterns of inheritance • 3.3.13: Deduce the genotypes or phenotypes of individuals in pedigree charts

25. Figure 14.4 Mendel’s law of segregation (Layer 2)

26. Terms from Crosses • Homozygous: Having 2 identical alleles for a given trait • True breeding, all gametes carry same allele • Heterozygous: Having 2 different alleles for a given trait • Not true breeding, half of gametes carry one allele, half carry the other • Phenotype: An organism’s expressed traits • F2 phenotypic ratio of 3:1 = 3 purple, 1 white • Genotype: An organism’s genetic make up • F2 genotypic ratio of 1:2:1 = 1PP, 2Pp, 1pp

27. Test Crosses Test Cross: Breeding an organism of unknown genotype with one that is known homozygous recessive e.g. P__ X pp : if all progeny purple, then ___ : if ½ purple, ½ white, then ___

28. Figure 14.9 Incomplete dominance in snapdragon color

29. Multiple Alleles • More than 2 alternate forms of a gene • E.g. = ABO blood group; produce 4 possible phentypes (A, B, AB, O) • A & B are 2 genetically determined polysaccharides (A & B antigens) found on the surface of RBC • Three alleles for this gene: IA (A antigen), IB (B antigen), and i (no antigen) • Alleles IA & IB are codominant, and dominant to i

30. Figure 14.10 Multiple alleles for the ABO blood groups

31. Pedigree Rules • By convention squares represent males • Circles represent females • Shaded symbols indicate individuals showing the trait • Horizontal line connecting male and female signifies reproduction (or marriage) • Offspring are listed below in birth order

32. Generally • If a mother & father do not show a trait but they have an offspring that does – usually this indicates recessive inheritance • If males show a trait more than females and / or mothers pass the trait to their sons it is generally sex linked inheritance • Otherwise its dominance inheritance – only specific example we discussed of this was huntington’s chorea

33. Practice Problem

35. Sex linkage • Sex linkage: In addition to determining sex, the X and Y chromosomes contain many genes not related to sex • In humans sex linkage is usually equivalant to X-linkage • Examples • Colorblindness: Xb presence relative to the normal XB • Hemophilia: Xh presence relative to the normal XH

36. Hemophilia • A man with hemophilia ( a recessive, sex linked condition Xh) has a daughter of normal phenotype. She marries a man who is normal for the trait • What is the probability that the daughter of this mating will be a hemophiliac? • What is the probability for a son? • If the couple has 4 sons, what is the probability that they will all be born with hemophilia?

37. Solution • P generation (man) = (XH XH) x (XhY) • F1 generation (daughter) = (Xh XH) x (XHY) • 0% • 50% • ½ * ½ * ½ * ½ = 1/16

38. Red-green Color Blindness • Red-green Color Blindness is caused by a sex-linked recessive allele (Xb). A color blind man marries a woman with normal vision whose father was color blind. • What is the probability that they will have a color blind daughter? (this is 2 events 1st daughter, then color blind) • What is the probability that thir first son will be color blind?

39. Solution • Color blind man = XbY • Woman = XBXb • P generation = (XBXb) x (XbY) • ¼ • ½

40. Sex Linkage: General Statements • Fathers pass X-linked alleles to only and all of their daughters • Fathers cannot pass X-linked traits to their sons • Females receive 2 X chromosomes, one from each parent • Mothers pass only 1 X chromosome (either maternal or paternal homologue) to every daughter and son • If a sex linked trait is due to a recessive allele, a female will express that trait if and only if she is homozygous

41. Syllabus Statements • 3.2.1 State that meiosis is a reduction division in terms of diploid and haploid numbers of chromosomes. • 3.2.2 Define homologous chromosomes • 3.2.3 Outline the process of meiosis, including pairing of chromosomes followed by two divisions, which results in four haploid cells. • 3.2.4 Explain how the movement of chromosomes during meiosis can give rise to genetic variety in the resulting haploid cells. • 3.2.5 Explain that non-disjunction can lead to changes in chromosome number, illustrated by reference to Down’s syndrome (trisomy 21). • 3.2.6 State Mendel’s law of segregation. (Done with genetics) • 3.2.7 Explain the relationship between Mendel’s law of segregation and meiosis (Done with genetics)

42. Asexual: 1 individual is sole parent Single parent passes on all its genes to offspring Offspring genetically identical to parent Results in clone; rarely, genetic differences occur as a result of mutation, a change in the DNA Sexual: 2 parents give rise to offspring Each parent passes on ½ its genes to offspring Offspring have unique combination of genes Results in greater variation; offspring vary genetically from sibs and parents Adaptive to changing environments Sexual vs. Asexual Reproduction

43. Definitions • Homologous Chromosomes: a pair of chromosomes that have the same size, centromere position and staining pattern. With one exception, homologues carry the same genetic loci. X and Y chromosomes pair during meiosis but Y is much smaller. • Autosome: chromosome that is not a sex chromosome. • Sex chromosome: dissimilar chromosomes that determine an individual’s sex. Chromosomal pairs in the human karyotype are a result of our sexual origins: 1 homologue is inherited from each parent

46. Mitosis vs. Meiosis

48. Chromosomal Non-disjunction and Down Syndrome: Errors in Meiosis • Sometimes chromosomes fail to disjoin during either Meiosis I or Meiosis II. Instead they move to the same pole. • Non-separation of chromosomes is called non-disjunction and results in gametes with either one chromosome too many or one chromosome too few (frequently lethal). • Gametes with 2 doses of a chromosome that join with a gamete with the normal dose of chromosomes now have 3 doses of this chromosome (trisomy).

49. Nondisjunction in Meiosis I and II