Exam 3 preview

1. Parents genotypes: Phil & Maggie must be Ee (heterozygous).

Probability second child has attached earlobes (recessive) = 1/4 (25%).

2, AaBb × AaBb →

(3/4) × (1/4) = 3/16.

3. Test cross (smooth = dominant) × rr (wrinkled)

if offspring show ~1:1 smooth:wrinkled, the unknown is Rr (heterozygous).

4.Law of Segregation

alleles separate during gamete formation so each gamete gets one allele.

5. If F1 all show one phenotype (e.g., purple only), that indicates the trait is

dominant

6. X-linked recessive rule:

the gene causing the trait or the disorder is located on the X chromosome. Females have two X chromosomes while males have one X and one Y chromosome. In X-linked recessive inheritance, males are more likely to express the trait since they have only one X chromosome.

7. B dominant / b recessive

dominant phenotype = BB or Bb; recessive phenotype = bb.

8. Incomplete dominance

occurs when the heterozygote shows an intermediate phenotype between the two homozygotes.

9. Homozygous

two identical alleles (AA or aa);

9. Heterozygous

two different alleles (Aa)

10. sex-linked (X-linked) inheritance.

Pattern where all females have one phenotype and all males the other

11. In that fruit-fly cross the P female is heterozygous (carrier) and

P male is hemizygous (single X; either mutant or wild-type depending on the cross).

12. Parents A (AO) × B (BO) can produce a child with

type O (if both parents are heterozygous AO and BO → child OO).

13. What determines how often linked genes recombine

physical distance (map distance) on the chromosome; closer genes recombine less often.

14. Chargaff’s rules: in DNA,

[A] = [T] and [G] = [C].

the amount of adenine (A) should be equal to the amount of thymine (T), and the amount of guanine (G) should be equal to the amount of cytosine (C)

15. Primary information source for determining DNA structure

X-ray diffraction

16. Semiconservative replication

each daughter DNA has one parental strand + one newly synthesized strand (Meselson–Stahl).

17. Primase (RNA primer) is required

to provide a free 3′-OH (an RNA primer) so DNA polymerase can begin synthesis.

18. Elongation:

DNA polymerase adds nucleotides to the 3′ end, synthesizing in the 5′→3′ direction; leading strand is continuous, lagging strand as Okazaki fragments.

19. Protein that acts during initiation:

helicase — it unwinds the double helix.

20. Role of helicase:

unwind (separate) the two DNA strands at the replication fork.

21. End-replication problem:

linear chromosomes shorten because DNA polymerase cannot fully replicate the 3′ ends, causing telomere shortening unless telomerase extends them.

22. Major chromosomal mutation types:

deletion, duplication, inversion, translocation

23. PCR is a laboratory technique used to

amplify (make many copies of) a specific DNA sequence.

24. In PCR, heat (high temperature) separates DNA strands (denaturation);

in cellular replication, helicase separates strands.

25. Point mutation:

a change in a single nucleotide base (usually a substitution)

26. Transcription

    Translation

DNA → RNA (making an RNA copy).

mRNA → polypeptide (protein synthesis).

27. tRNA during translation:

brings amino acids to the ribosome.

28. In eukaryotes: introns

long stretch of noncoding DNA found between exons (or coding regions) in a gene.

29. Start codon AUG signals the ribosome to

initiate translation (and codes for methionine).

30. RNA polymerase (specifically RNA polymerase II in eukaryotes) is responsible for

transcribing DNA into mRNA.

31. Transcription occurs in the nucleus (eukaryotes); translation occurs in the

cytoplasm (outside the nucleus)

32. Directly before translation comes

transcription of mRNA from DNA.

33. The first amino acid is always methionine because

AUG is the start codon and codes for methionine.

34. Involved in translation

the ribosome (with rRNA and proteins), mRNA, and tRNAs (all are required).

35. Nucleic acids (DNA/RNA)

are synthesized 5′→3′ as the template is read 3′→5′.
Polypeptides are synthesized N-terminus → C-terminus while mRNA is read 5′→3′. and are critical for protein synthesis.

36. Frameshift mutations

are most likely caused by insertions or deletions not in multiples of three nucleotides.

37. A tRNA carrying an amino acid is called

• Aminoacyl-tRNA (charged tRNA)

38. DNA replication is generally more accurate than transcription because

DNA polymerases have proofreading (3′→5′ exonuclease) activity and repair mechanisms

39.Release factors function to recognize stop codons and terminate translation

promoting polypeptide release.

40. A mutation changes the codon [codon withheld] to change to {codon withheld]. This is an example of a:

• codon → different amino acid = missense.

A mutation in DNA that changes a codon from UAU (tyrosine) to UAA (stop) is an example of:

nonsense mutation

Correct order of gene expressions

DNA→ transcription → translation → protein