1. Ch Review

2. 1. Compare and contrast true-breeding plants to hybrid plants.

3. The true-breeding parents are the P (parental) generation
He also used varieties that were true-breeding (plants that produce offspring of the same variety when they self-pollinate)
In a typical experiment, Mendel mated two contrasting, true-breeding varieties, a process called hybridization
The true-breeding parents are the P (parental) generation
The hybrid offspring of the P generation are called the F1(first filial) generation
When F1 individuals self-pollinate, the F2(second filial) generation is produced
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

4. 2. Differentiate between an allele and a locus on a chromosome.

5. Mendel’s Model: Four related concepts
1. Alternative versions of genes account for variations in inherited characters
These alternative versions of a gene are now called alleles
Each gene resides at a specific locus on a specific chromosome
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

6. 3. Summarize Mendel’s four hypotheses on genetics.

7. Mendel’s Model: Four related concepts
2. For each character an organism inherits two alleles, one from each parent
Mendel made this deduction without knowing about the role of chromosomes
3. If the two alleles at a locus differ, then one (the dominant allele) determines the organism’s appearance, and the other (the recessive allele) has no noticeable effect on appearance
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

8. Mendel’s Model: Four related concepts
4. Law of segregation, states that the two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes
This segregation of alleles corresponds to the distribution of homologous chromosomes to different gametes in meiosis
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

9. 4. Define the law of independent assortment
4. Define the law of independent assortment. Use homologous chromosomes in your definition.

10. The Law of Independent Assortment
Using a dihybrid cross, Mendel developed the law of independent assortment
The law of independent assortment states that each pair of alleles segregates independently of each other pair of alleles during gamete formation
The presence of a specific allele for one trait in a gamete has no impact on the presence of a specific allele for the second trait.
Strictly speaking this law applies only to genes located on different nonhomologus chromosomes.
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

11. 5. Use the multiplication and addition rule for the following problems: What is the probability of finding at least two recessive traits for a cross between PpRrYy X Pprryy?

12. 6. Compare and contrast incomplete dominance and codominance and give an example of each.

13. Degrees of Dominance
Complete dominance occurs when phenotypes of the heterozygote and dominant homozygote are identical
In incomplete dominance, the phenotype of F1 hybrids is somewhere between the phenotypes of the two parental varieties
In codominance, two dominant alleles affect the phenotype in separate, distinguishable ways (AB blood)
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

14. The allele for this unusual trait is dominant to the allele for the more common trait of five digits per appendage
In this example, the recessive allele is far more prevalent than the population’s dominant allele
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

15. 7. Define epistasis. How does it affect the phenotypic ratios in a dihybrid cross.

16. 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4  BbCc BbCc Sperm Eggs BBCC BbCC BBCc
Fig 
BbCc
BbCc
Sperm
1/4
1/4
1/4
1/4 BC bC Bc bc
Eggs
1/4 BC
BBCC
BbCC
BBCc
BbCc
1/4 bC
BbCC
bbCC
BbCc
bbCc
1/4 Bc
Figure An example of epistasis
BBCc
BbCc
BBcc
Bbcc
1/4 bc
BbCc
bbCc
Bbcc
bbcc 9
: 3
: 4

17. 7. Draw a pedigree that shows only an autosomal recessive trait
7. Draw a pedigree that shows only an autosomal recessive trait. Show three generations.

18. Figure 15.2 The chromosomal basis of Mendel’s laws
P Generation
Yellow-round
seeds (YYRR)
Green-wrinkled
seeds ( yyrr) Y y  r Y R R r y
Meiosis
Fertilization R Y y r
Gametes
All F1 plants produce
yellow-round seeds (YyRr)
F1 Generation R R y y r r Y Y
LAW OF SEGREGATION
The two alleles for each gene
separate during gamete
formation.
Meiosis
LAW OF INDEPENDENT
ASSORTMENT Alleles of genes
on nonhomologous
chromosomes assort
independently during gamete
formation. R r r R
Metaphase I Y y Y y 1 1 R r r R
Anaphase I Y y Y y
Figure 15.2 The chromosomal basis of Mendel’s laws R r
Metaphase II r R 2 2 Y y Y y y Y Y y Y Y y y
Gametes R R r r r r R R
1/4 YR
1/4 yr
1/4 Yr
1/4 yR
F2 Generation
An F1  F1 cross-fertilization 3 3 9
: 3
: 3
: 1

19. 8. Describe the importance of the SRY gene.

20. Inheritance of Sex-Linked Genes
The SRY gene on the Y chromosome codes for the development of testes
Researchers have sequenced the Y chromosome and identified 78 genes coding for about 25 proteins.
The sex chromosomes have genes for many characters unrelated to sex
A gene located on either sex chromosome is called a sex-linked gene
In humans, sex-linked usually refers to a gene on the larger X chromosome
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

21. 9. Explain why we don’t see male calico cats.

22. X chromosomes Allele for orange fur Early embryo: Allele for black fur
Fig. 15-8
X chromosomes
Allele for
orange fur
Early embryo:
Allele for
black fur
Cell division and
X chromosome
inactivation
Two cell
populations
in adult cat:
Active X
Inactive X
Active X
Black fur
Orange fur
Figure 15.8 X inactivation and the tortoiseshell cat

23. 10. Given the problem below determine the recombination frequency. 4
10. Given the problem below determine the recombination frequency. 4. A wild-type fruit fly (heterozygous for gray body color and normal wings was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162.

24. Black body, vestigial wings
Fig
Testcross
parents
Gray body, normal wings
(F1 dihybrid)
Black body, vestigial wings
(double mutant)
b+ vg+
b vg
b vg
b vg
Replication
of chromo-
somes
Replication
of chromo-
somes
b+ vg+
b vg
b+ vg+
b vg
b vg
b vg
b vg
b vg
Meiosis I
b+ vg+
Meiosis I and II
b+ vg
b vg+
b vg
Meiosis II
Recombinant
chromosomes
Figure Chromosomal basis for recombination of linked genes
b+ vg+
b vg
b+ vg
b vg+
Eggs
Testcross
offspring
965
Wild type
(gray-normal)
944
Black-
vestigial
206
Gray-
vestigial
185
Black-
normal
b vg
b+ vg+
b vg
b+ vg
b vg+
b vg
b vg
b vg
b vg
Sperm
Parental-type offspring
Recombinant offspring
Recombination
frequency
391 recombinants =
 100 = 17%
2,300 total offspring

25. 11. What can recombination frequency be used to determine? Explain.

26. RESULTS Recombination frequencies 9% 9.5% Chromosome 17% b cn vg
Fig
RESULTS
Recombination
frequencies 9%
9.5%
Chromosome
17%
Figure Constructing a linkage map b cn vg

27. 12. Define nondisjunction
12. Define nondisjunction. Explain how it affects gametes differently depending on when it happens during meiosis.

28. Meiosis I Meiosis II Gametes (a) Nondisjunction of homologous
Fig
Meiosis I
Nondisjunction
Meiosis II
Nondisjunction
Gametes
Figure Meiotic nondisjunction
n + 1
n + 1
n – 1
n – 1
n + 1
n – 1 n n
Number of chromosomes
(a) Nondisjunction of homologous
chromosomes in meiosis I
(b) Nondisjunction of sister
chromatids in meiosis II

29. 13. Give an example an aneuploid.

30. Down Syndrome (Trisomy 21)
Down syndrome is an aneuploid condition that results from three copies of chromosome 21
It affects about one out of every 700 children born in the United States
The frequency of Down syndrome increases with the age of the mother, a correlation that has not been explained
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

31. 14. List the four and explain different types of chromosomal mutations.

32. Reciprocal translocation
Fig
A B C D E F G H
A B C E F G H
Deletion
(a)
A B C D E F G H
A B C B C D E F G H
Duplication
(b)
A B C D E F G H
A D C B E F G H
(c)
Inversion
Figure Alterations of chromosome structure
A B C D E F G H
M N O C D E F G H
(d)
Reciprocal
translocation
M N O P Q R
A B P Q R

33. 15. What is genomic imprinting. Explain its effects.

34. Normal Igf2 allele is expressed Paternal chromosome Maternal
Fig
Normal Igf2 allele
is expressed
Paternal
chromosome
Maternal
chromosome
Normal Igf2 allele
is not expressed
Wild-type mouse
(normal size)
(a) Homozygote
Mutant Igf2 allele
inherited from mother
Mutant Igf2 allele
inherited from father
Normal size mouse
(wild type)
Dwarf mouse
(mutant)
Figure Genomic imprinting of the mouse Igf2 gene
Normal Igf2 allele
is expressed
Mutant Igf2 allele
is expressed
Mutant Igf2 allele
is not expressed
Normal Igf2 allele
is not expressed
(b) Heterozygotes

35. 16. Define and explain the central dogma theory
16. Define and explain the central dogma theory. What are some major differences between DNA and RNA.

36. Basic Principles of Transcription and Translation: An Overview
Nucleotides to Amino Acids: How an organism’s genotype produces its phenotype
The central dogma is the concept that cells are governed by a cellular chain of command: DNA RNA protein
RNA is the intermediate between genes and the proteins for which they code
How is RNA different from DNA?
DNA
RNA
Sugar = deoxyribose
Nitrogen bases: C,G,A,T
Double stranded
Sugar= ribose
Nitrogen bases: C,G,A,U
Single stranded
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

37. 17. Explain transcription including post transcriptional modifications.

38. http://vcell.ndsu.edu/animations/transcription/index.htm Promoter
Fig. 17-7a-4
Promoter
Transcription unit 5 3 3 5
DNA
Start point
RNA polymerase 1
Initiation 5 3 3 5
RNA
transcript
Template strand
of DNA
Unwound
DNA 2
Elongation
Rewound
DNA 5 3 3 3 5
Figure 17.7 The stages of transcription: initiation, elongation, and termination 5
RNA
transcript 3
Termination 5 3 3 5 5 3
Completed RNA transcript

39. http://vcell.ndsu.edu/animations/mrnasplicing/index.htm Pre-mRNA 1 30
Fig 5
Exon
Intron
Exon
Intron
Exon 3
Pre-mRNA
5 Cap
Poly-A tail 1 30 31
104
105
146
Introns cut out and
exons spliced together
Coding
segment
mRNA
5 Cap
Poly-A tail 1
146
Figure RNA processing: RNA splicing
5 UTR
3 UTR

40. Aminoacyl-tRNA Amino acid synthetase (enzyme) tRNA Aminoacyl-tRNA
Fig
Aminoacyl-tRNA
synthetase (enzyme)
Amino acid P P P
Adenosine
ATP P
Adenosine
tRNA P P i
Aminoacyl-tRNA
synthetase P i P i
tRNA
Figure An aminoacyl-tRNA synthetase joining a specific amino acid to a tRNA P
Adenosine
AMP
Computer model
Aminoacyl-tRNA
(“charged tRNA”)

41. 18. Explain translation. Include the EPA sites in your description.

42. (b) Schematic model showing binding sites
Fig b
P site (Peptidyl-tRNA
binding site)
A site (Aminoacyl-
tRNA binding site)
E site
(Exit site) E P A
Large
subunit
mRNA
binding site
Small
subunit
(b) Schematic model showing binding sites
Growing polypeptide
Amino end
Next amino acid
to be added to
polypeptide chain
Figure The anatomy of a functioning ribosome E
tRNA
mRNA 3
Codons 5
(c) Schematic model with mRNA and tRNA

43. GDP GDP Amino end of polypeptide E 3 mRNA Ribosome ready for
Fig
Amino end
of polypeptide E 3
mRNA
Ribosome ready for
next aminoacyl tRNA P
site A
site 5
GTP
GDP E E P A P A
Figure The elongation cycle of translation
GDP
GTP E P A

44. 19. Explain what occurs when a stop codon is reached.

45. Release factor Free polypeptide 5 3 3 3 2 5 5 Stop codon
Fig
Release
factor
Free
polypeptide 5 3 3 3 2 5 5
GTP
Stop codon
(UAG, UAA, or UGA)
2 GDP
Figure The termination of translation

46. Fig. 17-25 http://vcell.ndsu.edu/animations/translation/index.htm
DNA
TRANSCRIPTION 3
Poly-A
RNA
polymerase 5
RNA
transcript
RNA PROCESSING
Exon
RNA transcript
(pre-mRNA)
Intron
Aminoacyl-tRNA
synthetase
Poly-A
NUCLEUS
Amino
acid
AMINO ACID ACTIVATION
CYTOPLASM
tRNA
mRNA
Growing
polypeptide
Cap 3 A
Activated
amino acid
Poly-A P
Ribosomal
subunits
Figure A summary of transcription and translation in a eukaryotic cell E
Cap 5
TRANSLATION E A
Anticodon
Codon
Ribosome

47. 20. Differentiate between free and bound ribosomes.

48. Targeting Polypeptides to Specific Locations
Two populations of ribosomes are evident in cells: free ribsomes (in the cytosol) and bound ribosomes (attached to the ER)
Free ribosomes mostly synthesize proteins that function in the cytosol
Bound ribosomes make proteins of the endomembrane system and proteins that are secreted from the cell
Ribosomes are identical and can switch from free to bound
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

49. Ribosome mRNA Signal peptide ER membrane Signal peptide removed
Fig
Ribosome
mRNA
Signal
peptide ER
membrane
Signal
peptide
removed
Signal-
recognition
particle (SRP)
Protein
CYTOSOL
Translocation
complex
Figure The signal mechanism for targeting proteins to the ER
ER LUMEN
SRP
receptor
protein

50. 21. Differentiate between nonsense and missense mutations

51. Substitutions
Silent mutations have no effect on the amino acid produced by a codon because of redundancy in the genetic code
Missense mutations still code for an amino acid, but not necessarily the right amino acid
Nonsense mutations change an amino acid codon into a stop codon, nearly always leading to a nonfunctional protein
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

52. 22. What are two other types of DNA mutations. Explain their effect.

53. Insertions and Deletions
These mutations have a disastrous effect on the resulting protein more often than substitutions do
Insertion or deletion of nucleotides may alter the reading frame, producing a frameshift mutation
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings