Ch. 14-17 Review

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

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

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

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

3. Summarize Mendel’s four hypotheses on genetics.

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

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

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

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

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?

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

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

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

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

1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4  BbCc BbCc Sperm Eggs BBCC BbCC BBCc Fig. 14-12  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 14.12 An example of epistasis BBCc BbCc BBcc Bbcc 1/4 bc BbCc bbCc Bbcc bbcc 9 : 3 : 4

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.

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

8. Describe the importance of the SRY gene.

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

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

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

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.

Black body, vestigial wings Fig. 15-10 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 15.10 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

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

RESULTS Recombination frequencies 9% 9.5% Chromosome 17% b cn vg Fig. 15-11 RESULTS Recombination frequencies 9% 9.5% Chromosome 17% Figure 15.11 Constructing a linkage map b cn vg

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

Meiosis I Meiosis II Gametes (a) Nondisjunction of homologous Fig. 15-13-3 Meiosis I Nondisjunction Meiosis II Nondisjunction Gametes Figure 15.13 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

13. Give an example an aneuploid.

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

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

Reciprocal translocation Fig. 15-15 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 15.15 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

15. What is genomic imprinting. Explain its effects.

Normal Igf2 allele is expressed Paternal chromosome Maternal Fig. 15-18 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 15.18 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

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.

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

17. Explain transcription including post transcriptional modifications.

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 http://vcell.ndsu.edu/animations/transcription/index.htm 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

http://vcell.ndsu.edu/animations/mrnasplicing/index.htm Pre-mRNA 1 30 Fig. 17-10 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 17.10 RNA processing: RNA splicing 5 UTR 3 UTR http://vcell.ndsu.edu/animations/mrnasplicing/index.htm

Aminoacyl-tRNA Amino acid synthetase (enzyme) tRNA Aminoacyl-tRNA Fig. 17-15-4 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 17.15 An aminoacyl-tRNA synthetase joining a specific amino acid to a tRNA P Adenosine AMP Computer model Aminoacyl-tRNA (“charged tRNA”)

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

(b) Schematic model showing binding sites Fig. 17-16b 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 17.16 The anatomy of a functioning ribosome E tRNA mRNA 3 Codons 5 (c) Schematic model with mRNA and tRNA

GDP GDP Amino end of polypeptide E 3 mRNA Ribosome ready for Fig. 17-18-4 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 17.18 The elongation cycle of translation GDP GTP E P A

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

Release factor Free polypeptide 5 3 3 3 2 5 5 Stop codon Fig. 17-19-3 Release factor Free polypeptide 5 3 3 3 2 5 5 GTP Stop codon (UAG, UAA, or UGA) 2 GDP Figure 17.19 The termination of translation

Fig. 17-25 http://vcell.ndsu.edu/animations/translation/index.htm DNA http://vcell.ndsu.edu/animations/translation/index.htm 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 17.25 A summary of transcription and translation in a eukaryotic cell E Cap 5 TRANSLATION E A Anticodon Codon Ribosome

20. Differentiate between free and bound ribosomes.

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

Ribosome mRNA Signal peptide ER membrane Signal peptide removed Fig. 17-21 Ribosome mRNA Signal peptide ER membrane Signal peptide removed Signal- recognition particle (SRP) Protein CYTOSOL Translocation complex Figure 17.21 The signal mechanism for targeting proteins to the ER ER LUMEN SRP receptor protein

21. Differentiate between nonsense and missense mutations

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

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

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