AP Lecture Guide 09 – Respiration – Glycolysis

 

AP Biology: Chapter 9

 

RESPIRATION — GLYCOLYSIS

 

1. Identify some specific processes the cell does with ATP. _____________________________

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2. Explain why ATP is such a “high energy” molecule. __________________________________

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3. Sketch the ATP cycle:

 

 

4. How does ATP “couple reactions”? ______________________________________________

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5. What is the name of enzymes which phosphorylate molecules? ________________________

6. Define each of the following:

a. Oxidation _______________________________________________________________

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b. Reduction _______________________________________________________________

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7. What is the role of NAD+ & FAD+2 in respiration? ___________________________________

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8. Explain why respiration is considered exergonic. ____________________________________

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9. Glycolysis starts with _____________________ and produces _________________________

10. The Kreb’s cycle takes place in the: ______________________________________________

11. Pyruvate is converted to ___________________________________ before the Krebs cycle.

12. The Electron Transport Chain is located in the: _____________________________________

13. Describe the role of the Electron Transport Chain. What happens to the electrons and H+?

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14. What is chemiomosis and how is it generated?

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15. How does the mitochondrion generate ATP?

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16. Label the diagram. Include: CO2, organic compounds, O2, H2O, respiration, photosynthesis, light, heat, ATP

23. Write the summary equation for cellular respiration:

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a. Where did the glucose come from? ___________________________________________

b. Where did the O2 come from? _______________________________________________

c. Where did the CO2 come from? ______________________________________________

d. Where did the H2O come from? ______________________________________________

e. Where did the ATP come from? ______________________________________________

f. What else is produced that is not listed in this equation? ___________________________

24. What was the evolutionary advantage of the proto-eukaryotes that engulfed aerobic bacteria

but did not digest them?

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25. Why do we eat? _____________________________________________________________

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26. Why do we breathe? __________________________________________________________

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17. What happens to most of the energy released during cell respiration? ___________________

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18. Alcoholic fermentation converts glucose to ________________________________________

19. Alcoholic fermentation is utilized by what organisms? ________________________________

20. Lactic acid fermentation converts glucose to________________________________________

21. Lactic acid fermentation is utilized by what organisms? _______________________________

22. Identify examples of each of the following feedback mechanisms

a. Negative feedback ________________________________________________________

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b. Positive feedback _________________________________________________________

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AP Lecture Guide HGP

HUMAN GENOME PROJECT
INSIGHTS LEARNED FROM THE SEQUENCE

What has been learned from analysis of the working draft sequence of the human genome? What is still unknown?

By the Numbers

• The human genome contains 3164.7 million chemical nucleotide bases (A, C, T, and G).

• The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.

• The total number of genes is estimated at 30,000 to 35,000, much lower than previous estimates of 80,000 to 140,000 that had been based on extrapolations from gene-rich areas as opposed to a composite of gene-rich and gene-poor areas.

• The order of almost all (99.9%) nucleotide bases are exactly the same in all people.

• The functions are unknown for over 50% of discovered genes.

The Wheat from the Chaff

• Less than 2% of the genome encodes for the production of proteins.

• Repeated sequences that do not code for proteins (“junk DNA”) make up at least 50% of the human genome.

• Repetitive sequences are thought to have no direct functions, but they shed light on chromosome structure and dynamics. Over time, these repeats reshape the genome by rearranging it, thereby creating entirely new genes or modifying and reshuffling existing genes.

• During the past 50 million years, a dramatic decrease seems to have occurred in the rate of accumulation of repeats in the human genome.

How It’s Arranged

• The human genome’s gene-dense “urban centers” are predominantly composed of the DNA building blocks G and C.

• In contrast, the gene-poor “deserts” are rich in the DNA building blocks A and T. GC- and AT-rich regions usually can be seen through a microscope as light and dark bands on chromosomes.

• Genes appear to be concentrated in random areas along the genome, with vast expanses of noncoding DNA between.

• Stretches of up to 30,000 C and G bases repeating over and over often occur adjacent to gene-rich areas, forming a barrier between the genes and the “junk DNA.” These CpG islands are believed to help regulate gene activity.

• Chromosome 1 has the most genes (2968), and the Y chromosome has the fewest (231).

How the Human Genome Compares with That of Other Organisms

• Unlike the human’s seemingly random distribution of gene-rich areas, many other organisms’ genomes are more uniform, with genes evenly spaced throughout.

• Humans have on average three times as many kinds of proteins as the fly or worm because of mRNA transcript “alternative splicing” and chemical modifications to the proteins. This process can yield different protein products from the same gene.

• Humans share most of the same protein families with worms, flies, and plants, but the number of gene family members has expanded in humans, especially in proteins involved in development and immunity.

• The human genome has a much greater portion (50%) of repeat sequences than the mustard weed (11%), the worm (7%), and the fly (3%).

• Although humans appear to have stopped accumulating repeated DNA over 50 million years ago, there seems to be no such decline in rodents. This may account for some of the fundamental differences between hominids and rodents, although gene estimates are similar in these species. Scientists have proposed many theories to explain evolutionary contrasts between humans and other organisms, including those of life span, litter sizes, inbreeding, and genetic drift.

Variations and Mutations

• Scientists have identified about 1.4 million locations where single-base DNA differences (SNPs) occur in humans. This information promises to revolutionize the processes of finding chromosomal locations for disease-associated sequences and tracing human history.

• The ratio of germline (sperm or egg cell) mutations is 2:1 in males vs females. Researchers point to several reasons for the higher mutation rate in the male germline, including the greater number of cell divisions required for sperm formation than for eggs.

What We Still Don’t Know: A Checklist for Future Research

• Exact gene number, exact locations, and functions

• Gene regulation

• DNA sequence organization

• Chromosomal structure and organization

• Noncoding DNA types, amount, distribution, information content, and functions

• Coordination of gene expression, protein synthesis, and post-translational events

• Interaction of proteins in complex molecular machines

• Predicted vs experimentally determined gene function

• Evolutionary conservation among organisms

• Protein conservation (structure and function)

• Proteomes (total protein content and function) in organisms

• Correlation of SNPs (single-base DNA variations among individuals) with health and disease

• Disease-susceptibility prediction based on gene sequence variation

• Genes involved in complex traits and multigene diseases

• Complex systems biology, including microbial consortia useful for environmental restoration

• Developmental genetics, genomics

http://genome.gsc.riken.go.jp/hgmis/project/journals/insights.html

AP Genetics Problems

 

Genetics Problems

1. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white.

  • What is the simplest explanation for the inheritance of these colors in chickens?
  • What offspring would you predict from the mating of a gray rooster and a black hen?

2. In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: RR (red) x rr (white) —> Rr (pink). If flower position (axial or terminal) is inherited as it is in peas what will be the ratios of genotypes and phenotypes of the generation resulting from the following cross: axial-red (true-breeding) x terminal-white? What will be the ratios in the F2 generation?

3. Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:

 

Character Dominant Recessive
Flower position Axial (A ) Terminal (a )
Stem length Tall (T ) Dwarf (t )
Seed shape Round (R ) Wrinkled (r)

 

If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows: (Note – use the rules of probability (and show your work) instead of huge Punnett squares)

  1. homozygous for the three dominant traits
  2. homozygous for the three recessive traits
  3. heterozygous
  4. homozygous for axial and tall, heterozygous for seed shape

4. A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second one, 7 blacks and 5 albinos were obtained.

  • What is the best explanation for this genetic situation?
  • Write genotypes for the parents, gametes, and offspring.

5. In sesame plants, the one-pod condition (P ) is dominant to the three-pod condition (p ), and normal leaf (L ) is dominant to wrinkled leaf (l) . Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring:

  1. 318 one-pod normal, 98 one-pod wrinkled
  2. 323 three-pod normal, 106 three-pod wrinkled
  3. 401 one-pod normal
  4. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled
  5. 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod wrinkled

6. A man with group A blood marries a woman with group B blood. Their child has group O blood.

  • What are the genotypes of these individuals?
  • What other genotypes and in what frequencies, would you expect in offspring from this marriage?

7. Color pattern in a species of duck is determined by one gene with three alleles. Alleles H and I are codominant, and allele i is recessive to both. How many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles?

8. Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following?

  1. all three of their children will be of normal phenotype
  2. one or more of the three children will have the disease
  3. all three children will have the disease
  4. at least one child out of three will be phenotypically normal

(Note: Remember that the probabilities of all possible outcomes always add up to 1)

9. The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes?

  1. aabbccdd
  2. AaBbCcDd
  3. AABBCCDD
  4. AaBBccDd
  5. AaBBCCdd

10. In 1981, a stray black cat with unusual rounded curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the “curl” cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true breeding variety.

  • How would you determine whether the curl allele is dominant or recessive?
  • How would you select for true-breeding cats?
  • How would you know they are true-breeding?

11. What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs?

  1. AABbCc x aabbcc —-> AaBbCc
  2. AABbCc x AaBbCc —–> AAbbCC
  3. AaBbCc x AaBbCc —–> AaBbCc
  4. aaBbCC x AABbcc —-> AaBbCc

12. Karen and Steve each have a sibling with sickle-cell disease. Neither Karen, Steve, nor any of their parents has the disease, and none of them has been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple should have another child, the child will have sickle-cell anemia.

13. Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited. A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time. What is the probability that the second child will also have the disease? Assume both parents are heterozygous for the “disease” gene.

14. In tigers, a recessive allele causes an absence of fur pigmentation (a “white tiger”) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage will be white?

15. In corn plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant gene P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the F1 generation?

16. The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the filled-in circles and squares, are unable to break down a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant or recessive allele? Fill in the genotypes of the individuals whose genotypes you know. What genotypes are possible for each of the other individuals?

 
17. A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits (5). Extra digits is a dominant trait. What fraction of this couple’s children would be expected to have extra digits?

18. Imagine you are a genetic counselor, and a couple planning to start a family came to you for information. Charles was married once before, and he and his first wife had a child who has cystic fibrosis. The brother of his current wife Elaine died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has the disease)

19. In mice, black color (B ) is dominant to white (b ). At a different locus, a dominant allele (A ) produces a band of yellow just below the tip of each hair in mice with black fur. This gives a frosted appearance known as agouti. Expression of the recessive allele (a ) results in a solid coat color. If mice that are heterozygous at both loci are crossed, what will be the expected phenotypic ratio of their offspring?

20. The pedigree below traces the inheritance of a vary rare biochemical disorder in humans. Affected individuals are indicated by filled-in circles and squares. Is the allele for this disorder dominant or recessive? What genotypes are possible for the individuals marked 1, 2, and 3.

 

 

Solutions

AP Grades

 

AP Biology Grades    

August 20, 2011
First Nine Weeks Grades

2nd Period 4th Period
CHERYL MASSENGALE COMPLETE DETAILED RECORD – Quarter 1
2 AP BIOLOGY 1 2
Qtr. 1 Aug 18, 2011 100100 QUARTER Method: Weight/Categories
Student H T HOMEWORK LAB QUIZ TEST Qtr. 1
= = > 100100 A 100% A 100% A 100%
= = > 100100 A 100% A 100% A 100%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 80 94 B- 80% A 94% A- 92%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 87 A 100% B+ 87% B+ 89%
= = > 100 83 A 100% B- 83% B 85%
= = > 80 B- 80% B- 80%
= = > 100 72 A 100% C- 72% C 76%
Average = = > 98 91 A 98% A- 91% A- 92%
WT.10% WT.20% WT.70% Weight/Percentages
ASSIGNMENTS FOR 2 AP BIOLOGY Quarter 1
AUG 16 1 H 100 IS SAM ALIVE WKSH AUG 17 2 T 100 SAFETY TEST

 

 

CHERYL MASSENGALE COMPLETE DETAILED RECORD – Quarter 1
4 AP BIOLOGY 1 2
Qtr. 1 Aug 18, 2011 100100 QUARTER Method: Weight/Categories
Student H T HOMEWORK LAB QUIZ TEST Qtr. 1
= = > 100 A 100% A 100%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 80 94 B- 80% A 94% A- 92%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 83 A 100% B- 83% B 85%
= = > 100 83 A 100% B- 83% B 85%
= = > 100 72 A 100% C- 72% C 76%
Average = = > 99 89 A 99% B+ 89% A- 90%
WT.10% WT.20% WT.70% Weight/Percentages
ASSIGNMENTS FOR 4 AP BIOLOGY Quarter 1
AUG 16 1 H 100 IS SAM ALIVE WKSH AUG 17 2 T 100 SAFETY TEST

 

6th Period 7th Period
CHERYL MASSENGALE COMPLETE DETAILED RECORD – Quarter 1
6 AP BIOLOGY 1 2
Qtr. 1 Aug 18, 2011 100100 QUARTER Method: Weight/Categories
Student H T HOMEWORK LAB QUIZ TEST Qtr. 1
= = > 100100 A 100% A 100% A 100%
= = > 100100 A 100% A 100% A 100%
= = > 100100 A 100% A 100% A 100%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 83 A 100% B- 83% B 85%
= = > 100 83 A 100% B- 83% B 85%
= = > 100 83 A 100% B- 83% B 85%
= = > 100 83 A 100% B- 83% B 85%
= = > 100 78 A 100% C+ 78% B- 81%
= = > 100 72 A 100% C- 72% C 76%
= = > EX EX
Average = = > 100 89 A 100% B+ 89% A- 90%
WT.10% WT.20% WT.70% Weight/Percentages
ASSIGNMENTS FOR 6 AP BIOLOGY Quarter 1
AUG 16 1 H 100 IS SAM ALIVE WKSH AUG 17 2 T 100 SAFETY TEST

 

 

CHERYL MASSENGALE COMPLETE DETAILED RECORD – Quarter 1
7 AP BIOLOGY 1 2
Qtr. 1 Aug 18, 2011 100100 QUARTER Method: Weight/Categories
Student H T HOMEWORK LAB QUIZ TEST Qtr. 1
= = > 100100 A 100% A 100% A 100%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 94 A 100% A 94% A 95%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 89 A 100% B+ 89% A- 90%
= = > 100 83 A 100% B- 83% B 85%
= = > 100 83 A 100% B- 83% B 85%
= = > EX EX
Average = = > 100 90 A 100% A- 90% A- 91%
WT.10% WT.20% WT.70% Weight/Percentages
ASSIGNMENTS FOR 7 AP BIOLOGY Quarter 1
AUG 16 1 H 100 IS SAM ALIVE WKSH AUG 17 2 T 100 SAFETY TEST

 

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