AP Lecture Guide 25 – Phylogeny and Systematics

 

 

AP Biology: CHAPTER 25: PHYLOGENY AND SYSTEMATICS

 

1. What is phylogeny?

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2. How are fossils significant to our study of biology?

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3. Review these key points in the study of fossils:

a. Sedimentary rocks are the richest source of fossils.

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b. Paleontologists use a variety of methods to date fossils.

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c. The fossil record is a substantial, but incomplete, chronicle of evolutionary history.

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d. Phylogeny has a biogeographic basis in continental drift.

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e. The history of life is punctuated by mass extinctions.

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4. List examples of fossils. ______________________________________________________

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5. What techniques do relative dating use to place fossils in their place in geologic time?

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6. What marks the separation between the major eras in the geologic time scale?

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7. How does absolute dating compare to relative dating?

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8. Describe the two main characteristics of the Linnaean system of classification.

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9. What modern techniques are used as the basis for grouping creatures with modern

phylogenetic systematics?

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10. What does a phylogenic tree show?

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11. When classifying organisms in a cladistic diagram, identify three pitfalls scientists might

encounter classifying creatures.

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12. What do scientists use when placing an organism on a cladistic diagram?

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13. How have molecular clocks influenced our thoughts on evolutionary paths?

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14. Why is the four chamber heart a poor choice of structure to place creatures on a phylogenic

tree?

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15. Why are crocodiles now thought to be closer to birds than other reptiles?

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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:

 

CharacterDominantRecessive
Flower positionAxial (A )Terminal (a )
Stem lengthTall (T )Dwarf (t )
Seed shapeRound (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

Genetic Problems Solutions Campbell Ch14

 

Genetics Problems Campbell
1. A man with hemophilia (a recessive , sex-linked condition has a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? A son? If the couple has four sons, what is the probability that all four will be born with hemophilia?

Solution

 

2. Pseudohypertropic muscular dystrophy is a disorder that causes gradual deterioration of the muscles. It is seen only in boys born to apparently normal parents and usually results in death in the early teens. (a) Is pseudohypertrophic muscular dystrophy caused by a dominant or recessive allele? (b) Is its inheritance sex-linked or autosomal? (c) How do you know? Explain why this disorder is always seen in boys and never girls.

Solution

3. Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. (a) What is the probability that they will have a color-blind daughter? (b) What is the probability that their first son will be color-blind? (Note: the two questions are worded a bit differently.)

Solution

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. What is the recombination frequency between these genes for body color and wing type.

Solution

5. In another cross, a wild-type fruit fly (heterozygous for gray body color and red eyes) was mated with a black fruit fly with purple eyes. The offspring were as follows: wild-type, 721; black-purple, 751; gray-purple, 49; black-red, 45. (a) What is the recombination frequency between these genes for body color and eye color? (b) Following up on this problem and problem 4, what fruit flies (genotypes and phenotypes) would you mate to determine the sequence of the body color, wing shape, and eye color genes on the chromosomes?

Solution

6. A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns as those in humans. Three phenotypic characters are height (T = tall, t = dwarf), hearing appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures were not “intelligent” Earth scientists were able to do some controlled breeding experiments, using various heterozygotes in testcrosses. For a tall heterozygote with antennae, the offspring were tall-antennae, 46; dwarf-antennae 7; dwarf-no antennae 42; tall-no antennae 5. For a heterozygote with antennae and an upturned snout, the offspring were antennae-upturned snout 47; antennae-downturned snout, 2; no antennae-downturned snout, 48: no antennae-upturned snout 3. Calculate the recombination frequencies for both experiments.

Solution

7. Using the information from problem 6, a further testcross was done using a heterozygote for height and nose morphology. The offspring were tall-upturned nose, 40; dwarf-upturned nose, 9; dwarf-downturned nose, 42; tall-downturned nose, 9. Calculate the recombination frequency from these data; then use your answer from problem 6 to determine the correct sequence of the three linked genes.

Solution

8. Imagine that a geneticist has identified two disorders that appear to be caused by the same chromosomal defect and are affected by genomic imprinting: blindness and numbness of the limbs. A blind woman (whose mother suffered from numbness) has four children, two of whom, a son and daughter, have inherited the chromosomal defect. If this defect works like Prader-Willi and Angelman syndromes, what disorders do this son and daughter display? What disorders would be seen in their sons and daughters?

Solution

9. What pattern of inheritance would lead a geneticist to suspect that an inherited disorder of cell metabolism is due to a defective mitochondrial gene?

Solution

10. An aneuploid person is obviously female, but her cells have two Barr bodies. what is the probable complement of sex chromosomes in this individual?

Solution

11. Determine the sequence of genes along a chromosome based on the following recombination frequencies: A-B, 8 map units; A-C, 28 map units; A-D, 25 map units; B-C , 20 map units; B-D, 33 map units.

Solution

12. About 5% of individuals with Downs syndrome are the result of chromosomal translocation. In most of these cases, one copy of chromosome 21 becomes attached to chromosome 14. How does this translocation lead to children with Down syndrome?

Solution

13. Assume genes A and B are linked and are 50 map units apart. An individual heterozygous at both loci is crossed with an individual who is homozygous recessive at both loci. (a) What percentage of the offspring will show phenotypes resulting from crossovers? (b) If you did not know genes A and B were linked, how would you interpret the results of this cross?

Solution

14. In Drosophila, the gene for white eyes and the gene that produces “hairy” wings have both been mapped to the same chromosome and have a crossover frequency of 1.5%. A geneticist doing some crosses involving these two mutant characteristics noticed that in a particular stock of flies, these two genes assorted independently; that is they behaved as though they were on different chromosomes. What explanation can you offer for this observation?

Solution

 

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Campbell Chapter 14 Gen Prob 1

Molecular Genetics: Problem 1
A man with hemophilia (a recessive , sex-linked condition has a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? A son? If the couple has four sons, what is the probability that all four will be born with hemophilia?

Genotypes:

A man with hemophilia is XhY where h = hemophilia gene and H = the normal gene.
Any daughter with normal phenotype whose father has hemophilia will be a carrier.

Her genotype must be:

XhXH and NOT XHXH
We can use a Punnett square to show the probability of a daughter or son having hemophilia.

daughter x normal man
XhXH x XHY

A. If the daughter marries a normal male the probability of a daughter having hemophilia is zero.

B. About 50% of male children would have hemophilia (Boxes 2 and 4 above)

C. The probability that all 4 sons have inherited hemophilia would be: 1/2 x 1/2 x 1/2 x 1/2 or 1/16.

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