1. A monohybrid cross involves ONE PAIR of heterozygous genes for each parent. This cross can be expressed algebraically by squaring a binomial, such as the expression (P + p):

2. A dihybrid cross involves TWO PAIRS of heterozygous genes for each parent. This cross can be expressed algebraically by squaring the product of two binomials such as (P + p) x (S + s):

3. The Rh factor is an interesting example of polygenic inheritance. Unlike the A-B-O blood types where all the alleles occur on one pair of loci on chromosome pair #9, the Rh factor involves three different pairs of alleles located on three different loci on chromosome pair #1. In the following diagram, 3 pairs of Rh alleles (C & c, D & d, E & e) occur at 3 different loci on homologous chromosome pair #1. Possible genotypes will have one C or c, one D or d, and one E or e from each chromosome. For example: CDE/cde; CdE/cDe; cde/cde; CDe/CdE; etc.

In order to determine how many different genotypes are possible, you must first determine how many different gametes are possible for each parent, then match all the gametes in a genetic checkerboard. Although the three pairs of genes are linked to one homologous pair of chromosomes, there are a total of eight different possible gametes for each parent: CDE, CDe, CdE, Cde, cDE, cDe, cdE, and cde. This number of gametes is based on all the total possible ways these genes can be inherited on each chromosome of homologous pair #1. [It is not based on the random assortment of these genes during meiosis in the parents because all three genes are closely linked together on the same chromosome; therefore, all three genes tend to appear together in the same two gametes: CDE and cde.] The possible different genotypes are shown in the following table:

Because there are so many genotypic combinations (including many duplicate combinations), it is difficult to identify all of them in the above table (unless you can think like a computer). A much easier method to determine the total number of different genotypes is to use the following mathematical formula:

4. Any genotypic combination with either DD or Dd is phenotypically Rh positive. For example, the following genotypes are Rh positive: CDE/CDE, CDE/cde, cDe/cDe, CDe/Cde, cDE/cDE, cDe/cde, CDe/CDe, etc. The percentage of Rh positive phenotypes in the U.S. population can be found in your laboratory manual in exercise # 4.

5. Any genotypic combination with dd is phenotypically Rh negative. For example, the following genotypes are Rh negative: CdE/CdE, cde/cde, Cde/cde, cdE/cde, cdE/cdE, CdE/Cde, etc. The percentage of Rh negative phenotypes in the U.S. population can be found in your laboratory manual in exercise # 4.

6. In the following diagram, one of each of the 6 Rh alleles (C & c, D & d, E & e) may occur at a single locus on homologous chromosome pair number 1. This hypothetical situation is similar to the multiple alleles of the A-B-O blood groups. [This Is Not The Way The Rh Alleles Are Inherited]. Since there is only one locus per chromosome (rather than 3), there are only two letters in the genotypes (one from each chromosome). Possible genotypes will have many two-letter combinations of C's, Ds, and E's. For example: CC, Cc, CD, Cd, CE, Ce, DD, Dd, DE, De, EE, Ee, etc.

To calculate the total number of different possible genotypes, use the above formula in question # 3.

7-8. In normal spermatogenesis, X-bearing and Y-bearing sperm are produced. If an X-bearing sperm unites with an X-bearing egg, the resulting zygote is female (XX). If a Y-bearing sperm unites with an X-bearing egg, the resulting zygote is male (XY). Sometimes the X and Y chromosomes do not separate properly during the first division (Anaphase I) or the second division (Anaphase II) during spermatogenesis, a phenomenon known as nondisjunction. Nondisjunction may result in sperm that carry an extra X or an extra Y chromosome, such as XX-bearing sperm, XY-bearing sperm and YY-bearing sperm. If these sperm unite with an X-bearing egg, the result could be XXX (triple-X syndrome), XXY (Klinefelter's syndrome) or XYY (XYY-syndrome). The XXX and XXY chromosome anomalies can also result from an XX-bearing egg.

In the following diagram, normal spermatogenesis is compared with spermatogenesis with nondisjunction at meiosis I (anaphase I) and nondisjunction at meiosis II (anaphase II). If the doubled X and Y chromosomes move to the same cell at meiosis I, the resulting gametes will each contain single X and Y chromosomes. If meiosis I proceeds normally and nondisjunction occurs at meiosis II when the chromatids separate, it is possible to get gametes containing two single X chromosomes and gametes containing two single Y chromosomes:

9-13. In the following diagram, the cell contains 20 letters representing 20 chromosomes.

Since there are 4 of each letter, the cell is tetraploid and contains 4 haploid sets of chromosomes: ABCDE, ABCDE, abcde and abcde. A diploid cell would be AaBbCcDdEe. Since small case letters represent paternal chromosomes, a paternal haploid set would be represented as abcde.

14. The number of different possible gametes produced by the diploid genotype (AaBbCcDdEe) is 2 x 2 x 2 x 2 x 2 = 32 (2 for each pair of heterozygous genes).

15. The following diagram shows the tetraploid cell again with 20 letters representing 20 chromosomes.

For each of the 5 kinds of letters represented in 4's (i.e. tetraploid), there are 3 diploid combinations. For example, AAaa has 3 diploid combinations: AA, Aa and aa. Since the gametes will have 2 of each kind of letter and since there are 3 possibilities for each 2 letter combination, you must multiply 3 x 3 x 3 x 3 x 3 to get the total number of different possible diploid gametes.

In the flowering plant life cycle, the haploid generation is reduced to a germinated pollen grain containing three nuclei and a 7-celled embryo sac containing eight nuclei. Diploid microspore mother cells inside the anther undergo meiosis (microsporogenesis) forming haploid microspores (each mother cell dividing into four microspores). The microspores develop into binucleate pollen grains, each containing a tube nucleus and a generative nucleus. When the pollen grain lands on a receptive stigma it grows into an elongate pollen tube containing a tube nucleus and a generative nucleus, the latter of which divides into two sperm nuclei.

A diploid megaspore mother cell inside each ovule also undergoes meiosis (megasporogenesis) and forms four haploid megaspores, three of which abort leaving one functional megaspore. The functional megaspore (inside each ovule) undergoes nuclear division into a 7-celled, 8-nucleate embryo sac. At one end of the embryo sac are three antipodal cells. At the opposite end is an egg cell flanked by two synergid cells. A large binucleate cell in the center containing two polar nuclei is called the endosperm mother cell. During pollination, pollen grains land on the stigma where they form pollen tubes that penetrate the style and eventually the ovary of the flower. A separate sperm-bearing pollen tube must reach each ovule in order to fertilize the egg cell inside the embryo sac. During double fertilization two sperm are introduced into the embryo sac from the long pollen tube. One sperm nucleus fuses with the egg nucleus inside the egg cell to form a zygote which develops into the embryo of the seed. The other sperm nucleus fuses with the two polar nuclei inside the endosperm mother cell to form the endosperm of the seed.

After fertilization, the ovule enlarges and develops into a mature seed containing a diploid zygote and triploid endosperm. The seed coat is chromosomally identical to the female parent (ovary tissue) because it was derived from two outer layers of the ovule called the integument. On a mature seed the opening or pore in the seed coat is where the pollen tube once entered a gap in the integument layers called the micropyle. As the ovules develop into seeds, the outer ovary encasing the ovules develops and ripens into a fruit. Fruits that develop without double fertilization and without seeds are termed parthenocarpic. Examples of parthenocarpic fruits are navel oranges, bananas, seedless watermelons, and certain varieties of figs. Not all seedless fruits are parthenocarpic. In Thompson seedless grapes, fertilization does occur, but the ovules fail to develop within the fruit. Parthenocarpy can be induced artificially by the application of dilute growth hormone sprays (such as auxins) to the flowers, as in seedless tomatoes. Seedless watermelons come from triploid (sterile) plants; however, to set fruit they must be pollinated by a fertile diploid plant. Some embryos of seeds can develop apomictically without fertilization. A number of angiosperm families contain apomictic species, including figs, blackberries, hawthorns and dandelions. The embryo may develop from a diploid nutritive cell (nucellus tissue) surrounding the embryo sac or from the fusion of hapolid cells within the embryo sac. In general there are two main types of apomixis:

[1] Parthenogenesis (agamogenesis): A haploid or diploid egg within the embryo sac (or diploid cell from 2 fused haploid cells of embryo sac) develops into an embryo. [Formation of haploid cells may involve crossing over during Prophase I of meiosis resulting in some genetic variability.]

[2] Agamospermy: An embryo arises from tissue surrounding the embryo sac. If this involves cells of the nucellus or inner integument it is called a nucellar embryo. Nucellar embryos are chromosomally identical to the sporophyte parent. They are essentially clones of the female parent. In varieties of the edible fig (Ficus carica), apomictic seeds allow propagation of choice edible fig cultivars (female trees) without the transmission of viruses through cuttings. Apomixis also enables a pioneer seedling to colonize and become naturalized in a new habitat by reseeding itself without cross pollination.

16-20. If the diploid pollen parent is aabbcc, then the haploid sperm would be [abc]. If the diploid seed parent is AABBCC, then the haploid egg would be [ABC]. In double fertilization, one [abc] sperm unites with one [ABC] egg to form a diploid [AaBbCc] zygote. Another haploid sperm [abc] unites with two haploid polar nuclei [ABC] + [ABC] to form a triploid [AAaBBbCCc] endosperm within the seed. Since the seed coat originates from the outer wall of the ovule (called the integument), which was part of the original maternal seed parent, it is chromosomally identical with the original diploid seed parent.

Sperm #1 (abc) fuses with a haploid egg (ABC) resulting in a diploid zygote (AaBbCc) that grows into a diploid embryo (AaBbCc) within the seed. Sperm #2 (abc) fuses with the two haploid polar nuclei (ABC and ABC) within the endosperm mother cell resulting in a triploid endosperm cell (AAaBBbCCc) that develops into the nutritive endosperm tissue (AAaBBbCCc) surrounding the embryo. The following remarkable Wayne's Word image shows a minute diploid coconut embryo embedded in the triploid, meaty endosperm within the seed of a coconut palm.

Close-up view through the inside of a coconut seed showing a small, cylindrical embryo (A) embedded in the fleshy meat or endosperm (B). The wall of the endocarp (C) is a hard, woody layer that makes up the inner part of the fruit wall. The thick, fibrous husk (mesocarp) that surrounds the endocarp has been removed.

A Note For Biology 100 Students: In exalbuminous seeds, such as lima beans and walnuts, the endosperm has been completely absorbed by the embryo. The embryo of these seeds consists of two prominent halves called cotyledons. Attached between the cotyledons is a minute, primordial, leaf-bearing shoot called the plumule and an elongate primordial root called the radicle. See following photo: The embryo of a lima bean seed showing the embryonic shoot or plumule (A), the embryonic root or radicle (B) and two cotyledons (C). The two fleshy halves called cotyledons are actually part of the embryo. The seed coat (D) has been partially removed from the cotyledons. Since the seed coat originates from the outer wall of the ovule (called the integument), which was part of the original maternal seed parent, it is chromosomally identical with the original diploid seed parent.

See World's Largest Seed Embryo See The Structure Of A Coconut Fruit Read About Ocean Dispersal Of Coconuts Read About Coconuts And The Coconut Crab See The Double Coconut: Worlds Largest Seed

Chromosome Numbers In Polyploid Plants I prefer the terms "gametophyte" and "sporophyte" when dicussing polyploid chromosome numbers. Gametophyte refers to the chromosone number of gametes and sporophyte refers to the chromosome number of cells in adult plants. In humans, haploid (n) refers to the chromosome number of gametes, while diploid (2n) refers to the chromosome number after fertilization; however, in polyploid plants the chromosome numbers are very different. For example, I have studied a rare hybrid Brodiaea in San Marcos with a sporophyte chromosome number of 36. This is a hexaploid hybrid because the haploid number in Brodiaea species is 6 (6n = 36). The hybrid was derived from a cross between B. terrestris ssp. kernensis with an octoploid (8n) sporophyte number of 48, and B. filifolia with a tetraploid (4n) sporophyte number of 24. The gametes of these two parents are tetraploid (4n) and diploid (2n), resulting in a hexaploid (6n) hybrid: 4n + 2n = 6n. In this case, referring to the gametes as haploid (n) and the hybrid offspring as diploid (2n), as we do in humans, would be incorrect. Chromosome Numbers Of The Rare San Marcos Hybrid Beodiaea

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21-22. In marriages between normal parents who produce a PKU child, the parents must be carriers (heterozygous) for the recessive gene causing this disease. If the recessive gene is represented by (a), then the normal parents of a PKU child would be Aa x Aa. The probability of this couple having a PKU child (aa) can be shown with a simple Mendelian monohybrid cross resulting in 1/4 AA, 2/4 Aa and 1/4 aa; however, to calculate the total probability of a normal couple having a PKU child, you must also calculate the probability of each parent being heterozygous (Aa), and then multiply these two values by the 1/4 chance of having a PKU (aa) baby.

Use the following Punnet Square (genetic checkerboard) to calculate the chance of a parent being heterozygous (Aa) and the fractional ratio of the deleterious recessive gene (a) for PKU:

Don't forget to convert the fractional value for the frequency of the recessive gene, and the fractional probability of a couple having a PKU child, into percent values in order to find the correct answer on your ScanTron answer sheet. For example a fractional value of 1/100 = .01 = 1%.

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Angiosperm Male Gametophyte Questions:

23-28. A mature angiosperm pollen grain contains a tube nucleus and a generative nucleus, the latter of which divides into 2 sperm within the pollen tube. For this question, a hypothetical sperm-bearing angiosperm pollen tube contains a total of 12 chromosomes. Use the following answer choices for questions 23-28:

(a) 0       (b) 4       (c) 6       (d) 8       (e) 12

23. How many of these 12 chromosomes will be contributed to the zygote at fertilization?

24. How many of these 12 chromosomes will be contributed to the endosperm at fertilization?

25. What is the diploid chromosome number of this hypothetical plant (i.e. how many chromosomes per somatic cell)?

26. What is the haploid chromosome number of this hypothetical plant (i.e. how many chromosomes per gamete)?

27. What is the chromosome number of the endosperm tissue?

28. What is the chromosome number of a polar nucleus inside the endosperm mother cell within the embryo sac?

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29-32. A hypothetical angiosperm embryo sac (megagametophyte) contains a total of 32 chromosomes. Use the following answer choices for questions 29-32.

(a) 0       (b) 4       (c) 6       (d) 8       (e) 12

29. What is the chromosome number of an egg cell?

30. What is the chromosome number of a synergid cell?

31. How many total chromosomes are contained within the endosperm mother cell?

32. How many nuclei are contained within this embryo sac?

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Plants That Lived With Dinosaurs

References About Ancient Plants Fossils Of Ancient Plants Cycads And Continental Drift Ancient Plants Of Jurassic Park Living Fossils At Palomar College Amber: Nature's Transparent Tomb

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Symbiosis: An Intimate Association

Symbiosis may be defined as an intimate association between two or more organisms. There are three main types of symbiosis, including commensalism, parasitism and mutualism. In commensalism, one organism in the relationship is benefited while the other is neither benefited nor harmed. Some bird nests and trees form a commensal relationship. The birds obtain shelter and protection without harming the tree. Certain epiphytic orchids also form commensal relationships with trees. The dorsal fin of the remora is modified into a sucker which forms a temporary attachment to the shark. The shark does not seem to be inconvenienced by this and makes no attempt to remove the remora. When the shark feeds, the remora is in a good position to pick up scraps of food left by the shark. Marine mammals, including whales and manatees, often carry harmless hitchhikers called barnacles on their backs. The barnacles benefit from the ride through nutrient-rich waters. In parasitism, the host is usually harmed, while the parasite benefits from the association. There are many examples of parasitic plants and animals, including remarkable root parasites called broomrapes. The relationship between the jumping bean shrub and the jumping bean moth is certainly one-sided and probably slightly parasitic. The moth larva is clearly a seed predator. Although the moth doesn't appear to harm the host that much, it could decrease the percentage of viable seeds. In mutualism both members of the association (called symbionts) derive benefit from their relationship. In a sense, this mutually beneficial relationship is a type of "marriage." In lichens, the "marriage" is vital to the survival of both the algal and fungal symbionts, although some species have been grown separately in laboratories. There are many references in Wayne's Word about symbiosis, including the fig and fig wasp, yucca and yucca moth, acacia and acacia ant, azolla fern and cyanobacteria, and many others. See the table of Wayne's Word hyperlinks below:

Wayne's Word Hyperlinks About Symbiosis Fig And Fig Wasp Symbiosis           Gall Flowers & Fig Wasp Pollination Calimyrna Fig And Its Pollinator Wasp Evolution Of Dioecious Fig Species Pollination Patterns In Dioecious Figs Sex Determination In The Common Fig Strangler Figs & Banyans: Amazing Trees       Summary Of Common Fig Life Cycle The Amazing Fig/Fig Wasp Relationship Galls Caused By Insects           California's Amazing Jumping Galls Galls: Fasciated (Malformed) Stems) Galls: Strange & Growths On Plants Other Examples Of Symbiosis:           Sea Anemone & Green Zoochlorellae The Acacia Shrub & Acacia Ant The Cecropia Tree & Azteca Ant Azolla Fern And Its Cyanobacterium Cycads & N-Fixing Cyanobacteria Gunnera & N-Fixing Cyanobacteria The Pollination Of Coral Trees The Yucca And The Yucca Moth Lichens: Nature's Perfect Marriage Amazing Parasitic Flowering Plants The Jimsonweed And Its Junkie Moth The Mexican Jumping Bean & Its Moth

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Fig Questions:

See Biology 100 Laboratory Manual Pages 98-99.

Other Wayne's Word Articles About Figs & Wasps: Gall Controversy In Figs The Fig/Fig Wasp Relationship Pollination Patterns In Dioecious Figs Sexuality In Figs And Other Flowering Plants Figs Of The Holy Land (Their Role In World Religions)

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33. Name of remarkable fig structure that is lined on the inside with hundreds of tiny unisexual flowers (including male flowers, short-style female flowers and/or long-style female flowers)?

(a) fruit (b) ovary (c) syconium (d) ovule (e) synstigma

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34-45. The following answer choices are for questions 34-45:

(a) Short-Style Female Flower
(b) Long-Style Female Flower
(c) Male (Staminate) Flower
(d) Male "Caprifig" Tree
(e) Female Tree

34. Which flower produces a seed inside its ovary?

35. Which flower contains a wasp inside its ovary?

36. Which flower produces pollen?

37. Which flower does the female wasp oviposit in?

38. Which flower does the male wasp chew a hole in?

39. In which flower does the male wasp inseminate the female wasp?

40. In true dioecious figs, which tree perpetuates the symbiotic fig wasp?

41. In true dioecious figs, which tree perpetuates the fig by producing seeds?

42. In true dioecious figs, which tree produces pollen?

43. Which flowers produce the crunchy, nutty nutlets in your fig newton?

44. Which tree produces the crunchy nutlets of your fig newton?

45. Which flowers mature last within the syconium?

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46. Are figs (a) Protandrous or (b) Protogynous?

47. Is Wolffia (a) Protandrous or (b) Protogynous?

See Articles About Wolffia and Other Duckweeds: Lemnaceae On-Line The World's Smallest Fruit The World's Smallest Flowering Plant Weird Duckweeds From Far Away Lands Diversity Of Flowering Plants Of The World

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Honey Bee Questions:

48-49. In the honeybee (Apis mellifera, 2n=32), male bees (drones) are produced by parthenogenesis.

(a) 0       (b) 1       (c) 2       (d) 4       (e) 8      

48. How many sets of chromosomes in the somatic cells of a drone?

49. How many pairs of homologous chromosomes in somatic cells of a drone?

50. Since drone bees come from unfertilized eggs and receive only maternal genes, are they genetically identical clones?     (a) Yes      or      (b) No

See The Sad Saga Of A Drone Honey Bee

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Gender verification in the Olympic Games now employs sophisticated DNA testing rather than counting Barr bodies within the nuclei of cells. The test is designed to detect the presence of the SRY gene (Sex Region Y chromosome), a region of DNA on the short arm of the Y chromosome responsible for masculinization of the fetus. Cells from the buccal mucosa (squamous epithelial cells), often called "cheek cells" in general biology classes, are obtained by gently scraping the inside of the mouth with a toothpick. The DNA in the nuclei of these cells is amplified using the PCR tecnique (Polymerase Chain Reaction). If present, the SRY gene will show up as a unique banding pattern by electrophoresis on agar gels.

The following choices refer to questions 51-55. The gender of some of these chromosomal karyotypes and syndromes cannot be correctly identified using the Barr body technique.

(a) XY   (b) XX    (c) XXY    (d) XXX    (e) X_

51. A phenotypic male with one Barr body.

52. A phenotypic female with zero Barr bodies.

53. A phenotypic female with one Barr body.

54. A phenotypic male with no Barr bodies.

55. A phenotypic female with two Barr bodies.

See Explanation Of Gender Verification

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The following choices refer to questions 56-58. The gender of some of these chromosomal karyotypes and syndromes cannot be correctly identified using the Barr body or SRY techniques.

(a) androgen insensitivity syndrome
(b) adrenogenital syndrome
(c) the XYY Syndrome

56. A phenotypic male with twice as many SRY genes.

57. A chromosomal (XX) female with a male phenotype.

58. A chromosomal (XY) male with a female phenotype.

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A Genetic Cross Between Watermelons

In watermelons the gene for green rind (G) is dominant over the gene for striped rind (g), and the gene for short fruit (S) is dominant over the gene for long fruit (s). The alleles for rind color and fruit length occur on two different pairs of homologous chromosomes. For this question, assume that a gene for large melons (L) and and gene for many seeds (F) occur at opposite ends of another chromosome (linkage). The alleles for size and seed number, i.e. the genes for small melons (l) and few seeds (f), occur on a third homologous chromosome. A watermelon plant bearing large, green, short fruits containing many seeds was crossed with a plant bearing large, striped, long fruits containing many seeds. Some of the offspring from this cross produced small, striped, long fruits with few seeds.

Assuming no crossing over between homologous chromosomes, what is the fractional chance of producing the following offspring? Remember that there are three pairs of homologous chromosomes in this problem, and one of the homologous pairs exhibits autosomal linkage. The chromosomes of each parent are shown in the following illustration:

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Answer choices for questions 59-60:

(a) 2    (b) 4    (c) 8    (d) 16    (e) 32

59. How many different possible gametes can the green, short-fruit parent produce?

60. How many different possible gametes can the striped, long-fruit parent produce?

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Answer choices for questions 61-70:

(a) 1/16   (b) 2/16   (c) 3/16   (d) 3/8   (e) no chance

61. Large, striped, long watermelon containing few seeds.

62. Large, green, short watermelon containing many seeds.

63. Large, green, long watermelon containing many seeds.

64. Large, striped, short watermelon containing many seeds.

65. Large, striped, long watermelon containing many seeds.

66. Small, green, short watermelon containing few seeds.

67. Small, green, long watermelon containing few seeds.

68. Small, striped, short watermelon containing few seeds.

69. Small, striped, long watermelon containing few seeds.

70. Small, striped, long watermelon containing many seeds.

Hints On How To Solve The Watermelon Question

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