Vaccines Resulting In Active Immunity 2. Serums Resulting In Passive Immunity 3. Genetics Extra Credit Problems Questions 61 - Remember that the gene allele for taster T is dominant over the gene allele for nontaster t : Questions 63 - Human skin color is a good example of polygenic inheritance in people. The offspring contain seven different shades of skin color based on the number of capital letters in each genotype.
The words dominant and recessive are placed in quotation marks because these pairs of alleles are not truly dominant and recessive as in some of the garden pea traits that Gregor Mendel studied.
A genotype with all "recessive" small case genes aabbcc has the lowest amount of melanin and very light skin. Each "dominant" capital gene produces one unit of color, so that a wide range of intermediate skin colors are produced, depending on the number of "dominant" capital genes in the genotype. For example, a genotype with three "dominant" capital genes and three small case "recessive" genes AaBbCc has a medium amount of melanin and an intermediate skin color.
This latter genotype would be characteristic of a mulatto. In the above cross between two mulatto genotypes AaBbCc x AaBbCc , each parent produces eight different types of gametes and these gametes combine with each other in 64 different ways resulting in a total of seven skin colors. The skin colors can be represented by the number of capital letters, ranging from zero no capital letters to six all capital letters.
The approximate shades of skin color corresponding to each genotype are shown in the above table. Note: Skin color may involve at least four pairs of alleles with nine or more shades of skin color.
Each term in the expression represents the number of offspring with a specific skin color phenotype based on the number of capital letters in the genotype. For example, 20 offspring have three capital letters in their genotype and have a skin color that is intermediate between very dark with all caps AABBCC and very light with no caps aabbcc. Questions 69 - These questions refer to the Rh types of Chrissy and John, and their baby boy named Cinco. Questions 63 - Questions 78 - Remember that the A and B alleles are dominant over the O allele.
The type O blood phenotype must be homozygous for the O allele. Type AB blood phenotype must be heterozygous for the A and B alleles. For these questions, use the process of elimination. Then eliminate the only parents that could have an AB baby, and so forth.
Questions 87 - To determine the fractional probability for a taster boy with type B blood, you must make a cross between John and Mary using a genetic checkerboard Punnett square.
Questions 91 - Place only decimal values in the squares of your checkerboard because you can't multiply percentages. The total decimal value for gametes must add up to 1. In other words, 0. The total genotype values must also add up to 1. The following table using 5 coins illustrates these two questions.
Simply change the five coins to three coins or children. Remember that sex determination is much more complicated than tossing coins because many other factors are involved. There is only one permutation out of 32 refer to the top permutation, 3rd column from left. In this example you must consider all possible permutations with 3 Heads and 2 Tails. The 3rd column from left in the above Pascal's Triangle shows 10 permutations out of 32 with 3 Heads and 2 Tails.
This is also the probability of having 3 girls and 2 boys when all possible orders are considered. Alleles 0. In garden peas, the gene for round R is dominant over the gene for wrinkled r and the gene for tall T is dominant over the gene for short t. Serious complications may arise when the antibodies of the recipient clump the blood cells of the donor.
Clumping of the donor's blood is indicated by the word "Clump" in the red squares. No clumping of the donor's blood is indicated by the word "None" in the green squares.
None also denotes the lack of anti-A or anti-B antibodies in the type O recipient. It is clear from this chart that the "universal donor" is type O, while the "universal recipient" is type AB.
If you include the Rh factor, then the universal donor becomes O Negative while the universal recipient becomes AB Positive. Although it is much more complicated, the Rh blood factor can be explained by a pair of alleles on homologous chromosome pair 1. Like the type O gene, the recessive Rh negative gene - does not produce an antigen. Because of the time factor involved in building up a concentration titre of antibodies, the first transfusion may not cause any major problems; however, a subsequent transfusion of Rh positive blood could be very serious because the recipient will clump all of the incoming blood cells.
Based upon the above table, Rh positive recipients can theoretically receive positive or negative blood, and Rh negative donors can theoretically give to Rh positive and Rh negative recipients.
The following tables explain how to calculate the answers for questions - The data in the tables is slightly different from your exam, but the method of calculation is the same. These questions refer to a cross between two hypothetical watermelons with four pairs of fruit characteristics. 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. 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: There are several ways to solve this problem, but one way is to construct a 16 square checkerboard with eight rows and two columns. To the left of each row, put the eight gametes of the parental plant bearing large, green, short fruits containing many seeds.
At the top of each column, put the two gametes of the parental plant bearing large, striped, long fruits containing many seeds. The most difficult part of this problem is to figure out the exact gene combinations of the gametes. Once this is known, you can simply fill in the squares of the checkerboard with the correct gene combinations genotypes for each offspring.
There a total of 12 different genotypes in the checkerboard. The plant bearing large, striped, long fruits containing many seeds can produce only two different kinds of gametes shown in red in Table 1. The gametes must contain one of the LF or lf chromosomes, one of the g chromosomes, and one of the s chromosomes. Therefore, the two possible gametes are: LFgs and lfgs. The LF and lf genes always appear together because they occur on the same chromosomes. Without crossing over, you could never have Lf together or lF together.
The plant bearing large, green, short fruits containing many seeds can produce eight different kinds of gametes shown in green in Table 1.
The gametes must contain one of the LF or lf chromosomes, one of the G or g chromosomes, and one of the S or s chromosomes. Eventually it penetrates the micropyle of an ovule and releases its two sperm into the 8-nucleate embryo sac.
During double fertilization, one sperm fuses with the egg nucleus to form a zygote. The other sperm fuses with the two polar nuclei inside the endosperm mother cell to form the endosperm.
In corn, this process must occur for each grain that forms. Even more astonishing is the growth of separate pollen tubes down each strand of silk styles. 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. Microscopic view of the embryo sac megagametophyte of a lily Lilium. Three haploid antipodal cells 1 occur at the upper end of the emryo sac. A large endosperm mother cell containing two haploid polar nuclei 2 occupies the central portion of the embryo sac.
At the lower end nearest the micropyle and funiculus are two haploid synergid cells 3 and one haploid egg. The embryo sac contains a total of seven cells and eight nuclei. The sporophyte generation includes the stems, leaves, roots, flowers, fruits and seeds. 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. 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.
If the diploid pollen parent is aabbcc, then the haploid sperm would be [abc]. 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. The following diagram summarizes double fertilization in this question: 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.
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 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. 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 The hybrid was derived from a cross between B.
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. In marriages between normal parents who produce a PKU child, the parents must be carriers heterozygous for the recessive gene causing this disease.
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. Angiosperm Male Gametophyte Questions: 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 a 0 b 4 c 6 d 8 e 12 A steroid hormone which regulates glucose metabolism is - Chemical Coordination and Integration.
Amino acid sequence, in protein synthesis is decided by the sequence of- Molecular Basis of Inheritance. Golden Rice is a promising transgenic crop. When released for cultivation, it will help in Biotechnology and its Applications. Sertoli cells are regulated by the pituitary hormone known as- Human Reproduction. In which mode of inheritance do you expec more maternal influence among the offspring?
Molecular Basis of Inheritance. Treatment of seed at low temperature under moist conditions to break its dormancy called - Plant Growth and Development.
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