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    if flies that are heterozygous for all three traits are crossed, what proportion of the offspring would you expect to be heterozygous for all three traits?

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    get if flies that are heterozygous for all three traits are crossed, what proportion of the offspring would you expect to be heterozygous for all three traits? from EN Bilgi.

    Inheritance of Traits by Offspring Follows Predictable Rules

    Inheritance of Traits by Offspring Follows Predictable Rules

    Genes come in different varieties, called alleles. Somatic cells contain two alleles for every gene, with one allele provided by each parent of an organism. Often, it is impossible to determine which two alleles of a gene are present within an organism's chromosomes based solely on the outward appearance of that organism. However, an allele that is hidden, or not expressed by an organism, can still be passed on to that organism's offspring and expressed in a later generation.

    Tracing a hidden gene through a family tree

    Figure 1: In this family pedigree, black squares indicate the presence of a particular trait in a male, and white squares represent males without the trait. White circles are females. A trait in one generation can be inherited, but not outwardly apparent before two more generations (compare black squares).

    Figure Detail

    The family tree in Figure 1 shows how an allele can disappear or "hide" in one generation and then reemerge in a later generation. In this family tree, the father in the first generation shows a particular trait (as indicated by the black square), but none of the children in the second generation show that trait. Nonetheless, the trait reappears in the third generation (black square, lower right). How is this possible? This question is best answered by considering the basic principles of inheritance.

    Mendel's principles of inheritance

    Gregor Mendel was the first person to describe the manner in which traits are passed on from one generation to the next (and sometimes skip generations). Through his breeding experiments with pea plants, Mendel established three principles of inheritance that described the transmission of genetic traits before genes were even discovered. Mendel's insights greatly expanded scientists' understanding of genetic inheritance, and they also led to the development of new experimental methods.

    One of the central conclusions Mendel reached after studying and breeding multiple generations of pea plants was the idea that "[you cannot] draw from the external resemblances [any] conclusions as to [the plants'] internal nature." Today, scientists use the word "phenotype" to refer to what Mendel termed an organism's "external resemblance," and the word "genotype" to refer to what Mendel termed an organism's "internal nature." Thus, to restate Mendel's conclusion in modern terms, an organism's genotype cannot be inferred by simply observing its phenotype. Indeed, Mendel's experiments revealed that phenotypes could be hidden in one generation, only to reemerge in subsequent generations. Mendel thus wondered how organisms preserved the "elementen" (or hereditary material) associated with these traits in the intervening generation, when the traits were hidden from view.

    How do hidden genes pass from one generation to the next?

    Although an individual gene may code for a specific physical trait, that gene can exist in different forms, or alleles. One allele for every gene in an organism is inherited from each of that organism's parents. In some cases, both parents provide the same allele of a given gene, and the offspring is referred to as homozygous ("homo" meaning "same") for that allele. In other cases, each parent provides a different allele of a given gene, and the offspring is referred to as heterozygous ("hetero" meaning "different") for that allele. Alleles produce phenotypes (or physical versions of a trait) that are either dominant or recessive. The dominance or recessivity associated with a particular allele is the result of masking, by which a dominant phenotype hides a recessive phenotype. By this logic, in heterozygous offspring only the dominant phenotype will be apparent.

    The relationship of alleles to phenotype: an example

    Relationships between dominant and recessive phenotypes can be observed with breeding experiments. Gregor Mendel bred generations of pea plants, and as a result of his experiments, he was able to propose the idea of allelic gene forms. Modern scientists use organisms that have faster breeding times than the pea plant, such as the fruit fly (). Thus, Mendel's primary discoveries will be described in terms of this modern experimental choice for the remainder of this discussion.

    Figure 2: In fruit flies, two possible body color phenotypes are brown and black.

    The substance that Mendel referred to as "elementen" is now known as the gene, and different alleles of a given gene are known to give rise to different traits. For instance, breeding experiments with fruit flies have revealed that a single gene controls fly body color, and that a fruit fly can have either a brown body or a black body. This coloration is a direct result of the body color alleles that a fly inherits from its parents (Figure 2).

    In fruit flies, the gene for body color has two different alleles: the black allele and the brown allele. Moreover, brown body color is the dominant phenotype, and black body color is the recessive phenotype.

    Source : www.nature.com

    Genetics 314 Exam 1 Material (Chapters 1

    Start studying Genetics 314 Exam 1 Material (Chapters 1-5). Learn vocabulary, terms, and more with flashcards, games, and other study tools.

    Genetics 314 Exam 1 Material (Chapters 1-5)

    Prophase 1

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    During what stage does crossing over/recombination occur?

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    1/4 of all children

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    If a man in a couple has hemophilia, an X-linked recessive condition, and his his wife does not, but his wife's mother also has hemophilia, what is the probability out of all children that the couple would have a boy who is hemophiliac?

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    1/99 Created by paige_williams576

    Terms in this set (99)

    Prophase 1

    During what stage does crossing over/recombination occur?

    1/4 of all children

    If a man in a couple has hemophilia, an X-linked recessive condition, and his his wife does not, but his wife's mother also has hemophilia, what is the probability out of all children that the couple would have a boy who is hemophiliac?

    12

    Two parents EeFfgg x eeFfGg have how many possible genotype combinations in offspring?

    Half of males

    In birds, a male with speckled plumage Z^SP Z^sp is crossed to a female with plain plumage (ZP^sp W). What proportion of all offspring should be speckled males?

    e. Cell theory of organisms proposed.

    Which of the following events was the earliest to occur in the history of genetics?

    a. scientists crack the code on how to "read" DNA

    b. traits demonstrated to be passed generation to generation in peas

    c. chromosome theory of inheritance proposed

    d. Theory of natural selection proposed

    e. Cell theory of organisms proposed

    Two

    How many Barr bodies should be present in the nuclei of a human that is XXXY?

    G2/M

    During the cell cycle, the check to make sure that the chromosomes have been properly replicated occurs at the ________ checkpoint.

    Mutations in 2 different genes (complement)

    You have two mice, both of which have kinked tails (the mutant phenotype). If you cross the two mice and all offspring have wild-type (non-kinked) tail this means that __________.

    c. Incomplete dominance

    If you cross two carnations, both with pink flowers and in the offspring you routinely see a ratio of 1 red; 2 pink; 1 white. Which of the following is at work?

    a. codominance b. epistasis

    c. incomplete dominance

    d. incomplete penetrance

    e. variable expressivity

    All black

    In mealworms eye color is under maternal effect of a single gene, with black eyes dominant over red. A mother has red eyes and genotype Bb and a father has black eyes with genotype bb; what is the eye color of offspring?

    1/8

    In Drosophilia, white (X^r) vs. red (X^R) eyes is a sex-linked recessive trait and stubby bristles (B) are dominant over long bristles (b). For two parents X^RX^rBb x X^rYbb there is a ______ chance that offspring will be red eyed females with stubby bristles.

    AB

    If a mother is blood type A and a child is blood type O, ______ cannot be the blood type of the father.

    Segregation

    The fact that homologues separate during meiosis and offspring inherit only one allele of a gene from each parent is described by Mendel's principle of ________.

    Chromosomally male, Gonadally female, externally phenotypically female

    A human who is XY^sry- is...

    Chromosomally male, gonadally male, externally phenotypically male

    A Drosophilia that is XO is...

    Chromosomally female, gonadally female, externally phenotypically female

    A human who is X^tfmO

    a. Draw two cells, each with the mitotic spindles pulling apart the sister chromatids, there should be three in each cell (totaling to a total of 6 chromosomes per cell)

    b. haploid c. 6

    d. the sister chromatids are the two chromatids that are getting pulled apart from each other(same cell), while the homologues are the same size and have the same centromere location (one in each of the cells)

    For a starting cell where in G1 2n=6

    a. Diagram the cell(s), in particular, the chromosome arrangement, in anaphase II of meiosis

    b. Are the cell(s) you drew haploid or diploid?

    c.How many chromatids are in each of the cell(s)?

    d. Label one pair of homologues and one pair of sister chromatids on your diagram above.

    1/4

    In a family, what are the chances that a couple would have four children, 3 females and one male?

    a. white --W--> yellow --Y--> green

    b. Parent #1: white; Parent #2: yellow

    c. 2:1:1

    d. Gene W is epistatic to gene Y, so recessive epistasis.

    In a biochemical pathway, gene W allows conversion of a white precursor into a yellow intermediate. Gene Y allows conversion of the yellow intermediate to a green product.

    a. diagram the pathway

    b. What are the colors of the two parents, wwYy x Wwyy?

    c. What phenotypic ratios are predicted in the offspring from the parents in part b?

    d. Which gene is epistatic to the other? And is this dominant, recessive, or complementary gene action epistasis?

    a. Expected: 18 black and 18 brown; Observed: 20 black and 16 brown

    b. 0.444

    c. It supports the hypothesis

    Use a Chi-square test to determine if offspring numbers 20 black and 16 brown supports a hypothesized Bb x bb cross.

    a. What are your expected and observed numbers?

    b. Chi-square number?

    c. What does this say about the hypothesis?

    Source : quizlet.com

    Answer key to practice problems

    Answer key to practice problems--1999

    Week 1 Week 2 Week 3 Week 4 Week 5

    Week 6 Week 7 Week 8 Week 9 Week 10

    Week 11

    ..Week 11

    1. 1/2N = 1/800. 2.

    In the smaller population --

    Frequency of the recessive phenotype = (q1)2 = 4/400

    Frequency of the recessive allele = q1 = 1/10 = 0.1

    In the larger population --

    Frequency of the recessive phenotype = (q2)2 = 54/600

    Frequency of the recessive allele = q2 = 1/10 = 0.3.

    In the merged population --

    Frequency of recessive allele q = ((400 x 0.1) + (600 x 0.3))/1000 = 0.22

    Frequency of black cats in the next generation = q2 = (0.22)2 = 0.0484.

    A potential source of error in this problem is to simply add the number of recessive individuals from the two populations and to derive q from that -- i.e., take the square root of (4 + 54). However, doing so would ignore the contribution of recessive alleles from the heterozygotes in each population.

    3. (i) If only black cats are left standing after the virus goes through, then only the recessive (black) allele will be left in the population; the frequency of the black allele in the next generation will be 1.0 (= 100%).

    (ii)

    Before the virus comes through, the frequency of the three genotypes is:

    Homozygous dominant = p2 = 0.25

    Heterozygotes = 2pq = 0.5

    Homozygous recessive = q2 = 0.25

    After the viral epidemic, the only cats left are homozygous dominant and heterozygotes:

    Homozygous dominant = p2 = 0.25

    Heterozygotes = 2pq = 0.5

    Homozygous recessive = q2 = 0.25

    Now the heterozygotes make up 2/3 of the surviving population, so the recessive allele makes up 1/3 of the total alleles in the population. Therefore, in the next generation the frequency of black cats will be (1/3)2 = 1/9.

    4. (i) The d allele will be more frequent, as the forward mutation (D to d) occurs at a higher rate than the back mutation.

    (ii)

    Let the frequency of D = p, and the frequency of d = q, forward mutation rate = u, and back mutation rate = v

    Then the change in p would include loss from forward mutation and gain from back mutation; likewise, change in q would include gain from forward mutation and loss from back mutation:

    Change in p = vq - up

    Change in q = up - vq

    (iii)

    At equilibrium, change in p is exactly matched by change in q, so the change in p = 0 (as is the change in q)--

    vq - up = 0; vq = up

    Since q = 1 - p, we can substitute and solve for p--

    v(1 - p) = up v - vp = up up + vp = v p = v/(u + v)

    Therefore, at equilibrium, p = 0.00004/0.00016 = 0.25

    q = 1-0.25 = 0.75 5. (i) 250 ; 500

    (ii) df = 1. All we need to measure is the number of homozygous recessive and that lets us calculate the predicted number of the other classes (as was done in part i).

    6. (i) Probability of correct identification of each heterozygote = 0.7. Therefore, the probability that both members in a heterozygote/heterozygote couple will be correctly identified = 0.7 x 0.7 = 0.49. So the probability that both members will not be correctly identified = 1 - 0.49 = 0.51 (or 51%).

    (ii) 5% (= 0.05, the frequency of heterozygotes in the population).

    (iii) If one member is tested and not found to have a disease allele, that could either mean that the person is homozygous normal, or that the person is heterozygous (probability = 0.05) but is among the 30% false negatives (probability = 0.3). So the probability that the second person is in fact a heterozygote = 0.05 x 0.3 = 0.015.

    7. The premise of the resin treatment is that depletion of bile will cause liver cells to express more LDL receptors so as to increase the uptake of cholesterol. In this instance, since the cells are incapable of expressing LDL receptors anyway, depleting the body of bile acids will have no effect.

    8. Construct 2. The RNA transcribed from the construct has to be complementary to the target mRNA, so it has to be transcribed off the other strand of the template DNA (so the promoter has to be at the opposite end of the gene, as in Construct 2).

    Week 10

    1.

    Do a complementation test... the strain with the unknown mutation is crossed with the known mutant strain or the strain. If the unknown mutation (called in the diagram below) is in , the progeny of the cross will also have the same phenotype (tailless offspring) -- i.e., the unknown mutation fails to complement and therefore the unknown mutation is in Alternatively, if the unknown mutation fails to complement , the mutation must be in

    If the female progeny from Cross #1 have tailless offspring, the unknown mutation must be in ; if the female progeny from Cross #2 have tailless offspring, the unknown mutation must be in .

    There's a catch--how do we deal with the problem that the progeny from the cross are going to be inviable? If conditional alleles (--see Answer 4 in Problem set 5) are available, there's an easy solution: do the cross and allow development of the resulting embryos at the permissive condition, to let the embryos develop, and then shift the young animals to the restrictive condition to look at the phenotype of their progeny.

    If conditional alleles are not available, an alternative strategy is to cross heterozygotes and to ask if one quarter of the progeny show the phenotype:

    Source : depts.washington.edu

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