Combo with "Chapter 12" and 1 other – Flashcards

a plant cell in the process of cytokinesis

disruption of mitotic spindle formation

Cancer cell continue to divide even when they are tightly packed together

the degradation of cyclin

cells with more than one nucleus.

replication of DNA DOES occur in mitosis condensation of chromosomes, separation of sister chromatids, spindle formation, separation of the spindle poles

G1

cleavage furrow formation and cytokinesis

a sperm

multicellular haploid

meiosis I

sister chromatids separate during anaphase.

2x

x

16

a. Has no effect on phenotype in a heterozygote

b. A variant for a character

c. Having two identical alleles for a gene

d. A cross between individuals heterozygous for a single character

e. An alternative version of a gene

f. Having two different alleles for a gene

g. A heritable feature that varies among individuals

h. An organism's appearance or observable traits

i. A cross between an individual with an unknown genotype and a homozygous recessive individual

j. Determines phenotype in a heterozygote

k. The genetic makeup of an individual

l. A heritable unit that determines a character; can exist in different forms

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

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

1. All three children are of normal phenotype. 2. One or more of the three children have the disease. 3. All three children have the disease. 4. At least one child is phenotypically normal.

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

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

1. AABBCC 3 aabbcc → AaBbCc 2. AABbCc 3 AaBbCc → AAbbCC 3. AaBbCc 3 AaBbCc → AaBbCc 4. aaBbCC 3 AABbcc → AaBbCc

chapter 15

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.

(a) Pseudohypertropic muscular dystrophy is a recessive allele because if it were dominant then the heterozygous female would die before she was sexually mature, and thus could not pass on the trait (b) Pseudohypertropic muscular dystrophy is a sex-linked trait because it only occurs in males. (c) It is never seen in girls because for a girl to be homozygous recessive (and express the trait) XmXm, her father would have to be hemizygous recessive XmY) and her mother would have to be heterozygous or homozygous recessive (XmXM or XmXm). It is impossible for a hemizygous recessive male or a homozygous recessive female to have children because they will die before being old enough to have children. The Punnett square above shows that in a cross between two apparently normal parents, but the mother is a carrier, three children will be normal but one, always a boy will inherit the disease.

First count the total number of offspring 778+785+158+162 = 1883 In all dihybrid test crosses (a cross between a known heterozygote for two wild type traits and a homozygous recessive individual for both traits) the expected ratio of phenotypes if the genes are on separate chromosomes must be: wild type, 25%; black-vestigial, 25% black-normal, 25%; gray-vestigial, 25%. These results do not fit the experimental data above (778+785+158+162). In fact the black-normal (158) and gray-vestigial (162) offspring represent recombinant individuals. Calculation of recombination frequency: 778 - wild type 785 - black-vestigial 778/1883 = 41.3% 785/1883 = 41.7% 83% are non-recombinant 158 - black-normal 162 - gray-vestigial 158/1883 = 8.4% 162/1883 = 8.6% 17% are due to recombination Recombination frequency = 17% The generally accepted method of symbolizing the genotypes for a dihybrid cross of linked genes is as follows (pg. 264 - Campbell): b+ = wild-type (gray body) b = black body vg+ = normal wing shape vg = vestigial wings The testcross is symbolized as follows: If no cross over occurs then only two phenotypes should be seen. That is 50% of the offspring should be dominant for both traits ---- and the other 50% should be homozygous recessive ---- just as in the parents above. Cross over in the heterozygous parent results in 50% recombinants:

Experiment 1 (Frequency/Distance between T and A). Determine the recombination frequency for the genes controlling Tallness and Antennae: 46 tall-antennae = 46% expected 42 dwarf-no antennae = 42% expected 7 dwarf-antennae = 7% recombinant 5 tall-no antennae = 5% recombinant Total = 100 Therefore this recombination frequency between genes T and A is 12% Experiment 2. (Frequency/Distance between A and S) Determine the recombination frequency for the genes controlling Antennae and Snout: 47 antennae-upturned snout = 47% expected 48 no antennae-downturned snout = 48% expected 2 antennae-downturned snout = 2% recombinant 3 no antennae-upturned snout = 3% recombinant Total = 100 Therefore this recombination frequency between genes A and S is 5%

The disorder would always be inherited from the mother because the mother's mitochondrial gene is the only one that survives when the zygote is formed. The gamete from the mother contains all the information. The head of the father's sperm is the only part that survives during fertilization. The tail of the sperm containing the male's mitochondria (an their genes) is lost when the zygote begins development. Thus it is only from the mother that the disorder can be inherited.

A phenomenon in which expression of an allele in offspring depends on whether the allele is inherited from the male or female parent.

Complete dominance: Heterozygous phenotype same as that of homozygous dominate. Incomplete dominance: Heterozygous phenotype intermediate between the two homozygous phenotype. Codominance: Both phenotypes ezpressed in heterozygotes

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- An extra chromosome; down-syndrome (3 chromosome) - Klinefelter Syndrome -47 XXY - Turner Syndrome -45 X0 - Trisomy X -47 XXX

-used to describe the single copy of an x-linked gene in the male

-The Law of Segregation is dependent on the separation of members of homologous pairs. -While the Law of Independent assortment is dependent on the random arrangement of homolohous chromosomes at the metaphase plate.

-There is a 50% chance that i can reappear. 1:2:1 ratio

- Tumor suppressor gene is a gene that protects a cell from one step on the path to cancer. - The functions of tumor-suppressor proteins fall into several categories including the following: 1.Repression of genes that are essential for the continuing of the cell cycle. 2.Coupling the cell cycle to DNA damage. 3.If the damage cannot be repaired, the cell should initiate apoptosis (programmed cell death) to remove the threat it poses for the greater good of the organism. 4.Some proteins involved in cell adhesion prevent tumor cells from dispersing, block loss of contact inhibition, and inhibit metastasis. 5.DNA repair proteins are usually classified as tumor suppressors as well, as mutations in their genes increase the risk of cancer. - Apoptosis is the death of cells that occurs as a normal and controlled part of an organism's growth or development. - Example: p53 gene pRb, PTEN, pVHL, APC, CD95, ST5, YPEL3, ST7, and ST14.

The first meiotic division results in each pairing sorting its maternal and paternal homologs into daughter cells independently of every other pair. it occurs after metaphase 1.

During phase 1 in animal cells; plant cells can not pinch

Meiosis I in anaphase I because thats when they begin to head toward the spindle poles.

Second meiotic division is very similar to mitosis, except that meiosis II is not preceded by chromosomal replication. The chromosomes consisting of sister chromatids align at the equator. Then the sister chromatids separate, move to opposite poles and are surrounded by a re-formed nuclear membrane.

DNA is replicated - during S (Synthesis) Mature cells - Go The centrioles move apart - during prophase 1 Centromere the point on a chromosome by which it is attached to a spindle fiber during cell division. 2 chromatid : 1 centromere

Cleavage furrow: on the outer surface indicates that two new cells are forming. Cell Plate: Plant cells form a cell plate that separates the two new cells.

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