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87 Cards in this Set
- Front
- Back
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If an organism has a haploid number of 6 chromosomes how many chromatids are in a cell during prophase II of meiosis?
60 48 36 24 12 |
6 chromosomes=12 chomatids
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2) Given an individual who is heterozygous at 3 loci, how many different gametic genotypes are possible with independent assortment
2 4 8 16 32 |
8 |
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Given incomplete dominance among the pairs of alleles at two of four loci and complete dominance/recessiveness among the pairs of alleles at the other two loci a cross between two tetrahybrids will produce how many different phenotypes (assume independent assortment) among the offspring
18 27 36 49 81 |
36 |
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What phenotypic ratio would you expect from the testcross of an Fl individual that gives a 9:6:1 in theF2?
3:1 1:2:1 15:1 1:1:1:1 none
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1:2:1 |
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The gene specifying blood group determinants can occur in three alleles: A. R and O. A while B is regressive. if Luke Skywalker's mother was blood type A and Luke himself were type B, what blood type(s) - if any - would Darth Vader have to be to unequivocally rule him out as Luke's natural father?
A B O AB A or O B or AB |
A or O mom: I^A I^O Luke: I^B I^O |
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Which of the following has the potential to cause non mendelian phenotypic segregation ratios:
Epistasis co-dominant alleles lethal alleles all of the above two of the above |
all of the above |
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At what stage of meiosis do homologous chromosomes separate
prophase I anaphase I prophase II anaphase II sister chromatids do not separate, except by crossing over |
anaphase I |
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Mice with the genotypes BB and Bb are black, and those with the genotype bb are brown. At another locus the genotypes CC and Cc code for color while the homozygous recessive genotype cc codes for albinism. What phenotypic ratio would be expected from a cross between two mice with the genotype BbCc?
9 black, 3 brown, 4 albino 12 black, 3 brown, 1 albino 9 black 6 brown 1 albino 9 black 0 brown 7 albino not enough infomation |
9 black 3 brown 4 albino |
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The chromosomal theory of inheritance:
a) was proposed by Gleason and Thomas b) correlated Mendelian phenotypic segregation patterns with chromosomal segregation behavior during meiosis c)correlated Mendelian phenotypic segregation patterns with chromosomal segregation behavior during mitosis d) states that the haploid chromosome number correlates directly with the complexity of an organism e) none ofthe above |
correlated Mendelian phenotypic segregation patterns with chromosomal segregation behavior during meiosis |
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Two unlinked loci effect mouse hair color. CC or Cc mice have colored hair. Mice with genotype cc are albino because all pigment production and deposition of pigment in hair is blocked. At the second locus, the B allele (black coat) is dominant to the b allele (brown coat). A mouse with a black coat-is mated with an albino mouse of genotype bbcc. Half of the offspring are albino, one quarter are black, and one quarter are brown. What is the genotype of the black parent?
BBCC BbCc bbCC BbCC BBcc |
BbCc |
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A linear chromosome with the centromere exactly in the middle, so that the two arms are of equal length, is called:
acrocentric metacentric telocentric sub-metacentric none |
metacentric
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A test cross is used to determine if the genotype of a plant with the dominant phenotype is homozygous or heterozygous. If the unknown is homozygous, all of the offspring of the test cross have the ________ phenotype. If the unknown is heterozygous, half of the offspring will have the ____ phenotype.
a)dominant, incompletely dominant b) recessive, dominant c) dominant, epistatic d) codominant, complimentary e) dominant recessive |
dominant, recessive |
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The gametes of a plant of genotype SsYy should have the genotypes:
a) Ss and Yy b) SY and sy c) SY Sy sY and sy d) Ss Yy SY and sy e) SS ss YY and y |
c) SY Sy sY and sy |
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In radishes, red and white are the pure-breeding colors and long and round are the pure-breeding shapes- The hybrids are purple and oval. The cross of a white, oval, and a purple, oval, will produce more:
purple, oval than purple long purple round than white, long red, long than white, long purple, long than purple, round purple, round than white, oval |
purple, oval than purple, long |
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The probability of two independent events occurring together is the sum of the separate probabilities
True False |
False |
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Incompletely dominant alleles produce the traits of both alleles when heterozygous
True False |
False
co-dominant |
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A branch diagram can only be used to predict genotypic segregation ratios resulting from genetic crosses |
False |
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All organisms with identical phenotypes will have identical genotypes
True false |
False |
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Pleiotropy means that a single gene can determine multiple observable phenotypes.
true false |
True |
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What two mechanisms act during meiosis to maximize genetic variability among the gametes produced? |
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The amount of DNA per cell of a particular species is measured in cells found at various stages of meiosis. The following amounts are obtained: 3.7 pg, 7.4 pg, 14.8 pg
Match the amounts of DNA with the corresponding stages of the cell cycle below. G1- Prophase 1 of meiosis- following telophase II of meiosis and cytokinesis Metaphase I of meiosis Metaphase II of meiosis |
G1- 7.4 Prophase 1 of meiosis- 14.8 following telophase II of meiosis-3.7 Metaphase I of meiosis-14.8 Metaphase II of meiosis-7.4
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In dragons there are alleles for fire breathing (B) and for nonfire breathing (N) as well as alleles for red scales (R), green scales (G), short tail (S), and long tail (L) You are interested in breeding these dragons to save the species, which is endangered. You cross pure breeding, red,fire breathing dragons with short tails to pure breeding green, nonfire breathers with long tails. The F2 generation of this cross shows the phenotypes in the following table. The loci are not linked
6Green, fre breathing, short tail 30 green, fire breathing long tail 20 green nonfire breathing short tail 70 green nonfire breathing long tail 23 red fire breathing short tail 65 red fire breathing long tail 75 red nonfire breathing short tail 210 red nonfire breathing long tail
which alleles are dominant and which are recessive for each of the three genes?
You want to testcross a dragon exhibiting the THREE DOMINANT TRAITS and that has a triply heterozygoud genotype. What dragon genotype and phenotype would you use to testcross this dragon? (show alleles). What phenotypic segregation ratio would you expect as the outcome of the cross? |
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Nilsson-Ehle made crosses between two types of oats, one with white-hulled seeds and one black-hulIed. The F1 between them was black-hulled, and the F2 (Fl xFI) contained 560 plants as follows: 418 black,106 gray, and 36 white.
How many gene pairs are interacting to produce color differences in the oat hulls? Propose a complete genetic model (ie, define genes, alleles, allelic relationships and any other relevant information in your model) for these observations? Give generalized genotypes for all of the colors mentioned
b) Use a chi-square analysis to support your hypothesis. (A probability table is at the back of the exam.) Be sure to state the null hypothesis you are testing and your conclusion from the analysis. |
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Dr.Lucy Lepidoptera is interested in the traits of butterfly species X. vulgaris.One strain of -X vulgaris has gray wings conditioned by the recessive blk- allele. Another strain is homozygous for the allele blk+ that is needed to color wings black. The allele min for small body size is incompletely dominant to the allele max for large body size, and the allele rip- is a recessive lethal allele to rip + assuming the three genes exhibit independent arrortment, what is the expected phenotypic ratio of a cross between the X. vulgaris genotypes?
blk+blk- minmax rip+rip X blk-blk- minmax rip+rip |
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Chickens with shortened legs and wings are called creepers. When creepers are mated to normal birds they produce creepers and normals with equal frequency.When creepers are mated to creepers they produce 2 creepers to 1 normal. Crosses between normal birds produce only normal birds. Explain these results. |
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Both red/green color blindness (R:normal,r=colorblind) and Duchenne-type muscular dystrophy (D:normal, d:muscular dystrophy) are X-linked recessive traits that map close to each other. A woman has a father who is re/green color-blind. Her mother's family has a history of Duchenne's muscular dystrophy. This woman is apparently healthy with neither color blindness nor muscular dystrophy. She marries a healthy man and they have four sons and two daughters. Half the sons are healthy but color blind, the other half have normal color vision, but have Ducherure's musculal dystrophy. The daughters are both normal. What is the genotype of the woman?
RD/rd Rd/rD rd/rd RD/RD cannot determine |
Rd/rD |
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With non-disjunction, which parent leads to the sex chromosome aneuploid XYY?
mother father either parent both parent none of the above; non-disjunction only occurs in animal cloning |
father |
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A husband and wife have normal vision, although both of their fathers are red-green ' color-blind, which is inherited as an X-linked recessive condition. What is the probability that their first child will be a female with normal vision?
1 0.5 0.25 0 more than one of the above |
0.5 |
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Recombination frequencies are most often determined from:
F3 progeny testcross progeny F1 progeny F2 progeny none |
testcross progeny |
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Which of the following statements about the products (cells) produced when non-
Three products of the second meiotic division have too many chromosomes, while one is missing a chromosome for a particular set.
Both products of the first meiotic division contain dyads representing one chromosome of each homologous pair.
None of the above statements are true.
two products of the second meiotic division have both the maternal and paternal chromosomes of a set and the other two products have none for that set |
two products of the second meiotic division have both the maternal and paternal chromosomes of a set and the other two products have none for that set |
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How is sex determined in Drosophila?
By the ratio of the number of X chromosomes to the number of autosomes
By the ratio of the number of Y chromosomes to the number of autosomes
By the ratio of the number of Y chromosomes to the number of sets of autosomes
Exclusively by the Y chromosome.
By the ratio of the number of X chromosomes to the number of sets of autosomes |
By the ratio of the number of X chromosomes to the number of sets of autosomes |
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With cytoplasmic inheritance:
the genes inherited in this manner can be found in the nucleus
traits are inherited similar to a Mendelian fashion
the phenotype of the progeny is determined by the genotype of the mother
none of the above
more than one of the above
differences in organellar contributions from the mother (egg) and the father (sperm) is important |
differences in organellar contributions from the mother (egg) and the father (sperm) is important |
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Assume that a cross is made between AaBb and aabb plants and all of the offspring are either AaBb or aabb. These results are consistent with the following circumstance
Linkage with recombination codominance i) Independent assortment eJ Hemizygosity |
complete or absolute linkage |
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Tne sex of sorne birds, some insects, and other organisms is determined by aZW chromosornal arrangement in which the males have like sex chromosome (ZZ) and the females are ZW. Assume that a recessive lethal allele on the Z chromosome causes death of an embffi birds. What sex ratio would result in the offspring if a cross were made between a male heterozygous for the lethal allele and a normal female?
4:1 male to female 3:1 male to female 1:2 male to female 1:1 male to female |
2:1 |
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Suppose you discover a new variant in which hamsters have long tails instead of the usual stubby tails. You notice that this trait seems only to be present in males. to investigate this pattern, you cross a long-tail male with a true-breeding stubby tail female,and find that all of the Fl progeny of both sexes have stubby tails. You then interbleed the Fl and observe that all of the F2 females have stubby tails ,but 1/2 of the F2 males have long tails. You conclude that: |
stubby allele is an X-linked recessive the stubby allele is Y-linked c) the stubby allele is autosomal recessive d) the stubby alleie is an X-linked dominant e) the stubby allele is an autosomal dominant |
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What is the easiest way to determine if a particular trait of an organism is autosomally inherited or is sex-linked?
perform a test cross l perform selfing crosses any of the above simply look at the individuals perform reciprocal crosses |
perform reciprocal crosses |
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In humans, XXY individuals are males with Klinefelter's syndrome- Which of the following events could not give rise to a Klinefelter's male?
Nondisjunction at meiosis I in the mother. b) Nondisjunction at meiosis II in the mother. c) Nondisjunction at meiosis I in the father. nondisjunction at meiosis II in the father all of the above could give rise to a Klinefelter's male. |
nondisjunction at meiosis II in the fathe |
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the same as the number of functionally active X chromosomes in the cell one in every male and none in every female one more than the number of X chromosomes in that cell one less than the number of X chromosomes in that cell the same as the number of X chromosomes in that cell |
One less than the number of X chromosomes in that cell |
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Tortoise-shell cats have a patchwork cololation pattern of orange and black spots. This pattern is an example of:
somatic mutation b) mitotic segregation X-inactivation ) cytoplasmic inheritance e) matenral inheritance |
X inactivation (barr bodies) |
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Two practical benefits of polyploidy that have been exploited by breeders are:
sterility and enhanced flavor in fruits the associated bigger cell size and enhanced reproductive capacity sterility and enhanced nutritional content associated bigger cell size and sterility |
associated bigger cell size and sterility |
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Which of the follow'ing individuals will have the MOST trouble in producing functional gametes during meiosis?
tetraploid with 48 total chromosomes tetraploid with 72 total chromosomes diploid with 46 total chromosomes all of the above have equal difficulty triploid with 30 total chromosomes |
triploid with 30 total chromosomes |
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Individuals with three copies of most autosomes do not survive. Individuals with an extraX chromosome, however, survive with relatively mild (compared to autosomal J abnormalities) consequences. Why?
The X chromosome contains few genes while autosomes contain a large number
The extra X chromosomes are inactivated by conversion into Barr bodies
The X chromosome determines sex only
The X chromosome carries the Sry gene that mitigates developmental abnormalities
None of the above |
The extra X chromosomes are inactivated by conversion into Barr bodies |
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Which of the following represents a euploid state:
diploid haploid tetraploid all of the above none of the above |
all of the above |
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In humans, females are____since they have two morphologically identical sex chromosomes
homogametic heterogametic hermaphroditic pseudohermaphroditic all of the above |
homeogametic |
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In chickens, it is the females that have two different sex chromosomes (Z and W) while the males have two Z chromosomes. A Z linked gene controls the pattern of the feathers with the dominant B allele causing the barred pattern and the B allele causing non-barred feathers. You cross a barred female with a non-barred male. What do you expect for the phenotype of the progeny?
daughters and sons of both types sons of one type, daughters of both types daughters of one type, sons of both types none of the above daughters all one type, sons all the other type. |
daughters all one type, sons all the other type |
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Doubling the chromosomes of a sterile species hybrid with colchicine is a method used to produce a fertile species hybrid
true false |
true |
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non-disjunction is viewed as a major cause of aneuploidy but has not been implicated as a cause for polyploidy.
true false |
false |
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Odd-numbered autopolyploids exhibit a larger size compared to their diploid relatives and can be successfully propagated through sexual reproduction.
true false |
False |
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Assue that a species has a diploid number of 24. The term applied to an individual with 25 chromosomes would be trisomic.
true false? |
True |
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It is safe to say that a maternal effect is caused by the phenotype of the parent producing he egg as opposed to the genotype of the offspring.
true false |
false |
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Genes on the same chromosome that aremore than 50 map units aparl appear to be inherited independently even though they are really linked.
true false |
true |
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Interference is a measure of the degree of suppression of double crossover
true false |
true |
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the coefficient of coincidence is calculated by dividing the observed number of double crossover progeny with the expected number of double crossover progeny resulting from a trihybrid testcross.
true false |
true |
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an repulsed (or trans) linkage arrangement is when a recessive allele for one gene is associated with a dominant allele for a second gene.
true false
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true |
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Map distances based on recombination frequencies become more accurate as the distance between linked genes increases.
true false |
true |
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In the beetles golden body (go) is a recessive X-linked mutation, and brown eyes (bw) is a recessive autosomal mutation. A female homozygous for golden body and red eyes is mated to a male with brown eyes and a black body.
Predict the phenotypes of their F1 offspring
If the F1 progeny are intercrossed, what kinds of progeny will appear in the F2, and in what proportions?
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Species X has a diploid chromosome number of 30, whereas that of species Y is 22. After many attempts a highly infertile F1 is produced from a cross between these two species. From this F1, a fertile allopolyploid is eventually derived.
give the chromosome count of the F1 plant
why is the F1 sterile
Give the chromosome count of the fertile derivative |
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In Tasmanian devils, three recessive mutations are h (hairy nostrils), r (rat tail), and f (short fang). A pule breeding strain of a devil with hairy nostrils and rat tail was crossed with a pure breeding strain of a devil with short fangs. The Fl progeny were testcrossed. From the resulting data below, calculate the map and map distances between linked genes, the coefficient of coincidence, and the interference.
HRF 73 hrf 63 Hrf 96 hrF 110 HrF 2 hRf 2 hrF 306 HRf 348
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A human with Turner's syndrome would represent:
diploid condition euploid condition aneuploid condition monosomic condition more than one of the above |
more than one of the above |
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you observe the progeny of a Drosophila cross to consist of all red-eyed females and white-
red-eyed females and white-eyed males red eyed females and red eyed males white eyed females and red eyed males white eyed females and white eyed males b or c |
white eyed females and red eyed males |
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Why do genetic mapping experiments become less accurate when the distances between genes become large?
Recombination occurs less frequently in long chromosomes
None of the above
Interference is greater when the distance between genes is large
crossover gametes become less common
multiple crossovers are more common.
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multiple crossovers are more common |
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One Drosophila genotype PpMm is crossed with another genotype ppmm. If the genes are absolutely Iinked in trans in the dihybrid parent
all of the progeny will be phenotypically pM b) all of the progeny will be phenotypically pm 10% of the progeny will be phenotypically PM and 90% phenotypically pm 50% of the progeny will be phenotypically PM and 50% phenotypically pm |
50% of the progeny will be phenotypically PM and 50% phenotypically pm |
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Assume that, regarditg apartrcular gene, one scored 60 second division ascospore arrangement s (2:2:2:2 or 2:4:2) and 40 first division arrangements (4:4) in Neurospora. What would be the map distance between the gene and the centromere?
10 20 30 60 insufficient information |
30 |
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In humans, XXY individuals are males with Klinefelter's syndrome. Which of the following events could not give rise to a Klinefelter's male?
Nondisjunction at meiosis II in the mother. b) Nondisjunction at meiosis I in the mother. Nondisjunction at meiosis II in the father Nondisjunction at meiosis I in the father e) all of the above could give rise to a Klinefelter's male. |
Nondisjunction at meiosis II in the father |
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A husband and wife have normal vision, although both of their fathers are red-green color blind, which is inherited as an X-linked recessive condition. What is the probability that their first child will be normal vision and male?
1 0.5 0.25 0 more than one |
0.25 |
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With maternal effect:
the genes inherited in this manner can be found in the mitochondria b) traits are inherited similar to a Mendelian fashion differences in organellar contributions from the mother [egg) and father (spermJ is important the phenotype of the progeny is determined by the genotype of the mother none of the above more than one of the above |
the phenotype of the progeny is determined by the genotype of the mother |
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A plant of genotype C dlc d is crossed to cDlc D and an Fl testcrossed to c dlc d. If the genes are linked and20 map units apart, the percentage of C D/c d offspring will be:
10% 20% 30% 40% 50% |
10% |
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The Drosophila genes for white eyes (w), tan body (r) and vestigial wings (vg) lie atmap positions 2,12, and 42 respectively. The distance, therefore, between w and / is 10, / and vg is 30 and w and vg is 40. Out of 1000 progeny,2l exhibited double crossovers. What is the degree of interference?
0.10 0.20 0.30 0.40 none |
0.30 |
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Duplication of short segments of DNA are often found in the genomes of organisms. These F are thought to be caused most commonly by:
aJ gene conversion b] unequal crossing over cJ transversion dl reciprocal translocation more than one of the aboce |
unequal crossing over |
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In theWZ|ZZ sex determination system, females are the heterogametic sex.
true false |
true |
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If an individual is heterozygous for a reciprocal translocation, we would expect about a:
33% reduction in fertility 25% reduction in fertility 50% reduction in fertility 66% reduction in fertility no reduction in fertility
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50% |
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What is the probability that a man has inherited his Y chromosome from his maternal grandfather?
0.50 0.25 1.00 0.75 0.00 |
0.00 |
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Traits exhibiting Y-linked inheritance are called:
autosomal recessive d) sex-linked eJ epistasis holondric autosomal dominant |
holondric |
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the conversion of a diploid kayotype of 8 acrocentric chromosomes to a diploid karyotype of 4 metacentric chromosomes would be due to:
a) poluploidy b) deletions c) translocations d) centric fusions/robertsonian translocations e)more than one |
centric fusions/robertsonian translocations |
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Duplication of short segments of DNA are often found in the genomes of organisms. These events can be beneficial because:
they increase cell size b) they always induce sterility c) they lead to speciation events d) there is no benefit, gene duplications always have a negative effect on phenotypes e) the creation of gene families through duplication events provide large quantaties of essential gene products to the cell |
the creation of gene families through duplication events provide large quantaties of essential gene products to the cell |
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Two practical benefits of polyploidy that have been exploited by breeders are:
sterility and enhanced flavor in fruits sterility and enhanced nutritional content the associated bigger cell size and sterility the associated bigger cell size and enhanced reproductive capacity e) none ofthe above |
the associated bigger cell size and sterility |
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Triploids:
are usually sterile can be produced by the fusion of a diploid and a haploid gamete. c) have difficulty in meiosis because the non-paired chromosomes have no homologue. d) result in Down syndrome all of the above all except one of the above |
all except one of the above |
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Consider a species with a diploid (2n) number of 20 chromosomes. How many chromosomes would be found in a tetrasomic body cell?
10 20 22 24 40 |
22 |
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A triply heterozygous ABC/abc individual (ABC is not necessarily the gene order) is crossed to an abc/abc individual. The following progeny are recovered in the testcross population
ABC 48 abc 42 Abc 44 aBC 46 ABc 4 abC 6 AbC 5 aBc 5 Total 200
what are the linkage associations for the 3 genes. Draw the genetic map for all linked genes.
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Two mutant genes in Drosophila melanogaster produce the recessive traits crossveinless wings (cv) and singed bristles (sn). Given the following genetic map, if F1 dihybrid flies (produced from a cross of pure-breeding males with singed bristles and pure- expressed both recessive phenotypes, what outcome would you predict in terms of phenotypic classes seen and numbers of progeny in each class (assume a population size of 1000)?
cv________10 mu______sn |
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Consider three recessive traits [pb, ni and ag) in Drosophila specified by the X-
a) A female fly expressing all three traits is mated to a wild-type male. What fraction of the female progeny will have all three recessive traits? What fraction of the male progeny will have all three recessive traits? |
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The genes are linked in the following order: Pb-Ni-Ag. A female fly (produced from a different cross than described in a)l heterozygous for all three loci is mated with a male that is homozygous recessive for all three loci. Eight phynonotypic classes are found among the progeny. The rarest classes are pb-nl-Ag and Pb-M-af. The female fly was produced by crossing flies from two different pure-breeding strains. What were the phenotypes of these two parental strains? |
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A man with an X-linked dominant condition "brown tooth enamel" marries a normal produce a son with brown tooth enamel. Provide a genetic-explanation (be complete) |
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A man rnakes 0 units of an enzyme due to homozygous recessive genes on chromosome 21. His wife makes 100 units of enzyme due to aheterczygous pair of genes on cln'omosome 21. Their son has 3 copies of chromosome 21 and makes 200 units of enzyme. Provide a genetic explanation (be complete). |
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Yellow body [cJ is a recessive mutation located near the tip of the X chromosome in Drosophila: A wild-type [c+) male was irradiated with X rays to induce chromosome breaks and rearrangements, and then crossed to a yellow [ccJ female. Among the progeny, a single rare wild-t/pe [c+J son was recovered. This male was mated with yellow [ccJ females, and the numbers and phenotypes of the offspring were as follows:
256 yellow females 0 yellow males 0 wild-type females 231 wild-type males
The yellow females had a normal karyotype, and the wild-type [c+J males were found to breed in the same manner as their fathers.
a) Explain what type of chromosome abnormality could account for these results. Be precise in describing exactly what happened to the genome of the rare c+ son recovered in the first cross.
Use a Punnett square to diagram the results of the cross between the rare wild-type male and the yellow females
The normal sequence of nine genes on a certain Drosophila chromosome is: 12345-6789 -Some fruit flies were found to have anaberuant chromosome as shown below: 12435-6879
Name the specific chromosome rearrangement represented in this aberrant chromosome and then draw a diagram to show how the aberrant chromosome would synapse with the normal chromosome during prophase I of meiosis |
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DNA sample B melts or denatures into two separate helices at a lower temperature than DNA sample A. Why is this so?
sample A has greater A-T bp than sample B sample A has greater G-C bp than sample B; sample A is flom a prokaryote and sample B is flom a eukaryote sample A is flom abacteria and sample B is fi'om a virus e) none ofthe above |
sample A has greater G-C bp than sample B; |
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Which of the following statements correctly describes telomerase?
Telomerase carries its own DNA template. Inactivation of telomerase contributes to the extended life span of cancer cells. Telomerase extends the 3'-ends of the template strands of DNA in linear chromosomes- J1 f"to-"rase unr.vinds the DNA in the initiating stage of DNA replication' |
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