Cross C: Homozygote wild type flies (+/+) were crossed with white eye flies. for the F1 and F2 generations. Wild type White eye White eye females Wild type males females males 50 50 F1 phenotypes (observed) F2 phenotypes (observed) F2 - expected 60 60 60 60 phenotypes What does the F1 tell you about the trait? Is the gene on an autosome or sex chromosome? What was the genotypes of each parent, male and female? What are the genotypes of the F1 flies?

Cross C: Homozygote wild type flies (+/+) were crossed with white eye flies. for the F1 and F2 generations. Wild type White eye White eye females Wild type males females males 50 50 F1 phenotypes (observed) F2 phenotypes (observed) F2 - expected 60 60 60 60 phenotypes What does the F1 tell you about the trait? Is the gene on an autosome or sex chromosome? What was the genotypes of each parent, male and female? What are the genotypes of the F1 flies?

Name: Cross C: Homozygote wild type flies (+/+) were crossed with white eye
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Description: Cross C: Homozygote wild type flies (+/+) were crossed with white eye flies. The table shows the resultsfor the F1 and F2 generations.White eye femalesWild typefemalesWhite eyemalesWild type males50F1 phenotypes(observed)F2 phenotypes(observed)F2 -

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter14: Chromosomes And Human Inheritance

Section: Chapter Questions

Problem 3GP: Human females have two X chromosomes (XX); males have one X and one Y chromosome (XY). a. With...

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You are given a Drosophila female that looks wild-type but is heterozygous for mutations intan body (t), miniature wings (m), and white eyes (w). You test cross this female with a tanbodied, miniature winged, and white-eyed homozygous mutant male, and you obtain thefollowing 1400 progeny: Phenotype : number+ + + : 608t m w : 516+ m w : 2t + + : 6+ m + : 39t + w : 46+ + w : 81t m + : 102 Calculate the distance between each pair t-m, m-w, and t-w only using the number ofrecombinants between them (i.e. ignoring the gene in the middle). Draw a linear map with thedistances between genes.
The maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). How is it possible for a male homozygous for the mutationto exist?
In 1952, an article in the British Medical Journalreported interesting differences in the behavior ofblood plasma obtained from several people who suffered from X-linked recessive hemophilia. Whenmixed together, the cell-free blood plasma from certain combinations of individuals could form clots inthe test tube. For example, the following table showswhether clots could form (+) or not (−) in variouscombinations of plasma from four people withhemophilia:1 and 1 − 2 and 3 +1 and 2 − 2 and 4 +1 and 3 + 3 and 3 −1 and 4 + 3 and 4 −2 and 2 − 4 and 4 −What do these data tell you about the inheritance ofhemophilia in these individuals? Do these data allowyou to exclude any models for the biochemical pathway governing blood clotting?
The XG locus on the human X chromosome has twoalleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells,while the recessive XG allele does not allow antigento appear. The XG locus is 10 m.u. from the STSlocus. The STS+ allele produces normal activity ofthe enzyme steroid sulfatase, while the recessive STSallele results in the lack of steroid sulfatase activityand the disease ichthyosis (scaly skin). A man withichthyosis and no Xg antigen has a normal daughterwith Xg antigen. This daughter is expecting a child.a. If the child is a son, what is the probability he willlack Xg antigen and have ichthyosis?b. What is the probability that a son would have boththe antigen and ichthyosis?c. If the child is a son with ichthyosis, what is theprobability he will have Xg antigen?
In Drosophila, the gene loci for curved wings and purple eyes are 20 centimorgans (map units) apart from each other on an autosomal chromosome. Wild-type flies were mated with double mutant flies with cv cv p p genotypes and curved wing/purple eye phenotypes. AlIl of the F1 flies had wild type phenotypes. The F, flies were crossed with double mutant flies with cv cv p'p genotypes and curved wing/purple eye phenotypes. What are the predicted percentages of flies in the F2 generation with wild type wings and eyes, curved wings and wild type eyes, wild type wings and purple eyes, and curved wings and purple eyes?
In Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) What progeny phenotypes are missing? Why?
a. In Drosophila, crosses between F1 heterozygotes ofthe form A b / a B always yield the same ratio ofphenotypes in the F2 progeny regardless of the distance between the two genes (assuming completedominance for both autosomal genes). What is thisratio? Would this also be the case if the F1 heterozygotes were A B / a b? (Hint: Remember that inDrosophila, recombination does not take placeduring spermatogenesis.)b. If you intercrossed F1 heterozygotes of the formA b / a B in mice, the phenotypic ratio among the F2progeny would vary with the map distance betweenthe two genes. Is there a simple way to estimate themap distance based on the frequencies of the F2phenotypes, assuming rates of recombination areequal in males and females? Could you estimatemap distances in the same way if the mouse F1heterozygotes were A B / a b?
Monohybrid Cross: 1) In dogs, wire hair (S) is dominant to smooth (s). a. In a cross between a homozygous wire-haired dog with a smooth haired dog what are the genotypes and phenotypes of the F1 generation? b. If a brother and sister from the F1 generation are crossed what is the expected ratio of wire-haired to smooth-haired dogs produced from this cross?
n corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines sing the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspring
In Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) What is the genotypic arrangement of the alleles ofthese genes on the X chromosome of the female?
In Drosophila, a cross was made between females expressing thethree X-linked recessive traits, scute bristles (sc), sable body (s),and vermilion eyes (v), and wild-type males. All females were wildtype in the F1, while all males expressed all three mutant traits.The cross was carried to the F2 generation and 1000 offspringwere counted, with the results shown in the following table. Nodetermination of sex was made in the F2 data. Question:Calculate the coefficient of coincidence; does this represent positive or negative interference? Phenotype Offspringsc s v 314+ + + 280+ s v 150sc + + 156sc + v 46+ s + 30sc s + 10+ + v 14
In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman? HR/hr Hr/hr hr/hR Hr/hR HR/Hr