Bio CH 15 short answer Flashcards
(14 cards)
A man with hemophilia (a recessive, sex-linked condition) has a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? That a son will be a hemophiliac? If the couple has four sons, what is the probability that all four will be born with hemophilia?
There is no chance for a daughter to have hemophilia because she would inherit a normal X from her father. A son has a 1/2 chance of being hemophiliac because he could inherit the affected X from his carrier mother. The probability that all four sons have hemophilia is (1/2)^4 = 1/16. Since hemophilia is X-linked, sons are more affected. That is why the answer is 0; 1/2; 1/16.
Pseudohypertrophic muscular dystrophy is an inherited disorder that causes gradual deterioration of the muscles. It is seen almost exclusively in boys born to apparently normal parents and usually results in death in the early teens. Is this disorder caused by a dominant or a recessive allele? Is its inheritance sex-linked or autosomal? How do you know? Explain why this disorder is almost never seen in girls.
Recessive; sex-linked because it is seen only in boys. Since only boys are affected and the parents are normal, the allele must be recessive. A dominant disorder would show up in at least one parent. It is sex-linked because boys only have one X chromosome, making recessive traits more visible. Girls would need two affected X chromosomes, which is rare.
Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. What is the probability that they will have a color-blind daughter? What is the probability that their first son will be color-blind? (Note the different wording in the two questions.)
1/4 for a daughter; 1/2 for first son. There is a 1/2 chance the child is female and a 1/2 chance she is color-blind, so 1/4 total for a color-blind daughter. For a son, there’s only a 1/2 chance because he inherits one X chromosome. The mother is a carrier and the father is color-blind. This affects the inheritance differently for sons and daughters.
A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162. What is the recombination frequency between these genes for body color and wing size?
The recombination frequency shows how often crossing over happens between two genes. A 17% frequency means the genes are 17 map units apart. Higher frequency would suggest the genes are further apart. This percentage comes from adding up the recombinant offspring.
In another cross, a wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are as follows: wild type, 721; black-purple, 751; gray-purple, 49; black-red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 4, what fruit flies (genotypes and phenotypes) would you mate to determine the sequence of the body-color, wing-size, and eye-color genes on the chromosome?
Recombination frequency is calculated by adding recombinant offspring and dividing by total offspring. Here, the recombinants were 6% of the total. This shows the genes for wing and eye color are close together. A smaller percentage means tighter linkage.
A fruit fly that is true-breeding for gray body, vestigial wings (b* b* vg vg) is mated with one that is true-breeding for black body, normal wings (b b vg* vg*).
A) Draw the chromosomes for each P generation fly, showing the position of each allele.
B) Draw the chromosomes and label the alleles for an F, fly.
C) Suppose an F, female is testerossed. Draw the chromosomes of the resulting offspring in a Punnett square like the one at the bottom of Fig. 15.10 in your textbook.
D) Knowing that the distance between these two genes is 17 map units, predict the phenotypic ratios of these offspring.
This cross shows linked genes with recombination happening. The chromosomes are drawn to reflect which alleles are on which chromosome. The phenotypic ratios show most offspring inherit parental combinations, with fewer recombinants. The 17% recombination frequency matches the 8.5% seen in each recombinant group.
What pattern of inheritance would lead a geneticist to suspect that an inherited disorder of cell metabolism is due to a defective mitochondrial gene?
The disorder would always be inherited from the mother. Mitochondrial DNA is inherited only through the egg. The sperm’s mitochondria are not passed to offspring. Thus, if a disorder is mitochondrial, it must come from the mother. All children of an affected mother could inherit the disease.
Women born with an extra X chromosome (XXX) are healthy and phenotypically indistinguishable from normal XX women. What is a likely explanation for this finding? How could you test this explanation?
Inactivation of two X chromosomes would leave one active X; microscopy would show two Barr bodies. In women with XXX, two of the X chromosomes are inactivated. This leaves only one active X like in normal XX women. Barr bodies are condensed, inactive X chromosomes visible under a microscope. Two Barr bodies would confirm the presence of three X chromosomes.
Determine the sequence of genes along a chromosome based on the following recombination frequencies: A -B, 8 map units; -C, 28 map units; A—D, 25 map units; B C, 20 map units;
B-D, 33 map units.
D — A — B — C. Gene order is determined by recombination frequencies. Genes closer together have smaller recombination percentages. The map shows D is closest to A, A is closer to B, and B is closer to C. The numbers match this sequence.
Assume that genes A and B are linked and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show phenotypes resulting from crossovers? If you did not know that genes A and B were linked, how would you interpret the results of this cross?
50% of offspring would show crossover phenotypes; same results as if genes were unlinked. If two genes are 50 map units apart, they assort independently. This means crossing over happens so frequently it mimics unlinked inheritance. A test cross would show 50% recombinants. Only additional crosses could reveal if they were actually on the same chromosome.
A space probe discovers a planet inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t = dwarf), head appendages (4 = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures are not “intelligent,” Earth scientists are able to do some controlled breeding experiments, using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are: tall-antennae, 46; dwarf-antennae, 7; dwarf-no antennae, 42; tall-no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are: antennae-upturned snout, 47; antennae-downturned snout, 2; no antennae-downturned snout, 48; no antennae-upturned snout, 3. Calculate the recombination frequencies for both experiments.
Recombination frequency shows how often two traits are separated during meiosis. A 12% frequency between T and A means they are further apart than A and S. The 5% between A and S means they are closely linked. Smaller recombination means tighter physical linkage on the chromosome.
Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue-oval plant with a homozygous white-round plant. The resulting F1 progeny are crossed with homozygous white-oval plants, and 1,000 F2 progeny are obtained. How many F2 plants of each of the four phenotypes do you expect?
The majority of the offspring are parental types because the genes are closely linked. Recombinants are rarer due to the 10% recombination frequency. You would expect 90% parental and 10% recombinant distribution among 1000 progeny. This matches the predicted numbers.
You design Drosophila crosses to provide recombination data for gene a, which is located on the chromosome shown in Figure 15.12 in the textbook. Gene a has recombination frequencies of 14% with the vestigial-wing locus and 26% with the brown-eye locus. Where is a located on the chromosome?
Gene a has recombination frequencies closer to vestigial than to brown. 14% is about one-third of the way along a map where 0% is vestigial and 43% is brown. This suggests a is located closer to vestigial than to brown. Gene mapping uses recombination percentages to estimate position.
Bananas plants, which are triploid, are seedless and therefore sterile. Propose a possible explanation.
Triploid organisms have three sets of chromosomes, making proper pairing impossible during meiosis. Without even pairing, gametes cannot form correctly. This leads to sterility since viable seeds cannot develop. Therefore, triploid bananas are seedless.
