Genetics book questions - exam 2
Hemophilia is caused by an X-linked recessive allele. In a particular population, the frequency of males with hemophilia is 1/4000. What is the expected frequency of females with hemophilia?
(0.00025)2 = 6.25 x 10-8.
The relationship between the melting Tm and GC content can be expressed, in its much simplified form, by the formula Tm = 69 + 0.41 (% GC). (a) Calculate the melting temperature of E. coli DNA that has about 50% GC. (b) Estimate the % GC of DNA from a human kidney cell where Tm = 85C.
(a) 89.50 C (b) about 39%
Other chromosomes have sequences as follows: (a) 1 2 5 6 7 8; (b) 1 2 3 4 4 5 6 7 8; (c) 1 2 3 4 5 8 7 6. What kind of chromosome change is present in each? Illustrate how these chromosomes would pair with a chromosome whose sequence is 1 2 3 4 5 6 7 8.
(a) Deletion: (b) Duplication: (c) A terminal inversion:
(a) How did the transformation experiments of Griffith differ from those of Avery and his associates? (b) What was the significant contribution of each? (c) Why was Griffith's work not evidence for DNA as the genetic material, whereas the experiments of Avery and coworkers provided direct proof that DNA carried the genetic information?
(a) Griffith's in vivo experiments demonstrated the occurrence of transformation in pneumococcus. They provided no indication as to the molecular basis of the transformation phenomenon. Avery and colleagues carried out in vitro experiments, employing biochemical analyses to demonstrate that transformation was mediated by DNA. (b) Griffith showed that a transforming substance existed; Avery et al. defined it as DNA. (c) Griffith's experiments did not include any attempt to characterize the substance responsible for transformation. Avery et al. isolated DNA in "pure" form and demonstrated that it could mediate transformation.
In Drosophila, the genes sr (stripe thorax) and e (ebony body) are located at 62 and 70 cM, respectively, from the left end of chromosome 3. A striped female homozygous for e+ was mated with an ebony male homozygous for sr+. All the offspring were phenotypically wild-type (gray body and unstriped). (a) What kind of gametes will be produced by the F1 females, and in what proportions? (b) What kind of gametes will be produced by the F1 males, and in what proportions? (c) If the F1 females are mated with striped, ebony males, what offspring are expected, and in what proportions? (d) If the F1 males and females are intercrossed, what offspring would you expect from this intercross, and in what proportions?
(a) The F1 females, which are sr e+/sr+ e, produce four types of gametes: 46% sr e+, 46% sr+ e, 4% sr e, 4% sr+ e+. (b) The F1 males, which have the same genotype as the F1 females, produce two types of gametes: 50% sr e+, 50% sr+ e; remember, there is no crossing over in Drosophila males. (c) 46% striped, gray; 46% unstriped, ebony; 4% striped, ebony; 4% unstriped, gray. (d) The offspring from the intercross can be obtained from the following table. Summary of phenotypes striped, gray 0.25 unstriped, gray 0.50 striped, ebony 0 unstriped, ebony 0.252pq = 2(0.2)(0.8) = 0.32
The nucleic acids from various viruses were extracted and examined to determine their base composition. Given the following results, what can you hypothesize about the physical nature of the nucleic acids from these viruses? (a) 35% A, 35% T, 15% G, and 15% C. (b) 35% A, 15% T, 25% G, and 25% C. (c) 35% A, 30% U, 30% G, and 5% C.
(a) double-stranded DNA; (b) single-stranded DNA; (c) single-stranded RNA.
(a) What was the objective of the experiment carried out by Hershey and Chase? (b) How was the objective accomplished? (c) What is the significance of this experiment?
(a)The objective was to determine whether the genetic material was DNA or protein. (b) By labeling phosphorus, a constituent of DNA, and sulfur, a constituent of protein, in a virus, it was possible to demonstrate that only the labeled phosphorus was introduced into the host cell during the viral reproductive cycle. The DNA was enough to produce new phages. (c) Therefore DNA, not protein, is the genetic material.
If two loci are 10 cM apart, what proportion of the cells in prophase of the first meiotic division will contain a single crossover in the region between them?
20%.
The frequency of an allele in a large randomly mating population is 0.2. What is the frequency of heterozygous carriers?
2pq = 2(0.2)(0.8) = 0.32
In humans, a cytologically abnormal chromosome 22, called the "Philadelphia" chromosome because of the city in which it was discovered, is associated with chronic leukemia. This chromosome is missing part of its long arm. How would you denote the karyotype of an individual who had 46 chromosomes in his somatic cells, including one normal 22 and one Philadelphia chromosome?
46, XX, del(22)(q) or 46, XY, del(22)(q), depending on the sex chromosome constitution.
A yellow-bodied Drosophila female with attached-X chromosomes was crossed to a white-eyed male. Both of the parental phenotypes are caused by X-linked recessive mutations. Predict the phenotypes of the progeny.
All the daughters will be yellow-bodied and all the sons will be white-eyed.
Distinguish between allopatric and sympatric modes of speciation.
Allopatric speciation occurs when populations diverge genetically while they are geographically separated. Sympatric speciation occurs when populations diverge genetically while they inhabit the same territory.
What was some of the evidence that led Charles Darwin to argue that species change over time?
Among other things, Darwin observed species on islands that were different from each other and from continental species, but were still similar enough to indicate that they were related. He also observed variation within species, especially within domesticated breeds, and saw how the characteristics of an organism could be changed by selective breeding. His observations of fossilized organisms indicated that some species have become extinct.
What are the differences between DNA and RNA?
DNA has one atom less of oxygen than RNA in the sugar part of the molecule; the sugar in DNA is 2-deoxyribose, whereas the sugar in RNA is ribose. In DNA, thymine replaces the uracil that is present in RNA. (In certain bacteriophages, DNA-containing uracil is present.) DNA is most frequently double-stranded, but bacteriophages such as φX174 contain single- stranded DNA. RNA is most frequently single-stranded. Some viruses, such as the Reoviruses, however, contain double-stranded RNA chromosomes.
Darwin stressed that species evolve by natural selection. What was the main gap in his theory?
Darwin did not understand the mechanism of inheritance; he did not know of Mendel's principles.
A gene has three alleles, A1, A2, and A3, with frequencies 0.6, 0.3, and 0.1, respectively. If mating is random, predict the combined frequency of all the heterozygotes in the population.
Frequency of heterozygotes combined = 2[(0.6)(0.3) + (0.3)(0.1) + ((0.6)(0.1)] = 0.54
Human beings carrying the dominant allele T can taste the substance phenylthiocarbamide (PTC). In a population in which the frequency of this allele is 0.4, what is the probability that a particular taster is homozygous?
Frequency of tasters (genotypes TT and Tt): (0.4)2 + 2(0.4)(0.6) = 0.64. Frequency of TT tasters among all tasters: (0.4)2/(0.64) = 0.25.
Mendel did not know of the existence of chromosomes. Had he known, what change might he have made in his Principle of Independent Assortment?
If Mendel had known of the existence of chromosomes, he would have realized that the number of factors determining traits exceeds the number of chromosomes, and he would have concluded that some factors must be linked on the same chromosome. Thus, Mendel would have revised the Principle of Independent Assortment to say that factors on different chromosomes (or far apart on the same chromosome) are inherited independently.
Experimental evidence indicates that most highly repetitive DNA sequences in the chromosomes of eukaryotes do not produce any RNA or protein products. What does this indicate about the function of highly repetitive DNA?
It indicates that most highly repetitive DNA sequences do not contain structural genes specifying RNA and polypeptide gene products.
RNA was extracted from TMV (tobacco mosaic virus) particles and found to contain 20 percent cytosine (20 percent of the bases were cytosine). With this information, is it possible to predict what percentage of the bases in TMV are adenine? If so, what percentage? If not, why not?
No. TMV RNA is single-stranded. Thus the base-pair stoichiometry of DNA does not apply.
If a is linked to b, and b to c, and c to d, does it follow that a recombination experiment would detect linkage between a and d? Explain.
No. The genes a and d could be very far apart on the same chromosome—so far apart that they recombine freely, that is, 50 percent of the time.
A woman with X-linked color blindness and Turner syndrome had a color-blind father and a normal mother. In which of her parents did nondisjunction of the sex chromosomes occur?
Nondisjunction must have occurred in the mother. The color blind woman with Turner syndrome was produced by the union of an X-bearing sperm, which carried the mutant allele for color blindness, and a nullo-X egg.
How could it be demonstrated that the mixing of heat-killed Type III pneumococcus with live Type II resulted in a transfer of genetic material from Type III to Type II rather than a restoration of viability to Type III by Type II?
Purified DNA from Type III cells was shown to be sufficient to transform Type II cells. This occurred in the absence of any dead Type III cells.
In a Drosophila salivary chromosome, the bands have a sequence of 1 2 3 4 5 6 7 8. The homologue with which this chromosome is synapsed has a sequence of 1 2 3 6 5 4 7 8. What kind of chromosome change has occurred? Draw the synapsed chromosomes.
The animal is heterozygous for an inversion
From a cross between individuals with the genotypes Cc Dd Ee × cc dd ee, 1000 offspring were produced. The class that was C- D- ee included 351 individuals. Are the genes c, d, and e on the same or different chromosomes? Explain.
The class represented by 351 offspring indicates that at least two of the three genes are linked.
In humans, Hunter syndrome is known to be an X-linked trait with complete penetrance. In family A, two phenotypically normal parents have produced a normal son, a daughter with Hunter and Turner syndromes, and a son with Hunter syndrome. In family B, two phenotypically normal parents have produced two phenotypically normal daughters and a son with Hunter and Klinefelter syndromes. In family C, two phenotypically normal parents have produced a phenotypically normal daughter, a daughter with Hunter syndrome, and a son with Hunter syndrome. For each family, explain the origin of the child indicated in italics.
The daughter with Turner and Hunter syndrome in family A must have received her single X chromosome from her mother, who is heterozygous for the mutation causing Hunter syndrome. The daughter did not receive a sex chromosome from her father because sex chromosome nondisjunction must have occurred during meiosis in his germline. The son with Klinefelter syndrome in family B is karyotypically XXY, and both of his X chromosomes carry the mutant allele for Hunter syndrome. This individual must have received two mutant X chromosomes from his heterozygous mother due to X chromosome nondisjunction during the second meiotic division in her germline. The daughter with Hunter syndrome in family C is karyotypically XX, and both of her X chromosomes carry the mutant allele for Hunter syndrome. This individual received the two mutant X chromosomes from her heterozygous mother through nondisjunction during the second meiotic division in the mother's germline. Furthermore, because the daughter did not receive a sex chromosome from her father, sex chromosome nondisjunction must have occurred during meiosis in his germline too.
The temperature at which one-half of a double-stranded DNA molecule has been denatured is called the melting temperature, Tm. Why does Tm depend directly on the GC content of the DNA?
The value of Tm increases with the GC content because GC base pairs, connected by three hydrogen bonds, are stronger than AT base pairs connected by two hydrogen bonds.
In a survey of moths collected from a natural population, a researcher found 51 dark specimens and 49 light specimens. The dark moths carry a dominant allele, and the light moths are homozygous for a recessive allele. If the population is in Hardy- Weinberg equilibrium, what is the estimated frequency of the recessive allele in the population? How many of the dark moths in the sample are likely to be homozygous for the dominant allele?
Under the assumption that the population is in Hardy-Weinberg equilibrium, the frequency of the allele for light coloration is the square root of the frequency of recessive homozygotes. Thus, q = √0.49 = 0.7, and the frequency of the allele for dark color is 1 - q=p=0.3. From p2=0.09, we estimate that 0.09×100=9 of the dark moths in the sample are homozygous for the dominant allele.
In the human karyotype, the X chromosome is approximately the same size as seven of the autosomes (the so-called C group of chromosomes). What procedure could be used to distinguish the X chromosome from the other members of this group?
Use one of the banding techniques.
Identify the sexual phenotypes of the following genotypes in human beings: XX, XY, XO, XXX, XXY, XYY
XX is female, XY is male, XO is female (but sterile), XXX is female, XXY is male (but sterile), XYY is male.
DNA was extracted from cells of Staphylococcus afermentans and analyzed for base composition. It was found that 37 percent of the bases are cytosine. With this information, is it possible to predict what percentage of the bases are adenine? If so, what percentage? If not, why not?
Yes. Because DNA in bacteria is double-stranded, the 1:1 base-pair stoichiometry applies. Therefore, if 37% of the bases are cytosine, then 37% are guanine. This means that the remaining 26% of the bases are adenine and thymine. Thus, 26%/2 = 13% of the bases are adenine.
A phenotypically wild-type female fruit fly that was hetero-zygous for genes controlling body color and wing length was crossed to a homozygous mutant male with black body (allele b) and vestigial wings (allele vg). The cross produced the following progeny: gray body, normal wings 126; gray body, vestigial wings 24; black body, normal wings 26; black body, vestigial wings 124. Do these data indicate linkage between the genes for body color and wing length? What is the frequency of recombination? Diagram the cross, showing the arrangement of the genetic markers on the chromosomes.
Yes. Recombination frequency = (24 + 26)/(126 + 24 + 26 + 124) = 0.167. Cross:
A man has attached chromosomes 21. If his wife is cytologically normal, what is the chance their first child will have Down syndrome?
Zygotes produced by this couple will be either trisomic or monosomic for chromosome 21. Thus, 100% of their viable children will develop Down syndrome.
The incidence of recessive albinism is 0.0004 in a human population. If mating for this trait is random in the population, what is the frequency of the recessive allele?
q2 = 0.0004; q = 0.02.
