Chapter 5 Practice Problems Flashcards
a Drosophila male from a true breeding stock with scabrous eyes was mated with a female from a true breeding stock with javelin bristles. Both scabrous eyes and javelin bristles are autosomal recessive mutant traits. The F1 progeny all had normal eyes and bristles. F1 females from this cross were mated with males with both scabrous eyes and javelin bristles. Write all the possible phenotypic classes of the progeny that could be produced from the cross of the F1 females with the scabrous javelin males and indicate for each class whether it is a recombinant or parent.
b. the cross in part a yielded the following progeny: 77 scabrous eyes and normal bristles, 76 wild type; 74 normal eyes and javelin bristles; 73 scabrous eyes and javelin bristles. are the genes governing these traits likely to be linked or do they instead assort randomly? why?
c. suppose you mated the F1 females from the cross in part a to wild type males. why would this cross fail to inform you whether the two genes are linked?
d. suppose you mated females from the true breeding stock with javelin bristles to males with scabrous eyes and javelin bristles. why would this cross fail to inform you whether the two genes are linked?
a. This F1 female will make four different types of gametes: two Parental (P) types and two Recombinant (R) types – 1/4 sc+ j (P) : 1/4 sc j+ (P): 1/4 sc+ j+ (R): 1/4 sc j (R). The two different Recombinant types will be present at equal frequencies, as will the two different Parental types. If the two genes are unlinked, the four progeny types will be present at equal frequencies. If the two genes are linked, the Parental gamete types will outnumber the Recombinants. The proportion of Recombinants (recombination frequency = RF) will thus be less than 50%; the precise number will depend on the physical distance between the genes on the chromosome.
b. In these results the Parentals (77 scabrous + 74 javelin) are present in equal frequency to the Recombinants (76 wild type + 73 scabrous javelin). Thus, the genes assort independently – they are unlinked
c. all the progeny will be phenotypically wild type. Thus, you could not determine the frequency of parental and recombinant gametes from the F1 female.
d. The javelin female will make only sc+ j gametes regardless of whether the gametes are parental or recombinant types. Because her recombinant gametes will be the same genotypes as her parental gametes, it impossible to detect from the progeny of this cross whether the genes are linked
The punnett square shows how Mendel’s dihybrid cross results would have been altered had the two genes (A and B) been linked and had the P generation cross been AB/AB x ab/ab.
a. what would be the frequency of each F2 phenotypic class if 80% of the gametes produced by the F1s were parental?
b. answer part a assuming the original P generation cross was Ab/Ab x aB/aB
a. Among the F2s, 66% will be A– B–, 9% will be aa B–, 9% will be A– bb, and 16% will be aa bb.
b. Among the F2s, 51% will be A– B–, 24% will be aa B–, 24% will be A– bb, and 1% will be aa bb.
In mice, the dominant all Gs of the x linked gene Greasy produces shiny fur, while ether recessive wild type Gs+allele determines normal fur. the dominant allele Bhd of the x like broached gene causes skeletal abnormalities including broad heads and snouts, while the recessive wild type Bhd+ allele yields normal skeletons. Female mice heterozygous for the two alleles of both genes were mated with wild type males. Among 100 made progeny of this cross, 49 had shiny fur, 48 had skeletal abnormalities, 2 had shiny fur and skeletal abnormalities, and 1 was wild type
a. diagram the cross described and calculate the distance between the two genes
b. what would have been the results if you had counted 100 female progeny of the cross?
a. Gs Bhd+ / Gs+ Bhd ♀ × Gs+ Bhd+ / Y ♂ → 49 Gs Bhd+ ♂ : 48 Gs+ Bhd ♂ : 2 Gs Bhd ♂ : 1 Gs+ Bhd+ ♂.
The two genes are linked and the distance between these two genes is:
RF = 2 + 1 / 100 = 0.03 = 3 mu.
b. The daughters in this cross must inherit the Gs+ Bhd+ X chromosome from their fathers. This chromosome carries the recessive alleles of both genes, so this is a true testcross. Thus, the genotypes, phenotypes and frequencies of the female progeny would be the same as their brothers.
In Drosophila males from a true breeding stock with a raspberry colored every were mated to females from a true breeding stock with sable colored bodies. In the F1 generation all the females had wild type eye and body color, while all the males had wild type eye color but sable colored bodies. when F1 males and females were mated, the F2 generation was composed of 216 females with a wild type eyes and bodies, 223 females with wild type eyes and sable bodies, 191 males with wild type and sable bodies, 188 males with raspberry eyes and wild type bodies, 23 males with wild type eyes and bodies, 27 males with raspberry eyes and sable bodies. Explain these results by diagramming the crosses and calculating any relevant map distances.
RF between sable and raspberry = 23 (wild-type males) + 27 (raspberry sable males) / 429 (total males) = 0.117 x 100 = 11.7 cM.
If the a and b loci are 20 m.u. apart in humans and an AB/ab woman mates with an ab/ab man, what is the probability that their first child will be Ab/ab.
The probability that a child receives the A b gamete from the female (and is therefore A b / a b) = 10%.
CCDD and ccdd individuals were crossed to each other and the F1 generation was backcrossed to the ccdd parent. 997 CcDd, 999 cc dd, 1 Ccdd and 3 cc Dd offspring resulted.
a. how far apart are the c and d loci?
b. what progeny and in what frequencies would you expect to result from test crossing the F1 generation from a CC dd x x ccDD cross to cc dd?
c. in a typical meiosis, how many crossovers occur between genes C and D?
d. assume that the C and D loci are on the same chromosome but the offspring from the test cross described in part b were 498 CcDd, 502 ccdd, 504 Ccdd, and 496 ccDd. How would your answerer to part c change?
a. The number of recombinants divided by the total number of offspring × 100 gives the map distance: (1 + 3)/(997 + 999 + 1 + 3) = 4/2000 = 0.002 × 100 = 0.2% RF or 0.2 map units (mu) or 0.2 cM
b. 49.9% Cc dd, 49.9% cc Dd, 0.1% Cc Dd, 0.1% cc dd.
c. in a typical meiosis, no crossovers would occur between genes C and D.
d. in a typical meiosis, at least one crossover occurs between genes C and D.
In mice, the autosomal locus coding for the Beta globin of hemoglobin is 1 m.u. from the albino locus. Assume for the moment that the same is true in humans. The disease sickle cell anemia is the result of homozygosity for a particular mutation in the beta globin gene.
a. a son is born to an albino man and a woman with sickle cell amen. what kinds of gametes will the son form, and in what proportions?
b. a daughter is born to a normal man and a woman who has both albinism and sickle cell anemia. what kinds of games will the daughter form, and in what proportions?
c. if the son in part a grows up and marries the daughter in part b, what is the probability that a child of theirs will be an albino with sickle cell anemia?
a. The son’s gametes will consist of: 49.5% a HbßA, 49.5% A HbßS, 0.5% a HbßS and 0.5% A HbßA.
b. The daughter’s gametes will be: 49.5% a HbßS, 49.5% A HbßA, 0.5% a HbßA and 0.5% A HbßS.
c. The probability of an a HbßS / a HbßS child (sickle cell and anemic) = 0.005 (probability of a HbßS from son) × 0.495 (probability of a HbßS from daughter) = 0.0025.
Cinnabar eyes (cn) and reduced bristles (rd) are autosomal recessive characters in Drosophila. A homozygous wild type female was crossed to a reduced, cinnabar male, and the F1 males were then crossed to the F1 females to obtain the F2. Of the 400 F2 offspring obtained, 292 were wild type, 9 were cinnabar, 7 were reduced, and 92 were reduced and cinnabar. Explain these results and estimate the distance between the cn and rd loci.
Recombination occurs in Drosophila females but not in males. Thus, males can produce only the parental cn+ rd+ or cn rd gametes. The females produce both the parental gametes and the recombinant gametes cn+ rd and cn rd+. The genes are separated by 8 cM.
In Drosophila the autosomal recessive dp allele of the dumpy gene produces short curved wings while the autosomal recessive allele bw of the brown gene causes brown eyes. In a test cross using females heterozygous for both of these genes, the following results were obtained:
wild type wings & eyes: 178
wild type wings, brown eyes:185
dumpy wings, wild type eyes: 172
dumpy wings, brown eyes: 181
In a test cross using males heterozygous for both of these genes a different set of results was obtained:
wild type wings, wild type eyes: 247
dumpy wings, brown eyes: 242
a. what can you conclude from the first test cross?
b.what can you conclude from the second test cross?
c. how can you reconcile the data shown in parts a and b? can you exploit the difference between the two sets of data to devise a general test for a synteny in Drosophila?
d. the genetic distance between dumpy and brown is 91.5 m.u. How could this value be measured?
a. the two genes are assorting independently
b. the two genes are on the same chromosome. the genotype of the heterozygous male must have been dp+ bw+/dp bw
c. If the two genes are far enough apart on the same chromosome they will assort independently in the first cross in which the doubly heterozygous parents are females.Independent assortment does not occur in the second cross in which the doubly heterozygous parents are males because there is no recombination in Drosophila males.A cross between a Drosophila male heterozygous for the two genes of interest and a recessive female will clearly distinguish between genes on separate chromosomes and syntenic genes.
d. Large genetic distances can be measured accurately only by summing up the values obtained for smaller distances separating other genes in between those at the ends.
From a series of two point crosses, the following map distances were obtained for the syntenic genes A, B, C and D in peas:
B C 23 mu
A C 15 mu
C D 14 mu
A B 12 mu
B D 11 mu
A D 1 mu
Chi square analysis cannot reject the null hypothesis of no linkage for gene E with any of the other four genes.
a. draw a cross scheme that would allow you to determine the B C map distance
b. Diagram the best genetic map that can be assembled from this data set.
c. explain any inconsistencies or unknown features in your map
d. what additional experiments would allow you to resolve these inconsistencies of ambiguities?
a. Bb Cc × bb cc
b.
c. the relative order of A and D is not clear. Another uncertainty is the location of gene E
d. do a three-point cross with either B A D or A D C and order the genes. The location of the E gene can be determined by finding new genes to which it is linked.
