*17. In cucumbers, heart-shaped leaves (h) are recessive to normal leaves (HD) a
ID: 203106 • Letter: #
Question
*17. In cucumbers, heart-shaped leaves (h) are recessive to normal leaves (HD) and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and for number of spines are located on the same chromosome, findings from mapping experiments indicate that they are 32.6 mu apart. A cucumber plaut having heart-shaped leaves and mumerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The Fi are crossed with plants that have heart-shaped leaves and numerous spines What phenotypes and phenotypic proportions are expected in the progeny of this cross? 18. In tomatoes, tall (D) is dominant over dwarf (d) and smooth fruit (P) is dominant over pubescent fruit (p), which is covered with fine hairs. A farmer has two tall and smooth tomato plants, which we will call plant A and plant B. The farmer crosses plants A and B with the same dwarf and pubescent plant and obtains the following numbers of progeny Progeny of Plant A Plant B Dd Pp Dd pp dd Pp Pp 122 82 82 4 4 124 a. What are the genotypes of plant A and plant B? b. Are the loci that determine the height of the plant and pubescence linked? If so, what is the rate of recombination between them? c. Explain why different proportions of progeny are produced when plant A and plant B are crossed with the same dwarf pubescent plant.Explanation / Answer
17. In the problem, it has been stated that :
Heart-shaped leaves (hl) is recessive to normal leves (Hl)
Numerous spines (ns) is recessive to fewer spines (Ns)
The genes controlling leaf shape and number of spines are located on the same chromosome, which implies that they are not assorting independently. Hence there is some amount of linkage and some amount of recombination (not complete recombination, as would have happened in the case of independent assortment). Therefore there can be four phenotypes possible in the offspring population, out of which two will be recombinant and two would be parental (non-recombinant).
The genes controlling these two traits are said to be located 32.6 map units apart.
From recombination frequency calculations, we know :
distance between two genes in map units = percentage of recombinants in the offspring population
Therefore here, 32.6 cM (centimorgan) implies that the total percentage of recombinants are 32.6%.
Since there are two kinds of recombinanants present in the population, each recobinant phenotype will contribute
(32.6 / 2) % = 16.3 % of the population.
The parental population would be (100 - 32.6) % = 67.4 % of the population.
Parental types are also two, therefore, eah parental type would be : (67.4 / 2) = 33.7 %
Now, parental phenotypes are the ones which were present in the parental species before recombinantion, while the recombinant phenotypes were formed after recombination.
Here the parental phenotypes (as stated in the question) along with their phenotypic proportions are are :
hlns (heart shaped, numerous spines) : 33.7%
HlNs (normal leaves, few spines) : 33.7 %
The recombinant phenotypes would be :
Hlns (normal leaves, numerous spines) : 16.3 %
hlNs (heart shaped, few spines) : 16.3 %
18. In this problem it has been stated that,
the tall gene (D) is dominant over the dwarf gene (d), and
the gene for smooth fruit (P) is dominant over the gene for pubescent fruit (p)
a. It is also stated that the two parental organisms which are being crossed are both tall and bear smooth fruit. This implies that both the parental organisms have dominant phenotypes for both genetic traits. However, the genotypes of these parental plants can be homozygous dominant for both genes (DDPP), heterozygous dominant for both genes (DdPp), or homozygous for one gene and heterozygous for the other (DDPp and DdPP). All four of these genotypes would yield the same tall-smooth fruit phenotype. But as we can see from the recombination data in the f2 generation, each of the genes have undergone recombination to yield four genotypic categories. If either of the genes had homozygous dominant alleles in the parental organism, then the recessive alleles would not have appeared in the offspring generation. Since recessive alleles for both the traits can be found in the offsprings, therefore the parents in the parental generation must have been heterozygous dominant in nature.
Therefore it can be inferred that the genotypes of both the parental organisms, A and B, are : DdPp.
b. Now that we have obtained the genotype of the parental plants A and B, and it is already being stated in the question that they are being crossed to a recessive parent (ddpp), we can deduce the parental (non-recombinant) genotypes and recombinant genotypes from the offspring data. From the recombination data it can be seen that the genotypes of the parents in the cross (DdPp and ddpp) occur at a different frequency than the recombinant Ddpp and ddPp. If the genes were not linked, then frequencies of each of the genotypes would have been the same. Since the parental genes occur with a frequency different than the recombinants, it can be inferred that the genes are indeed in linkage.
Recombination frequency can be calculated by dividing number of recombinant offsprings by the total number of offsprings and multiplying the result with 100.
Total number of recombinant offsprings in the cross of Plant A are : (6 + 4) = 10
Total number of offsprings = (6 + 4 + 122 + 124) = 256.
Therefore recombination frequency in Plant A is : (10 / 256)* 100 = 3.9 %
In plant B, it can be seen from the recombination data that the phenotypes Ddpp and ddPp are in higher frequency. Therefore these must be the genotypes that got combined during the cross. Hence :
Total number of recombinant offsprings in the cross of Plant B are : (4 + 2) = 6
Total number of offsprings = (82 + 82 + 2 + 4) = 170.
Therefore recombination frequency in Plant B is : (6 / 170)* 100 = 3.5 %
c. Despite having carried out the same test cross (cross with a recessive parent) of the plants A and B (which also bear the same genotype), we can observe that different phenotypic proportions are produced in the offspring generation. This can be attributed to the fact that the coupling configurations were different in plants A and B. As we can see from the question and the explanation above, in plant A, the alleles D and P are coupled (coupling phase cross), implying that D and P recombine with d and p. But in plant B, it can be observed that Dp combines with dP. This was a repulsion phase cross, therefore they produced different progeny proportions than plant A.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.