Q7: For Fig. 5,6, what is the fraction of each of the four types of gametes prod
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Q7: For Fig. 5,6, what is the fraction of each of the four types of gametes produced by the Normal leave, tall plant (MD/md) parent?Q8: For the four genotypes of the gametes produced by the normal leave, tall plant (MD/md) in Fig. 5.6, which two are recombinants?
Q9: For Punnet’s experiment (Fig. 5.2 on p117), if we want to measure the RF between the gene that determines the color of the flower and the gene that determines the shape of the pollen, what mating should we set up? (Please write down the genotypes of the two strains you are going to use).
Q10: A plant of genotype Cd/Cd is crossed to cD/cD and an F1 testcrossed to cd/cd. If the distance between these two genes is 20 map units, the percentage of cd/cd recombinants will be Coupling and Repulsion In crosses for linked genes, the the homologous chromosomes is the outcome of the cross. For tance of two genes in the Austra rina. In this species, one locus de thorax: a purple thorax (p) is reces thorax (p). A second locus deo puparium: a black puparium (b) brown puparium (b"). The loci for ium color are located close together some. Suppose that we test- cross a t at both loci with a fly that is ho 122 CHAPTER 5 dwartIn leaves, tal Because these genes are linked, arrangements on the chro parent. The dominant alleles for -- brown puparium (b") might reside oo of the homologous pair, and the recessd thorax (p) and black puparium (b) other homologous chromosome M D ses with and without orossing nless than 50% on averige d over together resuiltm together resut This arrangement, in which wild-t on one chromosome and mutant alle other chromosome, is referred to as configuration. Alternatively, one chre the alleles for green thorax (p") and and the other chromosome might can thorax (p) and brown puparium (b"y Nonrecombinant Nonrecombinant Recombinant gametes (1000 gametes (5000 gametes (5000 Fertilization eaves, tat leives dw NormaiMottled eives Gwaf eaves, tall NormalMottled This arrangement, in which each one wild-type and one mutant allele,i trans configuration. Whether the al parent are in coupling or repulsion d types will be most common among t When the alleles are in the cou most numerous progeny types are rax and brown puparium and those black puparium (Figure 5.7a). Ho M D M d the heterozygous parent are in re Progeny ous progeny types are those with puparium and those with a p puparium (Figure 5.7b). Notice th ents in Figure 5.7a and 5.7b are th and that the dramatic difference of the progeny in the two crosse configuration-coupling or repuls 53 Nonrecombinant Recombinant progeny progeny Conclusion: With linked genes and some crossing inant progeny predominate. 5.6 Crossing over between linked genes produces nonrecombinant and recombinant offspring. In thisKnowledge of the arrangement testcross, genes are linked and there is some crossing over. mosomes is essential to accurate crosses in which genes are linked
Explanation / Answer
Answer 7.
According to figure 5.6, MD/md and md/md are non-recombinants while Md/mD and mD/Md are recombinants.
The given fraactions are:
MD/md = 55; md/md = 53; Md/md = 8; mD/md = 7
So, parental types = (55 + 53) = 108; recombinants = (8 + 7) = 15; Total = (108 + 15) = 123
Distance between ‘m’ and ‘d’ genes = (number of recombinants / total) * 100
= (15 / 123) * 100 = 12.2mu
Now, if the cross is between normal leaf and tall plant with genotype MmDd.
The cross will be: MmDd X MmDd
Four types of gametes will be produced: MD and md (parental types) and Md and mD will be recombinants.
12.2% recombinants are there. So, fraction of Md = mD = (12.2 / 2) = 6.1%.
Fraction of MD = md = (100 - 12.2) = (87.8 / 2) = 43.9%
**Kindly post rest of the questions separately.
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