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Reciprocal cross between two primrose varieties, A and B, produced the following results: AfemaleBmaleoffspringwith all green (non-variegated) leaves;BfemaleAmaleoffspring with patterned (variegated) leaves. Explain these results

Short Answer

Expert verified

Plastids are the structures responsible for colors in the leaf. The features of a mother or the maternal characters get transferred to the offspring. The B variety female produces variegated offspring.

From this, it can be concluded that B-type females possess a wild type and the mutant allele is the reason for producing variegated leaves.

Step by step solution

01

Description of reciprocal offspring

A crossing between a pair of parents such as A and B is reciprocated as B and A. In this condition, the first parent is kept as male and the second as female.

02

Description of variegated and non-variegated leaves

The leaf having more color is known as variegated leaves. The colors may appear in the border of the leaf; bicolored strips are also found.

The leaf filled with chlorophyll pigment and appears only in green color is known as non-variegated leaves.

03

Reciprocal cross between two primrose and its results

The characteristic of the mother gets inherited by the offspring as per the pattern of inheritance. The plastids determine the color present in the leaf. Chloroplasts are organelles that come under the category of plastids.

The plastid gene present in the female plant gets inherited by the offspring. In this reciprocal crossing, the B is the mutant variety and wild type to produce variegated leaves.In this condition, the pigment gene is mutated.

Hence, the reciprocated cross between the B female and A male results in the formation of variegated leaves.

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Most popular questions from this chapter

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 offspring plants are obtained. How many plants of each of the four phenotypes do you expect?

The\({\chi ^2}\)value means nothing on its own- it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests that the observed data are consistent with the hypothesis, and thus the hypothesis should be rejected, A standard cutoff point used by biologists is a probability of 0.05(5%). If the probability corresponding to the\({\chi ^2}\)value is 0.05or considered statistically significant, the hypothesis (that the genes are unlinked) should be rejected. If the probability is above 0.05, the results are not statistically significant: the observed data are consistent with the hypothesis.

To find the probability, locate your\({\chi ^2}\)value in the\({\chi ^2}\)Distribution table in Appendix F. The 鈥渄egree of freedom鈥 (pdf) of your data set is the number of categories (here,4 phenotypes), minus 1, so df=3.

(a). Determines which values on the df =3 line of the table your calculated\({\chi ^2}\)value lies between.

(b). The column headings for these values show the probability range for your\({\chi ^2}\)number. Based on whether there is non-significant (p\( \le \)0.05) or significant (p>0.05) difference between the observed and expected values, are the data consistent with the hypothesis that the two genes are unlinked and assorting independently, or is there enough evidence to reject this hypothesis?

Banana plants, which are triploid, are seedless and therefore sterile. Propose a possible explanation.

Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

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