Biology: The Dynamic Science
Biology: The Dynamic Science
4th Edition
ISBN: 9781337246422
Author: Peter J. Russell; Paul E. Hertz; Beverly McMillan
Publisher: Cengage Learning US
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Chapter 29, Problem 1TYK
Summary Introduction

Introduction:

Reproductive innovation is a phenomenon that involves the introduction of new devices or methods in the process of reproduction during the process of evolution. These innovations are also called as the adaptations, which were acquired by the organisms according to the changes in the surrounding environment.

Expert Solution & Answer
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Answer to Problem 1TYK

Correct answer:

The ovary.

Explanation of Solution

Justification/ Explanation for the correct answer:

Option (d) is, ‘the ovary’. The ovary only gives protection to the seed. It does not take part in the formation of a seed, so, it was not a part of the reproductive innovation related to seed evolution. Hence, option (d) is correct.

Explanation for the incorrect answer:

Option (a) is, ‘reduced female gametophyte’. The special characteristic of the non-vascular plants was the large and dominant gametophytes. These gametophytes became smaller and the sporophytes became larger and dominant as the plants evolved according to the variations present in the environment. They play a very important role in reproductive innovation and result in seed evolution. So, it is an incorrect option.

Option (b) is, ‘heterospory’. It involves the production of two different-sized spores, and from different sexes by the sporophytes of land plants. It is thought that the early plants (all bryophytes and all non-vascular plants) were homosporous. They produced bisexual spores, which can give rise to both the male and female gametophytes. The plants that are present now are heterosporous in nature. It indicates that plants have been evolved through time and are part of the reproductive innovation. So, it is an incorrect option.

Option (c) is “the ovule.” The ovule is an important structure of the plant, which is converted into the seed after fertilization. The gymnosperms had uncovered ovules on the scales of the cone. The gymnosperms had the ‘naked seeds’, that is, their seeds were not covered by the ovary. Then, as time passed, the seeds adapted to the surrounding environment and became covered with the help of ovary. The seeds of angiosperms were covered by two or more seed coats, which protected the seed from harsh conditions. The ovule took part in the reproductive innovation which led to the evolution of the seed. So, it is an incorrect option.

Option (e) is “the integument.” It is a protective layer which surrounds the ovule. The gymnosperms have only one integument present around the ovule but, in the angiosperms, there are two integumentary coatings present as an adaptation toward the changing environment. The integument later converts into the seed coat which provides protection to the seed to survive in unfavorable conditions. So, it is an incorrect option.

Hence, options (a), (b), (c) and (e) are incorrect.

Conclusion

Therefore, it can be concluded that the female gametophytes, heterospory, the ovule, and the integument participated in the reproductive innovation related to seed evolution. The ovary does not have any role in reproductive innovation.

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For short answer questions, write your answers on the line provided. To the right is the mRNA codon table to use as needed throughout the exam. First letter U บบบ U CA UUCPhe UUA UCU Phe UCC UUG Leu CUU UAU. G U UAC TV UGCys UAA Stop UGA Stop A UAG Stop UGG Trp Ser UCA UCG CCU] 0 CUC CUA CCC CAC CAU His CGU CGC Leu Pro CCA CAA Gin CGA Arg CUG CCG CAG CGG AUU ACU AAU T AUC lle A 1 ACC Thr AUA ACA AUG Mot ACG AGG Arg GUU GCU GUC GCC G Val Ala GAC Asp GGU GGC GUA GUG GCA GCG GAA GGA Gly Glu GAGJ GGG AACASH AGU Ser AAA1 AAG Lys GAU AGA CAL CALUCAO CAO G Third letter 1. (+7) Use the table below to answer the questions; use the codon table above to assist you. The promoter sequence of DNA is on the LEFT. You do not need to fill in the entire table. Assume we are in the middle of a gene sequence (no need to find a start codon). DNA 1 DNA 2 mRNA tRNA Polypeptide C Val G C. T A C a. On which strand of DNA is the template strand (DNA 1 or 2)?_ b. On which side of the mRNA is the 5' end (left or…
3. (6 pts) Fill in the boxes according to the directions on the right. Structure R-C R-COOH OH R-OH i R-CO-R' R R-PO4 R-CH3 C. 0 R' R-O-P-OH 1 OH H R-C-H R-N' I- H H R-NH₂ \H Name Properties
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