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Chapter 17: Problem 74

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

Short Answer

Expert verified
Entropy quantifies the uncertainty or randomness in data, affecting the transmission of information. Redundancy, through the inclusion of additional bits, helps maintain accuracy and reliability in information transmission by detecting and correcting errors. In DNA, redundancy aids in error correction during replication and provides flexibility for adaptation. In space probes like Voyager II and Hubble Space Telescope, redundancy improves the accuracy and reliability of the received data during the long-distance transmission, ensuring precise scientific observations and measurements.

Step by step solution

01

1. Understanding Entropy

Entropy is a measure of the randomness or disorder in a system. In the context of information theory, entropy quantifies the amount of uncertainty involved in predicting the value of a random variable. The more uncertain or random the data, the higher its entropy, and the more information it contains.
02

2. Transmission of information

Information transmission refers to the process of sending data from one location to another, typically through a communication channel. In these situations, the main goal is to ensure that the transmitted information can be accurately and efficiently recovered at the destination. However, during the transmission process, the data may get corrupted due to noise or interference, resulting in errors in the received information.
03

3. Redundancy in information transmission

Redundancy refers to the inclusion of additional bits of information in a communication system to help maintain accuracy and reliability of the transmitted information. Redundant bits do not carry new information, but they enable the detection and correction of errors caused by noise or interference in the transmission process.
04

4. Redundancy in DNA

In the case of human DNA, there is almost twice the amount of information necessary to code for all the required substances. This additional information is redundant, and it serves several purposes: 1. Correction of errors: The redundancy in DNA helps in error detection and correction during replication. This is crucial in maintaining the accuracy of genetic information being passed on to future generations. 2. Flexibility for adaptation: The redundant information in DNA can serve as a backup or buffer in case environmental factors or mutations cause damage to certain sections of the genetic code. This can allow an organism to adapt and survive changing conditions.
05

5. Redundancy in space probes

In the case of the Voyager II and Hubble space telescope probes, redundant bits are added to the transmitted digital data: 1. Voyager II: The probe transmits one redundant bit for every bit of information. This 50% redundancy can improve the accuracy and reliability of the received data by allowing the correction of any errors caused by noise or interference during the long-distance transmission. 2. Hubble Space Telescope: Hubble transmits three redundant bits for every bit of information. This higher redundancy ensures the accuracy and reliability of the received data, which is crucial for the precise scientific observations and measurements conducted by the telescope.
06

6. Conclusion

Entropy is related to the transmission of information by quantifying the amount of uncertainty or randomness in the data. Redundancy plays a significant role in the transmission of information both in DNA and space probes by allowing the correction of errors, providing flexibility for adaptation, and ensuring the accuracy and reliability of the received data. This helps to maintain the integrity of the genetic code for life and enhance the quality of scientific observations and measurements in space missions.

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Most popular questions from this chapter

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix 4, predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature dependent. Predict the sign of \(\Delta S_{\text {sarr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) at \(298 \mathrm{~K}\). Using these values and data in Appendix 4, calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the preceding reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\).

When the environment is contaminated by a toxic or potentially toxic substance (for example, from a chemical spill or the use of insecticides), the substance tends to disperse. How is this consistent with the second law of thermodynamics? In terms of the second law, which requires the least work: cleaning the environment after it has been contaminated or trying to prevent the contamination before it occurs? Explain.

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)$$ \(\Delta H^{\circ}\) for this reaction is \(79.14 \mathrm{~kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\). Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C}\), calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .{ }^{\circ} \mathrm{C}\). (Hint: Manipulate the equation in Exercise 71 .)

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) c. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).
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