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0.38 0.05 The rst assumption is random mating, which means that the probability that two genotypes will mate is the product of the frequencies (or probabilities) of the genotypes in the population. If the MM genotype makes up 90% of a population, then any individual has a 90% chance (probability 0.9) of mating with a person with an MM genotype. The probability of an MM by MM mating is (0.9)(0.9), or 0.81. Deviations from random mating come about for two reasons: choice or circumstance. If members of a population choose individuals of a particular phenotype as mates more or less often than at random, the population is engaged in assortative mating. If individuals with similar phenotypes are mating more often than at random, positive assortative mating is in force; if matings occur between individuals with dissimilar phenotypes more often than at random, negative assortative mating, or disassortative mating, is at work. Deviations from random mating also arise when mating individuals are either more closely related genetically or more distantly related than individuals chosen at random from the population. Inbreeding is the mating of related individuals, and outbreeding is the mating of genetically unrelated individuals. Inbreeding is a consequence of pedigree relatedness (e.g., cousins) and small population size. One of the rst counterintuitive observations of population genetics is that deviations from random mating alter genotypic frequencies but not allelic frequencies. Envision a population in which every individual is the parent of two children. On the average, each individual will pass on one copy of each of his or her alleles. Assortative mating and inbreeding will change the zygotic (genotypic) combinations from one generation to the next, but will not change which alleles are passed into

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a A fresh battery is being tested, and it is found that the voltage, Vout , is 164 V with the switch open and 163 V with the switch closed Find the internal resistance of the battery b The same battery is tested one year later, and Vout is found to be 16 V with the switch open but 017 V with the switch closed Find the internal resistance of the battery

We derive an expression for calculating p and q based on genotypic frequencies as follows: p f(M) 2 number of MM number of MN 2 total number number of MN 2 total number

Figure P255, consisting of an ideal ammeter in series with a 2-k resistor The meter sees a full-scale de ection when the current through it is 50 A If we

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FL = 795 lbf 354 N The friction power is obtained by multiplying the friction force with the sliding speed Pf = FLn Pf = 795 50 Pf = 398 in lbf s 1 449 W For the 10 pads, the total friction power is multiplied by the number of pads nPf = 398 10 nPp = 06 hp 449 W Criterion of the Operating Temperature The average temperature rise per pass of the oil through the system of bearing pads, neglecting heat losses is given by p1 + P f / Q t F 152

2 number of MM 2 total number f(MM) and, q f(N ) 2 2 2 (1 2)f(MN )

wished to construct a multirange ammeter reading full-scale values of 1 mA, 10 mA, or 100 mA, depending on the setting of a rotary switch, what should R1 , R2 , and R3 be

that the meter reads if VS = 10 V and RS has the following values: RS = 01rm , 03rm , 05rm , rm , 3rm , 5rm , and 10rm How large (or small) should the internal resistance of the meter be relative to RS

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number of NN total number (1 2)f(MN )

a resistive element in the circuit of Figure P259 The instrument is modeled by an ideal voltmeter in parallel with a 97-k resistor, as shown The meter is placed to measure the voltage across R3 Let R1 = 10 k , RS = 100 k , R2 = 40 k , and IS = 90 mA Find the voltage across R3 with and without the voltmeter in the circuit for the following values:

f(NN )

2 k

5 10 6 50 [79 16 + 0588 ] 2 10 3

practical ammeter is shown in Figure P256, where RS = 10,000 , VS = 10 V, and Rp is a variable resistor that can be adjusted at will

Thus, allelic frequencies can be calculated as the frequency of homozygotes, plus half the frequency of heterozygotes, as follows: p f(M ) 0.57 q f(N ) 0.05 or q 1 p 1 0.76 0.24 f(MM ) (1 2)0.38 f(NN ) (1 2)0.38 (1 2)f(MN ) 0.76 (1 2) f(MN ) 0.24

Although not the most ef cient in all applications, the tracking ADC is an easy starting point to illustrate the operation of an ADC, in that it is based on the DAC presented in the previous section The tracking ADC, shown in Figure 1527, compares the analog input signal with the output of a DAC; the comparator output determines whether the DAC output is larger or smaller than the analog input to be converted to binary form If the DAC output is smaller, then the comparator output will cause an up-down counter (see 14) to count up, until it reaches a level close to the analog signal; if the DAC output is larger than the analog signal, then the counter is forced to count down Note that the rate at which the up-down counter is incremented is determined by the external clock, and that the binary counter output corresponds to the binary representation of the analog signal A feature of the tracking ADC is that it follows ( tracks ) the analog signal by changing one bit at a time

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