1. In the ADABarka plant the synthesis of pink pigment in the petals .pdf

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1. In the ADABarka plant the synthesis of pink pigment in the petals is controlled by two genes, S and K. (Unless told otherwise, you can always assume that all other genes in a pathway are homozygous WT in these individuals.) The pathway is If a pure breeding plant was unable to catalyze the first reaction, what color of petals would you expect to see? If a pure breeding plant was unable to catalyze the second reaction, what color of petals would you expect to see? If the plants from parts .a and b were crossed, what color petals will these FI offspring plants have? What would be the expected phenotypic ratios in a cross between F1 plants? Make sure you include phenotypic class information with your ratios. Solution a. the colour would always be yellow b.the colour would be green c: the flowers would be pink coloured because on crossing a would have k geneintact and b would have s gene intact ... thus whole pathway is complimented and pink colour is the result ...... but genotype would be s1s0k1k0 kind (1 and 0 mean presence and absence of gene). d. since f1 will genotype s1s0k1ko on selfing (s1s0k1k0 x s1s0k1k0 ) gametes produced would be s1k1, s1k0, s0k1, s0k0 assuming 1 and 0 mean presence and absence of allele we will have ratio as 7 pink :3 green: 2yellow hope it answers your question should u need further clarification kindly tell.

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I don\'t understand numbers 14 and 15, please help.. PROBLEM BAS 12, A plant of genotype is testcrossed wil 18 If the two loci are 10 m apart, what proportion of progeny will be AB/ab 13. The A locus and the Dlocus nked that no are so tightly recombination is ever observed between if Ad 1. Ad is crossed with ab,aD d the F, is intercrossed, what phenotypes will be seen in the Fa and in what proportions? 14. The R and Sloci are 35 m.u. apart. If a plant of genotype is selfed, what progeny phenotypes will be seen and in what proportions 15. The cross EE. FF x efe fif s de, and the F1 is then backcrossed with the recessive parent. The progeny gen. ocypes are inferred from the phenotypes. The progeny genotypes, written as the gametic contributions of the heterozygous parent, are in the following proportions: E. F 2 Explain these results. Progeny Class phenotypes. Solution Answers: 14. R S/r s * R S/r s Gametes: 1/2 (1 – 0.35) R S 1/2 (1 – 0.35) r s 1/2 (0.35) R s 1/2 (0.35) r S F1 genotypes will be: 0.1056 R S/R S 0.1138 r s/r S 0.1056 r s/r s 0.1138 r s/R s 0.2113 R S/r s 0.0306 R s/R s 0.1138 R S/r S 0.0306 r S/r S 0.1138 R S/R s 0.0613 R s/r S F1 phenotypes will be: 0.6058 R S 0.1056 r s 0.1444 R s 0.1444 r S 15. The cross: E/e · F/f * e/e · f/f. The progeny should be in a 1:1:1:1 ratio if independent assortment takes place, which is not observed. So, linkage exists. E f and e F are recombinants and equals one-third of the progeny. The two genes are 33.3 map units (m.u.) apart. RF = 100% * 1/3 = 33.3%.

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Biology 103 Laboratory Exercise – Genetic Problems Introduction Although the science of genetics has become a highly sophisticated discipline dealing with the interactions of hereditary factors at the molecular level, it has its roots in the basic laws of heredity initially discovered and presented by Gregor Mendel more than one hundred years ago. Mendel's success in discovering these laws was due largely to his application of the simple rules of mathematical probability - the laws of chance - to his observations concerning the inheritance of certain characteristics in the garden pea plant. Reginald Punnett and the Punnett Square The Punnett square is a diagram used by biologists to determine genotypic probability within the offspring from a particular genetic cross. The Punnett square shows every possible genotypic combination of maternal alleles with the paternal alleles for a genetic cross. Punnett squares only give probabilities for genotypes, not phenotypes. The square diagram was designed by the British geneticist, Reginald Punnett (1865-1967) and first presented to the science community in 1905. Punnett’s Mendelism (1905) is considered the first popular science book to introduce genetics to the public. Solving Genetic Problems R R' R RR RR' R' RR' R'R' Maternal alleles A A a Aa Aa Paternal Alleles a Aa Aa The first step in solving a genetic problem is to establish the genetic symbols you will use in your problem solution. Stay consistent by using these same symbols throughout the problem solving process. Represent dominant and recessive alleles (different forms of a gene) using traditional genetic symbols. Dominant alleles should be represented with the capital version of an alphabetic letter while using the lower case version to show recessiveness. For example: B = black color, b = white color. Each individual gene or trait is diploid (2n) in nature and therefore, must be represented with two alleles. Continuing with the alleles mentioned previously, an individual may have the genetic makeup BB, Bb, or bb when using those alleles. Remember that gametes (sperm and egg) are haploid (n) and can only provide one allele per trait. For example: B or b An individual’s genotype contains the possible gametes that can be expected to be produced by that individual. Much of genetics revolves around the probability of the makeup of gametes. If the individual is homozygous, all of the gametes produced will possess the same kind of allele. For example, an individual with the genotype BB would be expected to produce only B gametes and individuals with genotype bb would produce only b gametes. If the individual is heterozygous, that is the individual’s genotype contains one dominant allele and one recessive allele (Bb), the gametes produced will possess one or the other of the two forms of the gene – B or b. ...

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(1) In the tubular flowers of foxgloves, wild-type coloration is red, while a mutation called white produces white flowers. Another mutation, called peloria, causes the flowers at the apex of the stem to be huge. Yet another mutation, called dwarf, affects stem length. You cross a white-flowered plant (otherwise phenotypically wild type) to a plant that is dwarf and peloria but has wild-type red flower color. All the F1 plants are tall with white, normal-sized flowers. You cross an F1 plant back to the dwarf and peloria parent, and you see the 543 progeny shown in the chart to the right. (Only mutant traits are noted.) a. White, tall, and normal sized flowers/are the dominant alleles because these are what were expressed in the F1 generation, meaning that these hybrid offspring are exhibiting the dominant trait. b. What were the genotypes of the parents in the original cross? How do you know? c. Draw a map showing the linkage relationships of these three loci. Why is the calculated distance value for the two outside genes smaller than the value you get when you sum the two shorter distances? How can you correct this mistake? Make sure to show all your work. d. Is there interference? If so, calculate the coefficient of coincidence and the interference value (show your work!), and state the interpretation of the interference value. .

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Write a C program that converts an NFA file to a DFA, where the language of the NFA consists of lowercase letters of the alphabet. Then have it be able to determine whether strings in a string file are in the language or not. The program must take the two files as command line arguments (do not prompt the user). Solution #include #include #define STATES 256 #define SYMBOLS 20 int N_symbols; int NFA_states; char *NFAtab[STATES][SYMBOLS]; int DFA_states; /* number of DFA states */ int DFAtab[STATES][SYMBOLS]; /*Print state-transition table.*/ void put_dfa_table( int tab[][SYMBOLS], /* DFA table */ int nstates, /* number of states */ int nsymbols) /* number of input symbols */ { int i, j; puts(\"STATE TRANSITION TABLE\"); /* input symbols: \'0\', \'1\', ... */ printf(\" | \"); for (i = 0; i < nsymbols; i++) printf(\" %c \", \'0\'+i); printf(\"\ -----+--\"); for (i = 0; i < nsymbols; i++) printf(\"-----\"); printf(\"\ \"); for (i = 0; i < nstates; i++) { printf(\" %c | \", \'A\'+i); /* state */ for (j = 0; j < nsymbols; j++) printf(\" %c \", \'A\'+tab[i][j]); printf(\"\ \"); } } /*Initialize NFA table.*/ void init_NFA_table() { NFAtab[0][0] = \"12\"; NFAtab[0][1] = \"13\"; NFAtab[1][0] = \"12\"; NFAtab[1][1] = \"13\"; NFAtab[2][0] = \"4\"; NFAtab[2][1] = \"\"; NFAtab[3][0] = \"\"; NFAtab[3][1] = \"4\"; NFAtab[4][0] = \"4\"; NFAtab[4][1] = \"4\"; NFA_states = 5; DFA_states = 0; N_symbols = 2; } /*String \'t\' is merged into \'s\' in an alphabetical order.*/ void string_merge(char *s, char *t) { char temp[STATES], *r=temp, *p=s; while (*p && *t) { if (*p == *t) { *r++ = *p++; t++; } else if (*p < *t) { *r++ = *p++; } else *r++ = *t++; } *r = \'\\0\'; if (*p) strcat(r, p); else if (*t) strcat(r, t); strcpy(s, temp); } /*Get next-state string for current-state string.*/ void get_next_state(char *nextstates, char *cur_states, char *nfa[STATES][SYMBOLS], int n_nfa, int symbol) { int i; char temp[STATES]; temp[0] = \'\\0\'; for (i = 0; i < strlen(cur_states); i++) string_merge(temp, nfa[cur_states[i]-\'0\'][symbol]); strcpy(nextstates, temp); } int state_index(char *state, char statename[][STATES], int *pn) { int i; if (!*state) return -1; /* no next state */ for (i = 0; i < *pn; i++) if (!strcmp(state, statename[i])) return i; strcpy(statename[i], state); /* new state-name */ return (*pn)++; } /* Convert NFA table to DFA table. Return value: number of DFA states. */ int nfa_to_dfa(char *nfa[STATES][SYMBOLS], int n_nfa, int n_sym, int dfa[][SYMBOLS]) { char statename[STATES][STATES]; int i = 0; /* current index of DFA */ int n = 1; /* number of DFA states */ char nextstate[STATES]; int j; strcpy(statename[0], \"0\"); /* start state */ for (i = 0; i < n; i++) { /* for each DFA state */ for (j = 0; j < n_sym; j++) { /* for each input symbol */ get_next_state(nextstate, statename[i], nfa, n_nfa, j); dfa[i][j] = state_index(nextstate, statename, &n); } } return n; /* number of DFA states */ } void main() { init_NFA_table(); DFA_states = nfa_to_dfa(NFAtab, N.

Write a C program that converts an NFA file to a DFA, where the lang.pdf

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Which of these hypotheses cannot be tested by an experiment? Biological interactions between barnacle species are responsible for their distribution patterns. Night-feeding fishes rely on their sense of smell to locate prey. The growth of salt marsh gross is limited by nitrogen availability. Certain humpback whale behaviors are important for attracting a mate. Goldfish prefer to eat brightly-colored foods. TRUE/FALSE The oceans cover less than half of the earth\'s surface. Ocean productivity is the amount of food produced by marine organisms and the number of organisms this food supports. Aristotle developed one of the first taxonomical classification schemes. Today it is generally believed that animals cannot live below the lit regions of the ocean. Ecology is: a type of life style. the study of the interaction of organisms and their environment. a place where organisms live. what happens when a predator consumes a prey item. the study of the physical environment in an ecosystem. Organisms that live under similar environmental conditions and location, and interact directly or indirectly are part of the same: ecosystem. biosystem. abiotic habitat. biotic habitat. population. An example of abiotic factors influencing marine organisms is: predation. competition for food. parasitism. salinity. bacterial disease. Habitat refers to where an organism: lives. feeds. reproduces. forages. All of the above. Solution Answers 15. False Oceans cover about 70%of the surface of the earth 16. True 17. True Aristotle developed the first classification system which divided all known organisms into two groups plants and animals 18. False Multiple choice 19. b study of interaction of organism\'s their environment Ecology is the scientific analysis and study of interactions among organisms and their environment . It is an interdisciplinary field that includes biology, geography and science . 20. a ecosystem Interactions describes the different ways organisms obtain their food and energy . Some organisms can make their own food and other organisms have to their food by eating other organism 22. a lives A habitat is an ecological or environmental area that is inhabited by a particular species of animals plants or other type of organism . It is the natural environment in which an organism lives ..

Which of these hypotheses cannot be tested by an experiment Biologi.pdf

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Which is the mode of inheritance that is most likely? [That is the mode of inheritance that would mean that the pattern would HAVE to be as described, not just that is could have come out this way.] sex-limited, autosomal dominant Y-linked X-linked recessive autosomal recessive X- linked dominant autosomal dominant autosomal sex-influenced Solution Non-autosomal mode of inheritance is the one related to sex-chromosomes. The alleles associated with sex-chromosomes represent a differential kind of inheritance pattern and display a differential phenotype response in offsprings based upon their sex. Majority of the human diseases related to sex-chromosomes are X-linked in nature such as color blindness, hemophilia, Duchenne muscular dystrophy, Vitamin D resistant rickets or X-linked hypophosphatemia, Rett syndrome, Incontinentia pigmention and Goltz syndrome. Since both parents, a male and a female carry an X chromosome, they both are potential donors of X-linked allele of a disease. Since the offsprings can acquire a differential X-chromosome profile either from mother or father, their genetic make up will be dependent upon that of their parents and thus, their disease condition (carrier, affected, unaffected). Thus, X-linked (dominant or recessive) mode of sex-chromosome related diseases need to be explained theoratically and observationally, and not only be described as general procedures of genetic segregation. Since X-linked recessive diseases are more likely to cause ambiguity among inheritance pattern, a more detailed explanation would be requried for X-linked recessive alleles..

Which is the mode of inheritance that is most likely [That is the mo.pdf

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Which of the following is a good example of homoplasy (convergent evolution)? a. Dolphins and humans b. Apes and monkeys c. Dolphins and fish d. Dolphins and whales e. Cats and bats Wings of bats are: a. Plesiomorphic (ancestral) feature for mammals b. Plesiomorphic feature for bats and birds c. Synapomorphic (shared-derived) feature for vertebrates d. Synapomorphic feature for bats e. Synapomorphic feature for bats and birds Solution 1). c. Dolphins and fish Analogous structures (convergent evolution) are not found in the common ancestors but the same structure or similar structure with the same function can be found in one or more groups. In such cases, it is not clear to tract the ancestry but the homologous structures are helpful to track the ancestry. For example, both the dolphins and sharks share the fins but the dolphins are mammals, while the shark are fish..

Which of the following is a good example of homoplasy (convergent ev.pdf

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