Principles of Inheritance & Variation (Genetics) Questions and Answers

3 In mice dwarfism is caused by an X linked recessive allele and the pink coat is caused by an autosomal dominant allele Coats are normally brownish Suppose a dwarf brown female from a pure breeding line is crossed with a normal sized pink male from a pure breeding line What are the genotypes of the parental mice Define your own gene and allele symbols What will the predicted phenotypic ratios be in the F1 generation for each sex
Biology
Principles of Inheritance & Variation (Genetics)
3 In mice dwarfism is caused by an X linked recessive allele and the pink coat is caused by an autosomal dominant allele Coats are normally brownish Suppose a dwarf brown female from a pure breeding line is crossed with a normal sized pink male from a pure breeding line What are the genotypes of the parental mice Define your own gene and allele symbols What will the predicted phenotypic ratios be in the F1 generation for each sex
What enzyme is responsible for covalently linking amino acids to the 3 end of the cognate tRNA peptidyltransferase O aminoacyl tRNA synthetase O glutamine synthetase RNA polymerase O ATP synthase QUESTION 35 Which of the following is not a normal property of eukaryotic mRNAs O They contain a continuous nucleotide sequence encoding a specific polypeptide O They are found in the cytoplasm and inside the Golgi complex O They are attached to ribosomes when they are translated O Most have a significant noncoding segment that does not direct assembly of amino acids O Eukaryotic mRNAs have special modifications at their 5 and 3 termini QUESTION 36
Biology
Principles of Inheritance & Variation (Genetics)
What enzyme is responsible for covalently linking amino acids to the 3 end of the cognate tRNA peptidyltransferase O aminoacyl tRNA synthetase O glutamine synthetase RNA polymerase O ATP synthase QUESTION 35 Which of the following is not a normal property of eukaryotic mRNAs O They contain a continuous nucleotide sequence encoding a specific polypeptide O They are found in the cytoplasm and inside the Golgi complex O They are attached to ribosomes when they are translated O Most have a significant noncoding segment that does not direct assembly of amino acids O Eukaryotic mRNAs have special modifications at their 5 and 3 termini QUESTION 36
0 a dominant gene with Huntington s disease a degenerative brain disorder caused by Mark and Karen are Rachel s parents and have 2 other children Sam and Dave who do not carry the gene Mark s family has no history of Huntington s disease and he does not carry the gene Karen s father and his family do not carry the gene however Karen s mother died from complications from the disease o Both of Karen s maternal grandparents were carriers of the gene Case Study 5 Questions 1 Construct a Punnett Square for Mark Karen Abwa 2 What is Karen s genotype 3 What is the genotype of Karen s father 4 What is the probability that Karen Mark will have another child who carries the Huntington gene
Biology
Principles of Inheritance & Variation (Genetics)
0 a dominant gene with Huntington s disease a degenerative brain disorder caused by Mark and Karen are Rachel s parents and have 2 other children Sam and Dave who do not carry the gene Mark s family has no history of Huntington s disease and he does not carry the gene Karen s father and his family do not carry the gene however Karen s mother died from complications from the disease o Both of Karen s maternal grandparents were carriers of the gene Case Study 5 Questions 1 Construct a Punnett Square for Mark Karen Abwa 2 What is Karen s genotype 3 What is the genotype of Karen s father 4 What is the probability that Karen Mark will have another child who carries the Huntington gene
Transcribe the following sequence of DNA into mRNA 3 TAC AAG TTT 5 O O O 5 AUG UUC AAA 3 3 ATG TTC AAA 5 3 AUG UUC AAA 5 3 AAA CUU GUA 5
Biology
Principles of Inheritance & Variation (Genetics)
Transcribe the following sequence of DNA into mRNA 3 TAC AAG TTT 5 O O O 5 AUG UUC AAA 3 3 ATG TTC AAA 5 3 AUG UUC AAA 5 3 AAA CUU GUA 5
hromosome A contains genes for an eye color of hazel Chromosome B contains genes for ha color of brown and chromosome C contains genes for an eye color of blue Which ones re homologous chromosomes a none of the chromosomes are homologous to one another b chromosomes A and C are homologous c chromosomes B and C are homologous d chromosomes A and B are homologous
Biology
Principles of Inheritance & Variation (Genetics)
hromosome A contains genes for an eye color of hazel Chromosome B contains genes for ha color of brown and chromosome C contains genes for an eye color of blue Which ones re homologous chromosomes a none of the chromosomes are homologous to one another b chromosomes A and C are homologous c chromosomes B and C are homologous d chromosomes A and B are homologous
Question 3 of 10 After being exposed to ionizing radiation an insect with red eyes has offspring that all have yellow eyes The offspring of the next generation all have yellow eyes as well Which claim is best supported by this information OA Inheritable genetic variations can form due to lateral gene transfers during conjugation OB Mutations caused by environmental factors can result in inheritable genetic variation OC Inheritable genetic variation can occur during meiosis due to independent assortment OD Errors can occur during replication causing new inheritable genetic variations
Biology
Principles of Inheritance & Variation (Genetics)
Question 3 of 10 After being exposed to ionizing radiation an insect with red eyes has offspring that all have yellow eyes The offspring of the next generation all have yellow eyes as well Which claim is best supported by this information OA Inheritable genetic variations can form due to lateral gene transfers during conjugation OB Mutations caused by environmental factors can result in inheritable genetic variation OC Inheritable genetic variation can occur during meiosis due to independent assortment OD Errors can occur during replication causing new inheritable genetic variations
5 A black male crossed with a calico female to YR Y XBXB XBY XB VR v B VR V
Biology
Principles of Inheritance & Variation (Genetics)
5 A black male crossed with a calico female to YR Y XBXB XBY XB VR v B VR V
and prevents the senescence of leaves Cytokinins Question 3 1 pts Identify the class of hormone that is growth inhibiting it promotes seed
Biology
Principles of Inheritance & Variation (Genetics)
and prevents the senescence of leaves Cytokinins Question 3 1 pts Identify the class of hormone that is growth inhibiting it promotes seed
Exercise 3 Questions Question 1 How are karyotypes used to screen for genetic disease E E T T B I U I 0 10000
Biology
Principles of Inheritance & Variation (Genetics)
Exercise 3 Questions Question 1 How are karyotypes used to screen for genetic disease E E T T B I U I 0 10000
In the following pedigree of an autosomal recessive disorder what is the probability that IV 1 will be affected I II III IV 2 3 O 1 2 O 3 4 1 12 Rr 1 1 Rr 2 R R R3 2 3 1 RR 3 R rr 1 5 Rr 4 2
Biology
Principles of Inheritance & Variation (Genetics)
In the following pedigree of an autosomal recessive disorder what is the probability that IV 1 will be affected I II III IV 2 3 O 1 2 O 3 4 1 12 Rr 1 1 Rr 2 R R R3 2 3 1 RR 3 R rr 1 5 Rr 4 2
1 61 GTCGACTGGG CACCAGCCGG ATGCGTCCGT CAACCCCGAC TGTTCCGCCA AGAACTCCG CCTTCTGTGG CGGGATGACG TTGAGCACAT CGGCACAGAA AGTGTCGAAC TCGGTTTCG 121 CGGTTCGCCC GGCGGCATCG ATCCGCTCCA GCCGTCCCTC TTTTTCCGAG GAAACCCAT 181 CCACCATGTC TCCGTAGAGT TCGCGCCAGG CCTGCTCGAA TGCCGGCTGC TTGGAAAACC 241 GGGTCTTGGC ATCGAGTATC AGGATCTTGG AGCGGGGTTT GTGCTTTTTC ATGTAATAGE 301 CGATGAGCGA GGCGCGTTCG TAGGGGCCAG GCGGACAACG ATAGGGAGAA GGTGGTGCGG 361 TGATCAGAAC CAGGCCGCCG TCGGGCATCG CGCGGATCTG CCGGGCGAGC AAAGCGGTCT 421 GGGGGCCGGC CTTCCAGGCA TGGGGAACGA ACCGGCTCGC CGCCTCGTCA TAGCCCATTA 481 TCGCCTCCCA GCGAAAGTCG ATGCCGGGGG AGAGGACTAG CCTGTCGTAG GTGACCTCGG 541 CACCATCGTT CAGTATCACA CGCCGTCGCT GCCGATCCAG GCGGGCGACC CGGGCGGTTA 601 CCTTTCCTAT ATCCAGCTCC CGCCGCAACC AGTCATAAGA CCGCGCGAGA GTCCCCATGT 661 CCCCGAGGCC GGCGACTGCT TCGTTCGATC CGGGGCAGGA AAGATAAGTC TCCTGCGGTT 721 CGATCAATGT GATCGTCAGA CCGGGATTCA TCTGCTTCAG ATAGCGGGCG GCCGTGGCAC 781 CGCCATAACC GCCACCGACG ACCACGACCG GCCGCCTGAG CGAAGACCTT GCCGGAAGCG 841 CAGCCACCCA GTCCCAGCCC ATGCCGCAGA CTGCCAGCAG GCGCAGGAAC CGCCGCCGGC 901 GGATCATGGC CTCTTTCCCA GAAAGCCGGC GATCGCTTCA ATGTCCTGGT TCGTGAGTCC 961 GGCAGCGATC CGGTTCATGA CCGTGCCCGA TCTTTTTCCT TCACGGTACT CCCGCAACAG 1021 AGATGCCATC TCCTTCGCAT CGAAGCGGCG TAACGATGCC GGTTCGGGAA TCTGCTCCTC 1081 CTCGTCGGCA TGGCAGCCGA GGCAACCGAG CGCAGCCAAA ACCATGTCCG GTTTTTCGGC 1141 CCGGGCCGGG AAACAGAGAG CGACAGCGAT CGTACCGATC TGGACGCGTC TCACAAACGA 1201 CAAAACGTCA CGATGGGTGT TCGGTAGCTG AGTCACGGGG ATTTGTAGAA GTATAGGACC 1261 GACGGATTTT ATGCAAGCAT GTCGCTTTGA CCAAGCCGGG ATTCCATGGA AGGGATGTCA 1321 TCGGGAGAGT TATTTATGTC GTTGATTTAT AAGAAACTAC CCCTGCGTCA AAATGTCGCA 1381 GATTTTTCTT GACAGTTTGG GGGAGGGTGA TAGATCCTCC ACCGATGGAC CGGTACCGCC 1441 TCTGTTGCGG GGTCCATGAA ATGCCCGTTA GAGGCAGAAC CGATAGGGAA TTAGAGAAGC 1501 GGGCGTCGGC GCCGAATGCC GGCCCCTGTC AACCATCACT TTAGGAGGAA CAAACAATGG 1561 CAGCAACAAC CATTGGTGGT GCAGCTGCGG CGTAAGCGCC GCTGCTGGAC AAGAAGTGGC 1621 TCACGTTCGC ACTGGCGATT TACACCGTGT TCTACCTGTG GGTGCGGTGG TACGAAGGTG 1681 TCTATGGCTG GTCCGCCGGA CTGGACTCGT TCGCGCCGGA GTTCGAGACC TACTGGATGA 1741 ATTTCCTGTA CACCGAGATC GTCCTGGAGA TCGTGACGGC TTCGATCCTG TGGGGCTATC 1801 TCTGGAAGAC CCGCGACCGC AACCTGGCCG CGCTGACCCC GCGTGAAGAG CTGCGCCGCA 1861 ACTTCACCCA CCTGGTGTGG CTGGTGGCCT ACGCCTGGGC CATCTACTGG GGCGCATCCT 1921 ACTTCACCGA GCAGGACGGC ACCTGGCATC AGACGATCGT GCGCGACACC GACTTCACGC 1981 CGTCGCACAT CATCGAGTTC TATCTGAGCT ACCCGATCTA CATCATCACC GGTTTTGCGG 2041 CGTTCATCTA CGCCAAGACG CGTCTGCCGT TCTTCGCGAA GGGCATCTCG CTGCCGTACC 2101 TGGTGCTGGT GGTGGGTCCG TTCATGATTC TGCCGAACGT GGGTCTGAAC GAATGGGGCC
Biology
Principles of Inheritance & Variation (Genetics)
1 61 GTCGACTGGG CACCAGCCGG ATGCGTCCGT CAACCCCGAC TGTTCCGCCA AGAACTCCG CCTTCTGTGG CGGGATGACG TTGAGCACAT CGGCACAGAA AGTGTCGAAC TCGGTTTCG 121 CGGTTCGCCC GGCGGCATCG ATCCGCTCCA GCCGTCCCTC TTTTTCCGAG GAAACCCAT 181 CCACCATGTC TCCGTAGAGT TCGCGCCAGG CCTGCTCGAA TGCCGGCTGC TTGGAAAACC 241 GGGTCTTGGC ATCGAGTATC AGGATCTTGG AGCGGGGTTT GTGCTTTTTC ATGTAATAGE 301 CGATGAGCGA GGCGCGTTCG TAGGGGCCAG GCGGACAACG ATAGGGAGAA GGTGGTGCGG 361 TGATCAGAAC CAGGCCGCCG TCGGGCATCG CGCGGATCTG CCGGGCGAGC AAAGCGGTCT 421 GGGGGCCGGC CTTCCAGGCA TGGGGAACGA ACCGGCTCGC CGCCTCGTCA TAGCCCATTA 481 TCGCCTCCCA GCGAAAGTCG ATGCCGGGGG AGAGGACTAG CCTGTCGTAG GTGACCTCGG 541 CACCATCGTT CAGTATCACA CGCCGTCGCT GCCGATCCAG GCGGGCGACC CGGGCGGTTA 601 CCTTTCCTAT ATCCAGCTCC CGCCGCAACC AGTCATAAGA CCGCGCGAGA GTCCCCATGT 661 CCCCGAGGCC GGCGACTGCT TCGTTCGATC CGGGGCAGGA AAGATAAGTC TCCTGCGGTT 721 CGATCAATGT GATCGTCAGA CCGGGATTCA TCTGCTTCAG ATAGCGGGCG GCCGTGGCAC 781 CGCCATAACC GCCACCGACG ACCACGACCG GCCGCCTGAG CGAAGACCTT GCCGGAAGCG 841 CAGCCACCCA GTCCCAGCCC ATGCCGCAGA CTGCCAGCAG GCGCAGGAAC CGCCGCCGGC 901 GGATCATGGC CTCTTTCCCA GAAAGCCGGC GATCGCTTCA ATGTCCTGGT TCGTGAGTCC 961 GGCAGCGATC CGGTTCATGA CCGTGCCCGA TCTTTTTCCT TCACGGTACT CCCGCAACAG 1021 AGATGCCATC TCCTTCGCAT CGAAGCGGCG TAACGATGCC GGTTCGGGAA TCTGCTCCTC 1081 CTCGTCGGCA TGGCAGCCGA GGCAACCGAG CGCAGCCAAA ACCATGTCCG GTTTTTCGGC 1141 CCGGGCCGGG AAACAGAGAG CGACAGCGAT CGTACCGATC TGGACGCGTC TCACAAACGA 1201 CAAAACGTCA CGATGGGTGT TCGGTAGCTG AGTCACGGGG ATTTGTAGAA GTATAGGACC 1261 GACGGATTTT ATGCAAGCAT GTCGCTTTGA CCAAGCCGGG ATTCCATGGA AGGGATGTCA 1321 TCGGGAGAGT TATTTATGTC GTTGATTTAT AAGAAACTAC CCCTGCGTCA AAATGTCGCA 1381 GATTTTTCTT GACAGTTTGG GGGAGGGTGA TAGATCCTCC ACCGATGGAC CGGTACCGCC 1441 TCTGTTGCGG GGTCCATGAA ATGCCCGTTA GAGGCAGAAC CGATAGGGAA TTAGAGAAGC 1501 GGGCGTCGGC GCCGAATGCC GGCCCCTGTC AACCATCACT TTAGGAGGAA CAAACAATGG 1561 CAGCAACAAC CATTGGTGGT GCAGCTGCGG CGTAAGCGCC GCTGCTGGAC AAGAAGTGGC 1621 TCACGTTCGC ACTGGCGATT TACACCGTGT TCTACCTGTG GGTGCGGTGG TACGAAGGTG 1681 TCTATGGCTG GTCCGCCGGA CTGGACTCGT TCGCGCCGGA GTTCGAGACC TACTGGATGA 1741 ATTTCCTGTA CACCGAGATC GTCCTGGAGA TCGTGACGGC TTCGATCCTG TGGGGCTATC 1801 TCTGGAAGAC CCGCGACCGC AACCTGGCCG CGCTGACCCC GCGTGAAGAG CTGCGCCGCA 1861 ACTTCACCCA CCTGGTGTGG CTGGTGGCCT ACGCCTGGGC CATCTACTGG GGCGCATCCT 1921 ACTTCACCGA GCAGGACGGC ACCTGGCATC AGACGATCGT GCGCGACACC GACTTCACGC 1981 CGTCGCACAT CATCGAGTTC TATCTGAGCT ACCCGATCTA CATCATCACC GGTTTTGCGG 2041 CGTTCATCTA CGCCAAGACG CGTCTGCCGT TCTTCGCGAA GGGCATCTCG CTGCCGTACC 2101 TGGTGCTGGT GGTGGGTCCG TTCATGATTC TGCCGAACGT GGGTCTGAAC GAATGGGGCC
1 GTCGACTGGG CACCAGCCGG ATGCGTCCGT CAACCCCGAC TGTTCCGCCA AGAACTCCGO 61 CCTTCTGTGG CGGGATGACG TTGAGCACAT CGGCACAGAA AGTGTCGAAC TCGGTTTCGA 121 CGGTTCGCCC GGCGGCATCG ATCCGCTCCA GCCGTCCCTC TTTTTCCGAG GAAACCCATT 181 CCACCATGTC TCCGTAGAGT TCGCGCCAGG CCTGCTCGAA TGCCGGCTGC TTGGAAAACC 241 GGGTCTTGGC ATCGAGTATC AGGATCTTGG AGCGGGGTTT GTGCTTTTTC ATGTAATAGG 301 CGATGAGCGA GGCGCGTTCG TAGGGGCCAG GCGGACAACG ATAGGGAGAA GGTGGTGCGG 361 TGATCAGAAC CAGGCCGCCG TCGGGCATCG CGCGGATCTG CCGGGCGAGC AAAGCGGTCT 421 GGGGGCCGGC CTTCCAGGCA TGGGGAACGA ACCGGCTCGC CGCCTCGTCA TAGCCCATTA 481 TCGCCTCCCA GCGAAAGTCG ATGCCGGGGG AGAGGACTAG CCTGTCGTAG GTGACCTCGG 541 CACCATCGTT CAGTATCACA CGCCGTCGCT GCCGATCCAG GCGGGCGACC CGGGCGGTTA 601 CCTTTCCTAT ATCCAGCTCC CGCCGCAACC AGTCATAAGA CCGCGCGAGA GTCCCCATGT 661 CCCCGAGGCC GGCGACTGCT TCGTTCGATC CGGGGCAGGA AAGATAAGTC TCCTGCGGTT 721 CGATCAATGT GATCGTCAGA CCGGGATTCA TCTGCTTCAG ATAGCGGGCG GCCGTGGCAC 781 CGCCATAACC GCCACCGACG ACCACGACCG GCCGCCTGAG CGAAGACCTT GCCGGAAGCG 841 CAGCCACCCA GTCCCAGCCC ATGCCGCAGA CTGCCAGCAG GCGCAGGAAC CGCCGCCGGC 901 GGATCATGGC CTCTTTCCCA GAAAGCCGGC GATCGCTTCA ATGTCCTGGT TCGTGAGTCC 961 GGCAGCGATC CGGTTCATGA CCGTGCCCGA TCTTTTTCCT TCACGGTACT CCCGCAACAG 1021 AGATGCCATC TCCTTCGCAT CGAAGCGGCG TAACGATGCC GGTTCGGGAA TCTGCTCCTC 1081 CTCGTCGGCA TGGCAGCCGA GGCAACCGAG CGCAGCCAAA ACCATGTCCG GTTTTTCGGC 1141 CCGGGCCGGG AAACAGAGAG CGACAGCGAT CGTACCGATC TGGACGCGTC TCACAAACGA 1201 CAAAACGTCA CGATGGGTGT TCGGTAGCTG AGTCACGGGG ATTTGTAGAA GTATAGGACC 1261 GACGGATTTT ATGCAAGCAT GTCGCTTTGA CCAAGCCGGG ATTCCATGGA AGGGATGTCA 1321 TCGGGAGAGT TATTTATGTC GTTGATTTAT AAGAAACTAC CCCTGCGTCA AAATGTCGCA 1381 GATTTTTCTT GACAGTTTGG GGGAGGGTGA TAGATCCTCC ACCGATGGAC CGGTACCGCC 1441 TCTGTTGCGG GGTCCATGAA ATGCCCGTTA GAGGCAGAAC CGATAGGGAA TTAGAGAAGO 1501 GGGCGTCGGC GCCGAATGCC GGCCCCTGTC AACCATCACT TTAGGAGGAA CAAACAATGG 1561 CAGCAACAAC CATTGGTGGT GCAGCTGCGG CGTAAGCGCC GCTGCTGGAC AAGAAGTGGC 1621 TCACGTTCGC ACTGGCGATT TACACCGTGT TCTACCTGTG GGTGCGGTGG TACGAAGGTG 1681 TCTATGGCTG GTCCGCCGGA CTGGACTCGT TCGCGCCGGA GTTCGAGACC TACTGGATGA 1741 ATTTCCTGTA CACCGAGATC GTCCTGGAGA TCGTGACGGC TTCGATCCTG TGGGGCTATC 1801 TCTGGAAGAC CCGCGACCGC AACCTGGCCG CGCTGACCCC GCGTGAAGAG CTGCGCCGCA 1861 ACTTCACCCA CCTGGTGTGG CTGGTGGCCT ACGCCTGGGC CATCTACTGG GGCGCATCCT 1921 ACTTCACCGA GCAGGACGGC ACCTGGCATC AGACGATCGT GCGCGACACC GACTTCACGC 1981 CGTCGCACAT CATCGAGTTC TATCTGAGCT ACCCGATCTA CATCATCACC GGTTTTGCGG 2041 CGTTCATCTA CGCCAAGACG CGTCTGCCGT TCTTCGCGAA GGGCATCTCG CTGCCGTACC 2101 TGGTGCTGGT GGTGGGTCCG TTCATGATTC TGCCGAACGT GGGTCTGAAC GAATGGGGCC 2161 ACACCTTCTG GTTCATGGAA GAGCTGTTCG TGGCGCCGCT GCACTACGGC TTCGTGATCT
Biology
Principles of Inheritance & Variation (Genetics)
1 GTCGACTGGG CACCAGCCGG ATGCGTCCGT CAACCCCGAC TGTTCCGCCA AGAACTCCGO 61 CCTTCTGTGG CGGGATGACG TTGAGCACAT CGGCACAGAA AGTGTCGAAC TCGGTTTCGA 121 CGGTTCGCCC GGCGGCATCG ATCCGCTCCA GCCGTCCCTC TTTTTCCGAG GAAACCCATT 181 CCACCATGTC TCCGTAGAGT TCGCGCCAGG CCTGCTCGAA TGCCGGCTGC TTGGAAAACC 241 GGGTCTTGGC ATCGAGTATC AGGATCTTGG AGCGGGGTTT GTGCTTTTTC ATGTAATAGG 301 CGATGAGCGA GGCGCGTTCG TAGGGGCCAG GCGGACAACG ATAGGGAGAA GGTGGTGCGG 361 TGATCAGAAC CAGGCCGCCG TCGGGCATCG CGCGGATCTG CCGGGCGAGC AAAGCGGTCT 421 GGGGGCCGGC CTTCCAGGCA TGGGGAACGA ACCGGCTCGC CGCCTCGTCA TAGCCCATTA 481 TCGCCTCCCA GCGAAAGTCG ATGCCGGGGG AGAGGACTAG CCTGTCGTAG GTGACCTCGG 541 CACCATCGTT CAGTATCACA CGCCGTCGCT GCCGATCCAG GCGGGCGACC CGGGCGGTTA 601 CCTTTCCTAT ATCCAGCTCC CGCCGCAACC AGTCATAAGA CCGCGCGAGA GTCCCCATGT 661 CCCCGAGGCC GGCGACTGCT TCGTTCGATC CGGGGCAGGA AAGATAAGTC TCCTGCGGTT 721 CGATCAATGT GATCGTCAGA CCGGGATTCA TCTGCTTCAG ATAGCGGGCG GCCGTGGCAC 781 CGCCATAACC GCCACCGACG ACCACGACCG GCCGCCTGAG CGAAGACCTT GCCGGAAGCG 841 CAGCCACCCA GTCCCAGCCC ATGCCGCAGA CTGCCAGCAG GCGCAGGAAC CGCCGCCGGC 901 GGATCATGGC CTCTTTCCCA GAAAGCCGGC GATCGCTTCA ATGTCCTGGT TCGTGAGTCC 961 GGCAGCGATC CGGTTCATGA CCGTGCCCGA TCTTTTTCCT TCACGGTACT CCCGCAACAG 1021 AGATGCCATC TCCTTCGCAT CGAAGCGGCG TAACGATGCC GGTTCGGGAA TCTGCTCCTC 1081 CTCGTCGGCA TGGCAGCCGA GGCAACCGAG CGCAGCCAAA ACCATGTCCG GTTTTTCGGC 1141 CCGGGCCGGG AAACAGAGAG CGACAGCGAT CGTACCGATC TGGACGCGTC TCACAAACGA 1201 CAAAACGTCA CGATGGGTGT TCGGTAGCTG AGTCACGGGG ATTTGTAGAA GTATAGGACC 1261 GACGGATTTT ATGCAAGCAT GTCGCTTTGA CCAAGCCGGG ATTCCATGGA AGGGATGTCA 1321 TCGGGAGAGT TATTTATGTC GTTGATTTAT AAGAAACTAC CCCTGCGTCA AAATGTCGCA 1381 GATTTTTCTT GACAGTTTGG GGGAGGGTGA TAGATCCTCC ACCGATGGAC CGGTACCGCC 1441 TCTGTTGCGG GGTCCATGAA ATGCCCGTTA GAGGCAGAAC CGATAGGGAA TTAGAGAAGO 1501 GGGCGTCGGC GCCGAATGCC GGCCCCTGTC AACCATCACT TTAGGAGGAA CAAACAATGG 1561 CAGCAACAAC CATTGGTGGT GCAGCTGCGG CGTAAGCGCC GCTGCTGGAC AAGAAGTGGC 1621 TCACGTTCGC ACTGGCGATT TACACCGTGT TCTACCTGTG GGTGCGGTGG TACGAAGGTG 1681 TCTATGGCTG GTCCGCCGGA CTGGACTCGT TCGCGCCGGA GTTCGAGACC TACTGGATGA 1741 ATTTCCTGTA CACCGAGATC GTCCTGGAGA TCGTGACGGC TTCGATCCTG TGGGGCTATC 1801 TCTGGAAGAC CCGCGACCGC AACCTGGCCG CGCTGACCCC GCGTGAAGAG CTGCGCCGCA 1861 ACTTCACCCA CCTGGTGTGG CTGGTGGCCT ACGCCTGGGC CATCTACTGG GGCGCATCCT 1921 ACTTCACCGA GCAGGACGGC ACCTGGCATC AGACGATCGT GCGCGACACC GACTTCACGC 1981 CGTCGCACAT CATCGAGTTC TATCTGAGCT ACCCGATCTA CATCATCACC GGTTTTGCGG 2041 CGTTCATCTA CGCCAAGACG CGTCTGCCGT TCTTCGCGAA GGGCATCTCG CTGCCGTACC 2101 TGGTGCTGGT GGTGGGTCCG TTCATGATTC TGCCGAACGT GGGTCTGAAC GAATGGGGCC 2161 ACACCTTCTG GTTCATGGAA GAGCTGTTCG TGGCGCCGCT GCACTACGGC TTCGTGATCT
Determine the inheritance pattern of each of the following pedigrees Then label the genotypes of each individual in the pedigrees Recessive Ar Here R
Biology
Principles of Inheritance & Variation (Genetics)
Determine the inheritance pattern of each of the following pedigrees Then label the genotypes of each individual in the pedigrees Recessive Ar Here R
their respective processes For 9 Explain why a fifteen nucleotide sequence that includes a stop codon would only code for four amino acids DNA play in translation How is it different from mRNA in its function and structure
Biology
Principles of Inheritance & Variation (Genetics)
their respective processes For 9 Explain why a fifteen nucleotide sequence that includes a stop codon would only code for four amino acids DNA play in translation How is it different from mRNA in its function and structure
Paragraph about result explanation trend of Figure 1 and Figure 2 1 2 about gel image Figure 1 1 2 sentences about standard curve Figure 2 2000 bp 1500 bp 1000 bp 700 bp 500 bp 400 bp 300 bp 200 bp 75 bp DNA Ladder GMO GMO Food sample GMO GMO Food sample
Biology
Principles of Inheritance & Variation (Genetics)
Paragraph about result explanation trend of Figure 1 and Figure 2 1 2 about gel image Figure 1 1 2 sentences about standard curve Figure 2 2000 bp 1500 bp 1000 bp 700 bp 500 bp 400 bp 300 bp 200 bp 75 bp DNA Ladder GMO GMO Food sample GMO GMO Food sample
is the condition presented in the pedigree below autosomal dominant autosomal recessive or sex linked OPKU heterozygote carrier PKU homozygote 100 66 O sex linked recessive O Not enough information to determine O autosomal recessive O autosomal dominant
Biology
Principles of Inheritance & Variation (Genetics)
is the condition presented in the pedigree below autosomal dominant autosomal recessive or sex linked OPKU heterozygote carrier PKU homozygote 100 66 O sex linked recessive O Not enough information to determine O autosomal recessive O autosomal dominant
List the geologic layers from oldest to youngest B 1 F H C E GI BI 3 J A I D
Biology
Principles of Inheritance & Variation (Genetics)
List the geologic layers from oldest to youngest B 1 F H C E GI BI 3 J A I D
You are called in to track outbreaks of a new disease called simpox in two neighboring towns Springfield and South Park Simpox is spread by direct contact between people in a similar manner as with influenza Over a 40 week period you track the number of people who are infected with the disease The curves for the two outbreaks are shown in the graph below Which of the following is plausible explanation for the difference in the course of the epidemic between the two towns Number of Infected Individuals Springfield South Park Time weeks A Springfield has a lower population density than South Park B Springfield residents wash their hands less frequently than do South Park residents C Springfield has better access to a new simpox vaccine than South Park has D Springfield s residents come into contact with each other less often than South Park s do
Biology
Principles of Inheritance & Variation (Genetics)
You are called in to track outbreaks of a new disease called simpox in two neighboring towns Springfield and South Park Simpox is spread by direct contact between people in a similar manner as with influenza Over a 40 week period you track the number of people who are infected with the disease The curves for the two outbreaks are shown in the graph below Which of the following is plausible explanation for the difference in the course of the epidemic between the two towns Number of Infected Individuals Springfield South Park Time weeks A Springfield has a lower population density than South Park B Springfield residents wash their hands less frequently than do South Park residents C Springfield has better access to a new simpox vaccine than South Park has D Springfield s residents come into contact with each other less often than South Park s do
1 20 points Many species of bacteria are becoming resistant to antibiotics Recently studies have analyzed the chromosomal DNA of antibiotic resistant bacteria One bacteria studied is methicillin resistant Staphylococcus aureus MRSA Researchers have found that its chromosomal DNA does not contain the genes for antibiotic resistance They also found that when MRSA is cultured with a non resistant bacteria the non resistant bacteria becomes antibiotic resistant a Identify the location of the antibiotic resistant genes h Discuss how the antibiotic resistant genes are being transmitted to non resistant bacteria when cultured together
Biology
Principles of Inheritance & Variation (Genetics)
1 20 points Many species of bacteria are becoming resistant to antibiotics Recently studies have analyzed the chromosomal DNA of antibiotic resistant bacteria One bacteria studied is methicillin resistant Staphylococcus aureus MRSA Researchers have found that its chromosomal DNA does not contain the genes for antibiotic resistance They also found that when MRSA is cultured with a non resistant bacteria the non resistant bacteria becomes antibiotic resistant a Identify the location of the antibiotic resistant genes h Discuss how the antibiotic resistant genes are being transmitted to non resistant bacteria when cultured together
critical immunization threshold The critical immunization threshold symbolized pe is the minimum proportion of a population that must be individually immune to a disease in order to achieve herd immunity If the entire population is susceptible pe is the proportion to vaccinate to prevent the spread of disease The formula for p is Pc 1 1 Ro where Ro is the basic reproductive number Click to see additional instructions Imagine that a new deadly coronavirus arises and starts a global pandemic Experts are worried because the disease spreads easily having a basic reproductive number Ro of 8 The good news is that an effective vace kly developed What minimum proportion of the population po would need to be vaccinated to ensure that the disease can no longer spread Round your answer to two decimal places
Biology
Principles of Inheritance & Variation (Genetics)
critical immunization threshold The critical immunization threshold symbolized pe is the minimum proportion of a population that must be individually immune to a disease in order to achieve herd immunity If the entire population is susceptible pe is the proportion to vaccinate to prevent the spread of disease The formula for p is Pc 1 1 Ro where Ro is the basic reproductive number Click to see additional instructions Imagine that a new deadly coronavirus arises and starts a global pandemic Experts are worried because the disease spreads easily having a basic reproductive number Ro of 8 The good news is that an effective vace kly developed What minimum proportion of the population po would need to be vaccinated to ensure that the disease can no longer spread Round your answer to two decimal places
homology and biogeography Sort the following examples into the correct categories The evidence exists in a variety of categories including direct observation of evolutionary change the fossil Drag each phrase to the appropriate bin View Available Hint s direct observation of evolutionary change development of drug resistance in bacteria fossil record discovery of shells of extinct species discovery of transitional forms of horses homology similarities in mammalian forelimbs same genetic code in fireflies and tobacco plants vestigial pelvis in right whales biogeography Reset Help the high concentration of marsupial species in Australia similarity of endemic island species to nearby mainland species
Biology
Principles of Inheritance & Variation (Genetics)
homology and biogeography Sort the following examples into the correct categories The evidence exists in a variety of categories including direct observation of evolutionary change the fossil Drag each phrase to the appropriate bin View Available Hint s direct observation of evolutionary change development of drug resistance in bacteria fossil record discovery of shells of extinct species discovery of transitional forms of horses homology similarities in mammalian forelimbs same genetic code in fireflies and tobacco plants vestigial pelvis in right whales biogeography Reset Help the high concentration of marsupial species in Australia similarity of endemic island species to nearby mainland species
12 Multiple Choice Questions 12 14 refer to the pedigree below The trait being represented is caused by a rare genetic mutation that results in severe weakening of muscles with aging males squares females circles shaded in affected The individuals in generation I have been labeled as 1 9 I I OTE III TO For this pedigree how does the tran seem to be Inherked AS a Dominant Trait As a Recessive Trait As an X Link Dominant Tran 19
Biology
Principles of Inheritance & Variation (Genetics)
12 Multiple Choice Questions 12 14 refer to the pedigree below The trait being represented is caused by a rare genetic mutation that results in severe weakening of muscles with aging males squares females circles shaded in affected The individuals in generation I have been labeled as 1 9 I I OTE III TO For this pedigree how does the tran seem to be Inherked AS a Dominant Trait As a Recessive Trait As an X Link Dominant Tran 19
10 For each of the following problems using the information below determine the parent genotypes draw the Punnett square and provide the possible genotypes and phenotypes of the offspring Each question is worth 5 points 1pt for parent genotypes 1pt for offspring genotypes 1pt for offspring phenotypes 2pts for correct and complete Punnett square In mice the ability to run normally is a dominant trait Mice with this trait are called running mice R The recessive trait causes mice to run in circles only Mice with this trait are called waltzing mice r Hair color is also inherited in mice Black hair B is dominant over brown hair b a Cross a heterozygous running brown mouse with a heterozygous running homozygous black mouse Parents Genotypes Phenotypes b Cross a heterozygous running heterozygous black mouse with a heterozygous running heterozygous black mouse Parents Genotypes Phenotypes
Biology
Principles of Inheritance & Variation (Genetics)
10 For each of the following problems using the information below determine the parent genotypes draw the Punnett square and provide the possible genotypes and phenotypes of the offspring Each question is worth 5 points 1pt for parent genotypes 1pt for offspring genotypes 1pt for offspring phenotypes 2pts for correct and complete Punnett square In mice the ability to run normally is a dominant trait Mice with this trait are called running mice R The recessive trait causes mice to run in circles only Mice with this trait are called waltzing mice r Hair color is also inherited in mice Black hair B is dominant over brown hair b a Cross a heterozygous running brown mouse with a heterozygous running homozygous black mouse Parents Genotypes Phenotypes b Cross a heterozygous running heterozygous black mouse with a heterozygous running heterozygous black mouse Parents Genotypes Phenotypes
9 In horses black is dependent upon a dominant gene B and chestnut upon its recessive allele b The trotting gait is due to a dominant gene T the pacing gait to its recessive allele t If a homozygous black pacer mates to a homozygous chestnut trotter what will be the genotypes and phenotypes of the offspring BB x bb TT CON CER Parents Genotypes Phenotypes white for nich brown byt for 11 BLU BLU 16 0 40 FD 50 4 5 re similar write b
Biology
Principles of Inheritance & Variation (Genetics)
9 In horses black is dependent upon a dominant gene B and chestnut upon its recessive allele b The trotting gait is due to a dominant gene T the pacing gait to its recessive allele t If a homozygous black pacer mates to a homozygous chestnut trotter what will be the genotypes and phenotypes of the offspring BB x bb TT CON CER Parents Genotypes Phenotypes white for nich brown byt for 11 BLU BLU 16 0 40 FD 50 4 5 re similar write b
lete the following dihybrid crosses Using the information provided in each problem and the steps d below answer the questions of this worksheet to the best of your ability Each question is worth 5 ints 1 pt for parent genotypes 1pt for offspring genotypes 1pt for offspring phenotypes 2pts for correct and complete Punnett square 8 In mammoths assume that spotted skin N is dominant over non spotted skin n and that wooly hair H is dominant over non wooly hair h If a heterozygous spotted non wooly mammoth is crossed with a non spotted heterozygous wooly haired mammoth what will be the genotypes and phenotypes of the offspring hh nh Parents N Genotypes Phenotypes x nn Hh Nh Nh nn nn nn N Wh Wh An h NN hb NN hh n h NnhH h NWNhh NNhh NNth Nahh n NN Hh Nnhh NnHh Nnth Nnth W nhn Nnh h hnith nn hn 11 F 2 4 2 2
Biology
Principles of Inheritance & Variation (Genetics)
lete the following dihybrid crosses Using the information provided in each problem and the steps d below answer the questions of this worksheet to the best of your ability Each question is worth 5 ints 1 pt for parent genotypes 1pt for offspring genotypes 1pt for offspring phenotypes 2pts for correct and complete Punnett square 8 In mammoths assume that spotted skin N is dominant over non spotted skin n and that wooly hair H is dominant over non wooly hair h If a heterozygous spotted non wooly mammoth is crossed with a non spotted heterozygous wooly haired mammoth what will be the genotypes and phenotypes of the offspring hh nh Parents N Genotypes Phenotypes x nn Hh Nh Nh nn nn nn N Wh Wh An h NN hb NN hh n h NnhH h NWNhh NNhh NNth Nahh n NN Hh Nnhh NnHh Nnth Nnth W nhn Nnh h hnith nn hn 11 F 2 4 2 2
The purpose of the reaction buffer Tris HC1 KC1 and MgCl in the PCR mixture is to O a Provide the DNA template for Taq polymerase to bind to Ob Provide a stable environment pH ions cofactors for Taq polymerase O C Allow conductivity of DNA to move from the cathode to the anode d It includes the dNTPS
Biology
Principles of Inheritance & Variation (Genetics)
The purpose of the reaction buffer Tris HC1 KC1 and MgCl in the PCR mixture is to O a Provide the DNA template for Taq polymerase to bind to Ob Provide a stable environment pH ions cofactors for Taq polymerase O C Allow conductivity of DNA to move from the cathode to the anode d It includes the dNTPS
An amplicon is a PCR product and the DNA to be amplified is the template True False Question 4 1 point Listen Positive and negative controls are necessary in experiments because they O a provide a standard for the outcomes of the experimental sample being tested Ob add additional samples to the experiment O c make the experiment easier to complete
Biology
Principles of Inheritance & Variation (Genetics)
An amplicon is a PCR product and the DNA to be amplified is the template True False Question 4 1 point Listen Positive and negative controls are necessary in experiments because they O a provide a standard for the outcomes of the experimental sample being tested Ob add additional samples to the experiment O c make the experiment easier to complete
A garden pea with a purple flower is crossed with a garden pea with a white 1 poin flower and produces offspring all of which have purple flowers The offspring then self pollinate Based on the results of Mendel s experiment out of 1000 garden pea plants in the F2 generation approximately how many plants would have white flowers 250 plants 750 plants 0 plants 1000 plants
Biology
Principles of Inheritance & Variation (Genetics)
A garden pea with a purple flower is crossed with a garden pea with a white 1 poin flower and produces offspring all of which have purple flowers The offspring then self pollinate Based on the results of Mendel s experiment out of 1000 garden pea plants in the F2 generation approximately how many plants would have white flowers 250 plants 750 plants 0 plants 1000 plants
Listen This figure showcases haplodiploidy in honeybees In this case the female worker bee shown is noted on the figure as Self Use the figure below to select which of the following apply to the figure Brother 25 Full sister 75 Nephew 375 Mother Father Mate Niece 375 Self Son 5 Mate Daughter 5 Each daughter of a malentains all his genes Each father will contribute at least half of its genes to its son that will then transport one fourth of its genes to its grandson The relatedness of the female worker bee Self to a brother is only 25 because the brother is fatherless Full sisters relatedness is 0 75 to the female worker bee Self because full sisters receive 75 of their genome from the same father Females are diploid and come from fertilized eggs while males are haploid and come from unfertilized eggs
Biology
Principles of Inheritance & Variation (Genetics)
Listen This figure showcases haplodiploidy in honeybees In this case the female worker bee shown is noted on the figure as Self Use the figure below to select which of the following apply to the figure Brother 25 Full sister 75 Nephew 375 Mother Father Mate Niece 375 Self Son 5 Mate Daughter 5 Each daughter of a malentains all his genes Each father will contribute at least half of its genes to its son that will then transport one fourth of its genes to its grandson The relatedness of the female worker bee Self to a brother is only 25 because the brother is fatherless Full sisters relatedness is 0 75 to the female worker bee Self because full sisters receive 75 of their genome from the same father Females are diploid and come from fertilized eggs while males are haploid and come from unfertilized eggs
8 In pea plants purple P flower color is dominant to white p flowers and yellow Y seeds are dominant to green y seeds Flower color and seed color are inherited independently A cross is done between two plants of unknown genotypes and the results of the F1 generation are analyzed Purple yellow 0 Purple green Purple yellow 0 155 You notice right away that all offspring are white You then notice that the yellow green ratio is approximately 3 1 Knowing this determine the parental genotypes White yellow I 9 A second cross is done between two plants of unknown genotypes following the same traits as in problem 8 The results of the F1 generation are in the table below Purple green 321 146 Show work here Py PY py py Py Pypy PyPy pypy pypy Whitesgroen By By PY PYPY by PY PYPY 101 White yellow 52 White green 49 The results for this appear to be more complicated but you know that the ratios between purple white flowers and yellow green seeds are the key to solving the parental genotypes Use problem 8 to help you if you get stuck py Pypy Pypy pypy pley by Tryp YLPYPY HEAR YpYp Ple Show work here pY4y
Biology
Principles of Inheritance & Variation (Genetics)
8 In pea plants purple P flower color is dominant to white p flowers and yellow Y seeds are dominant to green y seeds Flower color and seed color are inherited independently A cross is done between two plants of unknown genotypes and the results of the F1 generation are analyzed Purple yellow 0 Purple green Purple yellow 0 155 You notice right away that all offspring are white You then notice that the yellow green ratio is approximately 3 1 Knowing this determine the parental genotypes White yellow I 9 A second cross is done between two plants of unknown genotypes following the same traits as in problem 8 The results of the F1 generation are in the table below Purple green 321 146 Show work here Py PY py py Py Pypy PyPy pypy pypy Whitesgroen By By PY PYPY by PY PYPY 101 White yellow 52 White green 49 The results for this appear to be more complicated but you know that the ratios between purple white flowers and yellow green seeds are the key to solving the parental genotypes Use problem 8 to help you if you get stuck py Pypy Pypy pypy pley by Tryp YLPYPY HEAR YpYp Ple Show work here pY4y
One of the most common genetic inherited diseases in America is Cystic Fibrosis CF CF is a chronic disease that affects the lungs and digestive systems for which there currently is no cure Every person has two copies of the CFTR Cystic Fibrosis Transmembrane conductance Regulator gene one from each parent For a person to have CF both parents must be carriers of mutated CFTR The CFTR gene contains the instructions for making the CFTR protein which is produced in many organs among them the lungs and the pancreas The CFTR protein creates channels in the cell membrane to allow the movement of chloride ions in and out of the cell When the CFTR protein functions properly the balance of chloride and fluid at the cell surface remains normal The mutated version of the CFTR gene causes the CFTR protein to malfunction leading to a buildup of thick mucus especially in the lungs This leads to lung infections and eventually respiratory failure caused by a microbial biofilm Pseudomonas aeruginosa is especially problematic in CF patients Scientists have found more than 1 700 different mutations in the CFTR gene that can cause CF which accounts for the fact that this genetic disease is so common The FDA recently approved the drug Ivacaftor for treatment of splice mutations in CFTR but people with these types of mutations make a small amount of normal CFTR Ivacaftor forces the gate on the normal CFTR protein to stay open for longer so that the channels can function with a reduced amount of CFTR protein in the membrane 1 Explain how a nonsense mutation in CFTR gene can cause CF 2 What would be the effect of a mutation in the intron of the CFTR gene 3 Speculate on the role of Ivacaftor in the treatment of CF
Biology
Principles of Inheritance & Variation (Genetics)
One of the most common genetic inherited diseases in America is Cystic Fibrosis CF CF is a chronic disease that affects the lungs and digestive systems for which there currently is no cure Every person has two copies of the CFTR Cystic Fibrosis Transmembrane conductance Regulator gene one from each parent For a person to have CF both parents must be carriers of mutated CFTR The CFTR gene contains the instructions for making the CFTR protein which is produced in many organs among them the lungs and the pancreas The CFTR protein creates channels in the cell membrane to allow the movement of chloride ions in and out of the cell When the CFTR protein functions properly the balance of chloride and fluid at the cell surface remains normal The mutated version of the CFTR gene causes the CFTR protein to malfunction leading to a buildup of thick mucus especially in the lungs This leads to lung infections and eventually respiratory failure caused by a microbial biofilm Pseudomonas aeruginosa is especially problematic in CF patients Scientists have found more than 1 700 different mutations in the CFTR gene that can cause CF which accounts for the fact that this genetic disease is so common The FDA recently approved the drug Ivacaftor for treatment of splice mutations in CFTR but people with these types of mutations make a small amount of normal CFTR Ivacaftor forces the gate on the normal CFTR protein to stay open for longer so that the channels can function with a reduced amount of CFTR protein in the membrane 1 Explain how a nonsense mutation in CFTR gene can cause CF 2 What would be the effect of a mutation in the intron of the CFTR gene 3 Speculate on the role of Ivacaftor in the treatment of CF
Leucine Zipper shown below is an example of a coiled coil A typical Leucine Zipper consists of two a helixes with leucine residues in about 7 residue intervals Shown below is one of those intervals In Biochemistry mutations often are done to study the effect of changing residues in a protein Electrostatic reactions are where a positively charged side chain interacts with a negatively charged side chain like the one shown below lysine s positive side chain interacting with glutamate s negative side chain 0 Potential cons Potential electrostatic interactions Gly f Instructions Prompt 1 Explain what principles are involved in holding the two a helices together Suggest one possible residue that could replace leucine as indicated by the red star and maintain the hydrophobic core shown below and suggest one possible residue that would destabilize the core Please explain your reasoning Glu b Glu Class ww d Hydrophobic core Potential electrostatic interactions Potential electrostatic Lys Glu Gly
Biology
Principles of Inheritance & Variation (Genetics)
Leucine Zipper shown below is an example of a coiled coil A typical Leucine Zipper consists of two a helixes with leucine residues in about 7 residue intervals Shown below is one of those intervals In Biochemistry mutations often are done to study the effect of changing residues in a protein Electrostatic reactions are where a positively charged side chain interacts with a negatively charged side chain like the one shown below lysine s positive side chain interacting with glutamate s negative side chain 0 Potential cons Potential electrostatic interactions Gly f Instructions Prompt 1 Explain what principles are involved in holding the two a helices together Suggest one possible residue that could replace leucine as indicated by the red star and maintain the hydrophobic core shown below and suggest one possible residue that would destabilize the core Please explain your reasoning Glu b Glu Class ww d Hydrophobic core Potential electrostatic interactions Potential electrostatic Lys Glu Gly
Summer squash plants with the dominant allele C bear white fruit whereas plants homozygous for the recessive allele c bear colored fruit Summer squash have a second locus that acts as a modifier gene if the plant is colored IF the plant is colored and is G at the second locus they will be yellow If they are colored and gg at the second locus they will be green You cross a double heterozygous plant and a double homozygous recessive plant CcGg x ccgg What will be the percentage of white yellow and green squash plants in this case 20 20 10 O 50 25 25 O 40 25 25 75 30 30
Biology
Principles of Inheritance & Variation (Genetics)
Summer squash plants with the dominant allele C bear white fruit whereas plants homozygous for the recessive allele c bear colored fruit Summer squash have a second locus that acts as a modifier gene if the plant is colored IF the plant is colored and is G at the second locus they will be yellow If they are colored and gg at the second locus they will be green You cross a double heterozygous plant and a double homozygous recessive plant CcGg x ccgg What will be the percentage of white yellow and green squash plants in this case 20 20 10 O 50 25 25 O 40 25 25 75 30 30
Can you please make dna tree from olive babion Papio anybis mouse Gerbil meriones unguiculatus brown rat Dog chicken Striped hyena hyena hvena domestic cat delis catus western lowland gorilla human
Biology
Principles of Inheritance & Variation (Genetics)
Can you please make dna tree from olive babion Papio anybis mouse Gerbil meriones unguiculatus brown rat Dog chicken Striped hyena hyena hvena domestic cat delis catus western lowland gorilla human
purple P flower color is dominant to white p flowers and yellow Y seeds are dominant to green y seeds Flower color and seed color are inherited independently A cross is done between two plants of unknown genotypes and the results of the F generation are analyzed Purple yellow 0 Purple green White yellow 0 321 White green 101 You notice right away that all offspring are white You then notice that the yellow green ratio is approximately 3 1 Knowing this determine the parental genotypes
Biology
Principles of Inheritance & Variation (Genetics)
purple P flower color is dominant to white p flowers and yellow Y seeds are dominant to green y seeds Flower color and seed color are inherited independently A cross is done between two plants of unknown genotypes and the results of the F generation are analyzed Purple yellow 0 Purple green White yellow 0 321 White green 101 You notice right away that all offspring are white You then notice that the yellow green ratio is approximately 3 1 Knowing this determine the parental genotypes
Brown eyes are a dominant eye color allele and blue eyes are recessive A brown eyed woman whose father had blue eyes and whose mother had brown eyes marries a brown eyed man whose parents are also brown eyed They have a son who is blue eyed Draw a pedigree info above showing all four grandparents the two parents and the son Indicate each individuals possible genotypes
Biology
Principles of Inheritance & Variation (Genetics)
Brown eyes are a dominant eye color allele and blue eyes are recessive A brown eyed woman whose father had blue eyes and whose mother had brown eyes marries a brown eyed man whose parents are also brown eyed They have a son who is blue eyed Draw a pedigree info above showing all four grandparents the two parents and the son Indicate each individuals possible genotypes
1 Select on the response boxes and drag them to the correct position in the list 2 Select Check Answer 3 Select Next to proceed Arrange the steps in the order in which you proceed through the lesson Review all content before completing the Evaluation section Read all procedures in the experiment to ensure you are prepared for your hands on tasks Check that you have gathered all materials required to perform your experiment Review all data tables to ensure you have completed the quired areas of the experimentation section tructions and questions
Biology
Principles of Inheritance & Variation (Genetics)
1 Select on the response boxes and drag them to the correct position in the list 2 Select Check Answer 3 Select Next to proceed Arrange the steps in the order in which you proceed through the lesson Review all content before completing the Evaluation section Read all procedures in the experiment to ensure you are prepared for your hands on tasks Check that you have gathered all materials required to perform your experiment Review all data tables to ensure you have completed the quired areas of the experimentation section tructions and questions
Set up a Punnett square using the following information Dominate allele for normal coat color in wolves N Recessive allele for black coat color in wolves n Dominant allele for brown eyes B Recessive allele for blue eyes b Cross a heterozygous dominant parent with a homozygous recessive parent X 150
Biology
Principles of Inheritance & Variation (Genetics)
Set up a Punnett square using the following information Dominate allele for normal coat color in wolves N Recessive allele for black coat color in wolves n Dominant allele for brown eyes B Recessive allele for blue eyes b Cross a heterozygous dominant parent with a homozygous recessive parent X 150
Set up a Punnett square using the following information Dominate allele for tall plants D Recessive allele for dwarf plants d Dominate allele for purple flowers W Recessive allele for white flowers w Cross a homozygous dominant parent with a homozygous recessive pare X
Biology
Principles of Inheritance & Variation (Genetics)
Set up a Punnett square using the following information Dominate allele for tall plants D Recessive allele for dwarf plants d Dominate allele for purple flowers W Recessive allele for white flowers w Cross a homozygous dominant parent with a homozygous recessive pare X
The Hardy Weinberg law is a logical consequence of Mendel s first law of segregation af expresses the tendency toward equilibrium inherent in Mendelian heredity Let us select for our example a population having a single locus bearing just two alleles T and r The phenotypic expression of this gene might be for example the ability to taste a chemical compound called phenylthiocarba mide Individuals in the population will be of three genotypes for this locus T T T t both tasters and t t nontasters In a sample of 100 individuals let us suppose that we have 20 of T T genotype 40 of T t genotype and 40 of t t genotype We could then make a table showing the allelic frequencies remember that every individual has two copies of the gene Genotype T T T t 1 1 Total Number of Individuals 20 40 40 100 Sperm T 0 4 1 0 6 Copies of the T Allele 40 40 0 80 Of the 200 copies the proportion of the T allele is 80 200 0 4 40 and the proportion of the rallele is 120 200 0 6 60 It is customary to use p and q to represent the two allelic frequencies The genetically dominant allele is represented by p and the genetically recessive by q Thus T 0 4 TIT 0 16 Tlt 0 24 p frequency of 7 0 4 q frequency of 1 0 6 Therefore p q 1 Copies of the t Allele 0 40 80 120 Having calculated allelic frequencies in the sample let us determine whether these frequencies will change spontaneously in a new genera tion of the population Assuming that mating is random gametes are sampled independently in pairs each individual contributes an equal number of gametes to the common pool from which the next genera tion is formed Frequencies of gametes in the pool then equal the al lelic frequencies in the sample 40 of the gametes are T and 60 are r ratio of 0 4 0 6 Both ova and sperm of course show the same frequen cies The next generation is formed hic05193 cho vg3 136 indd 129 1 0 6 Tlt 0 24 tt 0 36 contain copies of the allele The term inclusive fitness pertains to cases where the average effect of an allele on its own propagation in the gene pool would be calculated incorrectly if its effects on sur vival of its possessors relatives were ignored For example natural selection might favor an allele for a behavior through which an indi vidual dies in saving the lives of many relatives Some traits and combinations of traits are advantageous for Collecting genotypes we have frequency of T T 0 16 frequency of T t 0 48 frequency of t t 0 36 Next we determine the values of p and q from the randomly mated pop ulations From the table above we see that the frequency of T is the sum of genotypes T T which is 0 16 and one half of the genotype T t which is 0 24 T p 0 16 5 0 48 0 4 Similarly the frequency of t is the sum of genotypes 1 t which is 0 36 and one half the genotype T t which is 0 24 t q 0 36 5 0 48 0 6 The new generation bears exactly the same allelic frequencies as the parent population Note that there has been no increase in the frequency of the genetically dominant allele T Thus in a freely interbreeding sexually reproducing population the frequency of each allele would re main constant generation after generation in the absence of natural se lection migration recurring mutation and genetic drift see text A mathematically minded reader should recognize that the genotype fre quencies T T T t and t t are actually a binomial expansion of p q p q p 2pq q 1 Note that the equilibrium calculations give expected frequencies which are unlikely to be realized exactly in a population of finite size For this reason finite population size is a cause of evolutionary change Most genes have more than just a single pair of alleles especially when we measure genetic variation at the DNA sequence level The bi nomial expansion shown above can be used for any number of alleles Suppose that we have three alleles T T T whose frequencies are denoted p q and r respectively We now have six possible genotypes with the following Hardy Weinberg equilibrium frequencies O TT TT TT TT TT TT p q r p 2pg q 2 pr 2gr As the number of alleles at a gene increases the proportion of the popu lation having heterozygous genotypes also increases disadvantageous for others Darwin used the term sexual selection to denote the selection of traits that are advantageous for obtaining mates but not for survival Bright colors and elaborate feathers can enhance a male bird s competitive ability in obtaining mates while simultaneously increasing his visibility to predators see Figure 6 31 Environmental changes such as extinction of a predator population can alter the selective values of alternative traits The action of selec
Biology
Principles of Inheritance & Variation (Genetics)
The Hardy Weinberg law is a logical consequence of Mendel s first law of segregation af expresses the tendency toward equilibrium inherent in Mendelian heredity Let us select for our example a population having a single locus bearing just two alleles T and r The phenotypic expression of this gene might be for example the ability to taste a chemical compound called phenylthiocarba mide Individuals in the population will be of three genotypes for this locus T T T t both tasters and t t nontasters In a sample of 100 individuals let us suppose that we have 20 of T T genotype 40 of T t genotype and 40 of t t genotype We could then make a table showing the allelic frequencies remember that every individual has two copies of the gene Genotype T T T t 1 1 Total Number of Individuals 20 40 40 100 Sperm T 0 4 1 0 6 Copies of the T Allele 40 40 0 80 Of the 200 copies the proportion of the T allele is 80 200 0 4 40 and the proportion of the rallele is 120 200 0 6 60 It is customary to use p and q to represent the two allelic frequencies The genetically dominant allele is represented by p and the genetically recessive by q Thus T 0 4 TIT 0 16 Tlt 0 24 p frequency of 7 0 4 q frequency of 1 0 6 Therefore p q 1 Copies of the t Allele 0 40 80 120 Having calculated allelic frequencies in the sample let us determine whether these frequencies will change spontaneously in a new genera tion of the population Assuming that mating is random gametes are sampled independently in pairs each individual contributes an equal number of gametes to the common pool from which the next genera tion is formed Frequencies of gametes in the pool then equal the al lelic frequencies in the sample 40 of the gametes are T and 60 are r ratio of 0 4 0 6 Both ova and sperm of course show the same frequen cies The next generation is formed hic05193 cho vg3 136 indd 129 1 0 6 Tlt 0 24 tt 0 36 contain copies of the allele The term inclusive fitness pertains to cases where the average effect of an allele on its own propagation in the gene pool would be calculated incorrectly if its effects on sur vival of its possessors relatives were ignored For example natural selection might favor an allele for a behavior through which an indi vidual dies in saving the lives of many relatives Some traits and combinations of traits are advantageous for Collecting genotypes we have frequency of T T 0 16 frequency of T t 0 48 frequency of t t 0 36 Next we determine the values of p and q from the randomly mated pop ulations From the table above we see that the frequency of T is the sum of genotypes T T which is 0 16 and one half of the genotype T t which is 0 24 T p 0 16 5 0 48 0 4 Similarly the frequency of t is the sum of genotypes 1 t which is 0 36 and one half the genotype T t which is 0 24 t q 0 36 5 0 48 0 6 The new generation bears exactly the same allelic frequencies as the parent population Note that there has been no increase in the frequency of the genetically dominant allele T Thus in a freely interbreeding sexually reproducing population the frequency of each allele would re main constant generation after generation in the absence of natural se lection migration recurring mutation and genetic drift see text A mathematically minded reader should recognize that the genotype fre quencies T T T t and t t are actually a binomial expansion of p q p q p 2pq q 1 Note that the equilibrium calculations give expected frequencies which are unlikely to be realized exactly in a population of finite size For this reason finite population size is a cause of evolutionary change Most genes have more than just a single pair of alleles especially when we measure genetic variation at the DNA sequence level The bi nomial expansion shown above can be used for any number of alleles Suppose that we have three alleles T T T whose frequencies are denoted p q and r respectively We now have six possible genotypes with the following Hardy Weinberg equilibrium frequencies O TT TT TT TT TT TT p q r p 2pg q 2 pr 2gr As the number of alleles at a gene increases the proportion of the popu lation having heterozygous genotypes also increases disadvantageous for others Darwin used the term sexual selection to denote the selection of traits that are advantageous for obtaining mates but not for survival Bright colors and elaborate feathers can enhance a male bird s competitive ability in obtaining mates while simultaneously increasing his visibility to predators see Figure 6 31 Environmental changes such as extinction of a predator population can alter the selective values of alternative traits The action of selec
Match the term with the correct definition Genotype Phenotype Homozygous genotype Heterozygous genotype Dominant phenotype Recessive phenotype Testcross Choose two sets of chromosomes two different alleles make up the genotype Crossing of a pure breeding recessive with an unknown genotype phenotype observed in a heterozygous organism that is identical to the phenotype observed in a homozyg two alleles carried by an organism that are identical observable traits of an organism genetic constitution of an organism phenotype observed in an organism that is homozygous for the recessive allele Choose Choose Choose Choose
Biology
Principles of Inheritance & Variation (Genetics)
Match the term with the correct definition Genotype Phenotype Homozygous genotype Heterozygous genotype Dominant phenotype Recessive phenotype Testcross Choose two sets of chromosomes two different alleles make up the genotype Crossing of a pure breeding recessive with an unknown genotype phenotype observed in a heterozygous organism that is identical to the phenotype observed in a homozyg two alleles carried by an organism that are identical observable traits of an organism genetic constitution of an organism phenotype observed in an organism that is homozygous for the recessive allele Choose Choose Choose Choose
Use this information to answer questions 3 7 In the mystical Unicorn Dragon plant the red allele R is dominant to the blue allele r A plant that is homozygous recessive will be blue In the parental generation a pure breeding red plant was crossed with a pure breeding blue plant r r R P1 cross R
Biology
Principles of Inheritance & Variation (Genetics)
Use this information to answer questions 3 7 In the mystical Unicorn Dragon plant the red allele R is dominant to the blue allele r A plant that is homozygous recessive will be blue In the parental generation a pure breeding red plant was crossed with a pure breeding blue plant r r R P1 cross R
2 What kind of approach to phylogeny reconstruction is ribosomal RNA sequencing cladistic or phenetic Explain You may have to consult your lecture notes for the descriptions of these methods
Biology
Principles of Inheritance & Variation (Genetics)
2 What kind of approach to phylogeny reconstruction is ribosomal RNA sequencing cladistic or phenetic Explain You may have to consult your lecture notes for the descriptions of these methods
Phenotype ABC abc a BC Abc O 0 02 CM O 0 2 CM O2 CM Number of offspring for each phenotype Offspring Phenotype 353 ABC 352 20 CM C A 2 3 ab C AbC a Bc B Offspring 49 Based on the data provided what is the distance between genes C and A 46 97 98
Biology
Principles of Inheritance & Variation (Genetics)
Phenotype ABC abc a BC Abc O 0 02 CM O 0 2 CM O2 CM Number of offspring for each phenotype Offspring Phenotype 353 ABC 352 20 CM C A 2 3 ab C AbC a Bc B Offspring 49 Based on the data provided what is the distance between genes C and A 46 97 98
The area responsible for DNA s negative charge is the O Amino end O Carboxyl end O 5 end O 3 end
Biology
Principles of Inheritance & Variation (Genetics)
The area responsible for DNA s negative charge is the O Amino end O Carboxyl end O 5 end O 3 end
Besides the thymine and uracil base difference which DNA strand s sequence is complementary to the mRNA sequence used in translation select all that apply Anti Coding Strand Template Strand Coding Strand Non Template Strand
Biology
Principles of Inheritance & Variation (Genetics)
Besides the thymine and uracil base difference which DNA strand s sequence is complementary to the mRNA sequence used in translation select all that apply Anti Coding Strand Template Strand Coding Strand Non Template Strand
DNA base pairings follow Chargaff s rule therefore adenine pairs with pairs with Othymine guanine guanine uracil O guanine thymine Ouracil guanine and cytosine
Biology
Principles of Inheritance & Variation (Genetics)
DNA base pairings follow Chargaff s rule therefore adenine pairs with pairs with Othymine guanine guanine uracil O guanine thymine Ouracil guanine and cytosine
Half of the gametes produced by an organism with the genotype Aa will receive the A allele while half will receive the a allele This is a demonstration of O independent assortment O mutation random fertilization segregation
Biology
Principles of Inheritance & Variation (Genetics)
Half of the gametes produced by an organism with the genotype Aa will receive the A allele while half will receive the a allele This is a demonstration of O independent assortment O mutation random fertilization segregation
When homologous chromosomes fail to separate during meiosis occurs Your answer 1
Biology
Principles of Inheritance & Variation (Genetics)
When homologous chromosomes fail to separate during meiosis occurs Your answer 1
The offspring of two different true breeding plants that differ in only one characteristic is called Your answer is a portion of a DNA molecule that carries the information that helps to produce a particular trait of an organism A Your answer A n remains attached to the original chromosome at the centromere 1 point 1 point is the identical copy of a single chromosome that 1 point
Biology
Principles of Inheritance & Variation (Genetics)
The offspring of two different true breeding plants that differ in only one characteristic is called Your answer is a portion of a DNA molecule that carries the information that helps to produce a particular trait of an organism A Your answer A n remains attached to the original chromosome at the centromere 1 point 1 point is the identical copy of a single chromosome that 1 point