Question

Match the following terms with the appropriate description below:

a. alleles b. autosomes c. dominant allele d. genotype e. heterozygous f. homozygote g. phenotype h. recessive allele i. sex chromosomes

1. ________________ genetic make-up
2. ________________ how genetic make-up is expressed
3. ________________ chromosomes that dictate most body characteristics
4. ________________ alternative forms of the same gene
5. ___________an individual bearing two alleles that are the same for a particular trait 6. ________________ an allele that is expressed, whether in single or double dose
7. ________________ an individual bearing two alleles that differ for a particular trait 8. ________________ an allele that must be present in double dose to be expressed

9. Diagram by Punnett square a cross between individuals with the following genotypes:(on a separate sheet)

a. Aa x Aa b. Bb x BB c. Cc x cc

10. Brown hair (B) is dominant to red hair (b). What is the phenotypic and genotypic

    ratio of a cross between a homozygous recessive parent to a heterozygous parent: genotypic ratio:___________________ phenotypic ratio:________________

11. A brown-eyed man whose mother was blue-eyed marries a brown-eyed woman whose father had blue eyes. Brown is dominant to blue. What are the chances that this couple will have a blue-eyed child? ___________________% chance

12. True/False = A type O child may have two type A parents

13. True/False = A type O child may have one type AB parent

14. True/False = A type B child may have two type AB parents

15. True/False = A type AB child may have one type O parent

16. True/False = An individual heterozygous for sickle cell shows no ill effects.

17. True/False = An individual homozygous for sickle cell shows no ill effects.

18. Mrs. Doe and Mrs. Roe had babies at the same hospital at the same time. Mrs. Doe took home a girl and named her Nancy. Mrs. Roe received a boy and named him Larry. However, she was sure that she had a girl and brought a suit against the hospital. Blood tests showed that Mr. Roe was type O, Mrs. Roe was type AB, Mr. Doe and Mrs. Doe were both type B, Nancy was type A and Larry was type O. Had an exchange occurred?

19. In rabbits, black is due to a dominant gene (B) brown to its recessive allele (b). Short hair is due to the dominant gene (S) and long hair to its recessive allele (s). In a cross between a homozygous black long individual with a homozygous brown short one, what would be the nature of the F1 generation?

    (percentage of each):__________

4

20. Segregation and independent assortment (as defined by Mendel) occur during:_____

21. A couple who are both carriers for the gene for cystic fibrosis have two children who have cystic fibrosis. What is the probability (percentage) that their next child will have cystic fibrosis?__________________________________________

22. A couple who are both carriers for the gene for cystic fibrosis have two children who have cystic fibrosis. What is the probability that their next child will be phenotypically normal?_______________________________________________

    Albinism (lack of skin pigmentation) is caused by a recessive autosomal allele. A

man and woman, both normally pigmented, have an albino child together.

23. For this trait, what is the genotype of the albino child?_________________________

24. The couple decides to have a second child. What is the probability that this child will be albino?_______________________________________________

25. The mother is now pregnant for a third time and her doctor tells her she is carrying fraternal twins. What is the probability that both children will have normal pigmentation?___________________________________________________

26. A 9:3:3:1 phenotypic ratio is characteristic of what type of cross?________________

27. A 1:2:1 phenotypic ratio in the F2 generation of a monohybrid cross is a sign of?__________________________________________________________

28. In a dihybrid cross, the expected proportion of offspring showing both recessive traits is:______________________________________________________

29. In snapdragons, heterozygotes have pink flowers, whereas homozygotes have either red or white flowers. When plants with red flowers are crossed with plants with white flowers, what proportion of the offspring will have pink flowers?___________________

30. Every somatic cell contains _________pairs of chromosomes (in humans)

Mendelian Inheritance in Humans Monohybrid Crosses in Humans Several human traits may be used to demonstrate Mendel's law of segregation, The traits you will study are controlled by a single gene that can occur in two forms (alleles). One allele is dominant, the other is recessive. A person's phenotype is determined by the combination of alleles present. An individual can be homozygous dominant, heterozygous, or homozygous recessive. Objectives Describe the genetic basis for several human traits evident in the classroom population. -Determine the extent to which individuals are different from one another based on the presence or absence or selected genetic traits. Procedure For each of the following 10 traits, determine your phenotype and enter it in the table provided. Based on your phenotype, record your possible gentotypes for each of the traits. Remember that if you have a recessive characteristic you must have both recessive alleles, but if you have a dominant characteristic you may be either homozygous dominant or heterozygous. If you have a dominant phenotype, you may have no way of knowing whether you carry a recessive allele. In this case, use a dash (-) to represent the unknown second gene, Compile the data for your section. Calculate the percentage for each trait, 1. Dimpled chin. A cleft in the chin is a dominant trait. (D, allele for dimpling, d, allele for absence of a dimple) 2. Free ear lobe. In most people the ear lobes hang free (dominant allele, E), but in a person with two recessive alleles (e), the ear lobes are attached directly to the head. Use a mirror or the opinion of your classmates to determine your phenotype. 3. Widow's peak. The action of the dominant allele (W) results I a hairline that has a distinct point, known as a widow's peak, in the center of the forehead. The recessive (W) produces a continuous hairline, Omit this tabulation if a gene for baldness has an effect on your hairline 4. Interlocking fingers. When the fingers are interlocked, some people will almost invariably place the left thumb on top of the right (dominant allele F), whereas others will place the right over the left (recessive allele ). 5. Bent little finger. A dominant allele B causes the last joint of the little finger to bend inward toward the fourth finger (b is the recessive allele for a straight finger). Lay both hands flat on the table, relax your muscles, and note whether you have a bent or a straight little finger

Answer 1

Answers PAGE NO DATE 4) 7 enotype a Alleles bills) f. homozygole 6) c. e. Heteropgaur 8) b. Recessive allele. 2) g. phenolype 3) b. autosomes 5 c. dominant allele Ans-g Diagram by Punnelt square FITAA () Cc x cc 7 a Aa x Aa 6) Bb x BB AAA 7 fl je Aaaa (Bb/Bb aa 22 BB L /C/Cc7 cclcc And to let Brawn hair (B) - dominant NI Red hair (b) - Recessive Cross u between Homory gaus Recessive e (66) он n ggaur (Bb) Bb Bb Genotypic Ratio ! - 2207 02 C Bb ) : 2 (bb) bbbb Phenolypie Ratio: 2 Brown hais) : 2 cled Hall Ans-11) Brown eye - dominant Brown CB Blue – eye - Recessive (6 let cross Brown eye man whose mother was blue-eye therefore he must be Heterozyg to pail) & Boond eye female (whose to father had blue -eye re she must be hetawaygeus to trait Pin B6 X Bb a BB (4) Bb B62) 66 (14) chances of Blue-eye child = 25% (14)

PAGE DUATE Ans -12 ) – Type o True True may have to Type o (i) child may have two parents type A Type O crij child ald may have may have one one parents type AB ( IA IB) - False Type B Cai) child may have two type AB (II) parents - True Type B - thi or IB IB) A Tape AB child may have on type o cii) porents - False An individual Heterouygaus for Sie CHbA HOB) shaus noill effect - Tou An Individual homozygaur for sicked cell CHBA HbA) Shows no ill effdet - False cougaus for Sickel froit Ano-18) Given : Me Doe - Type B A Mis Doe- Type B Mr-Roe: - Type 0 (ii) Mrs- Roe - Type AB (1A 1B) (B & or I & IB) oor 1 1 B) Nancy Larry - - Type A Type o & possible Blood groups of childs of Mr & Mrs Roe - Pt ii X AIB FI JAI , Ai, Bi, Ibi jie either Blood Type A or Blood type B possible Blood paups of childs of Mr & Mr Due - P1 , I i Ox TB i Ise either Blood type 'B'or Blood type o es exchange has accuphalecarling Mona Barce IB IB i, ii. FL, IB IB Scu егошпо МК dan not had Blood type A