Genetic crosses Lab

[pic] |[pic] |1.3.5 Practice: DNA and Heredity

What percentage of Couple 3’s male and female offspring will be color blind? The percentage of male offspring that will be colorblind is 22%. The percentage of female offspring that will be colorblind is 19%. Part D Review the results you obtained for the female offspring of the three couples. Based on your results for the female offspring, predict whether color blindness is a dominant or recessive trait. Explain your reasoning.

In meiosis the pairs of chromosomes (that code for possible outcomes of characteristics) temporarily join and exchange information (crossing over) creating different combinations of gene types (alleles). For example; a pair of chromosomes could be a dominant allele (gene type) and a recessive allele which might code for brown hair. After crossing over it might be recessive which could be blond hair.

F1 Cross Offspring is F2 A a A AA Aa A Aa aa This Punnet Square represents the F1 offspring breeding with each other to create more offspring. This second set of offspring is the F2 generation. If both parents are heterozygous dominant, then the offspring expected would be: 50% heterozygous dominant, 25% homozygous dominant and 25% homozygous recessive.

Week 5 – Homework – Answer Key Due Feb. 23, 2013 A total of 20 points are possible for this homework 1. A black guinea pig is crossed with an albino guinea pig, producing 12 black offspring. When the very same albino is crossed with another black guinea pig, 7 black and 5 albinos

Your Journal 1.1.5 Journal: Can Science Answer It? Journal Psychology (S2958282) Faith Eke Points possible: 30 Date: ____________ Answer the questions below. When you are finished, turn in your assignment for grading. Use complete sentences and answer each part of the assignment. 1. Think of a time in your life when someone you know was having difficulty. Using that example, choose

Name: Date: 03.05.13 Student Exploration: Hardy-Weinberg Equilibrium Vocabulary: allele, genotype, Hardy-Weinberg equation, Hardy-Weinberg principle, heterozygous, homozygous, Punnett square Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Suppose the feather color of a bird is controlled by two alleles, D and d. The D allele results in dark feathers, while the d allele results in lighter feathers.

b. What forms the “steps of the ladder”? c. What kind of bonds hold the nitrogen bases together? d. What is meant by anti-parallel strands? 12. What is Chargaff’s rule (what binds with what?) 13. How does DNA fold into a chromosome? (the 6 steps) 14. What is a nucleosome? 15. What are histones? 16. What is the difference between chromatin and chromosomes? When do cells use these different forms? REPLICATION, TRANSCRIPTION, and TRANSLATION 1. What is the difference between REPLICATION, TRANSCRIPTION, and TRANSLATION? d. allele/trait? e. genetics/heredity? f. Genotype/phenotype? ---If given traits and parents, be able to use a Punnett square or patterns to predict the probability of offspring for a given cross and express it as a fraction, percent, or ratio.---

a. What was the phenotypic frequency from Step 1? (2 points) 38P/33Y = 1.15:1 b. What was the phenotypic frequency from Step 2? (2 points) 49P/11Y = 4.45:1 c. What was the phenotypic frequency from Step 3? (2 points) 5PW/16PS =0.28:1 2YW/4YS=0.5:1 12. Recall from the background information that purple corn kernels are dominant and yellow kernels are recessive. The second ear of corn was the result of crossing two heterozygous ears of male purple corn (Pp x Pp). This is represented by the Punnett square below. Complete the Punnett square by writing the correct letters that correspond to each number indicated in the table. (4 points)

Part B- Dihybrid Cross (F2 progeny) #purple starchy (smooth)|#purple sweet (wrinkled) | #yellow starchy (smooth)|#yellow sweet(wrinkled) 182 67 115 43 Genotypes of purple starchy | of purple sweet | of yellow starchy | of yellow sweet PPSs, PpSS, PpSs,PPSs PPSs, Ppss, PpSS, PPss PpSs, ppSs, PpSS, ppSS PpSs, PPSs, Ppss, ppss Possible genotype of P1 | genotype of F1 | ratio of F2 phenotypes| ratio of F2 genotypes PPSS PpSs 9:3:3:1 1:2:2:1 This table helps show all the possible genotypes from one set of parents. The table shows that the genotypes purple and starchy are dominant, and the genotypes yellow and sweet are recessive.(stallsmith)

Genetics 382 Lab Report 1 Robert Kamil (Student #11) TA: Camille English, Section 4 Rutgers University, Spring 2017 Laboratory 4- Recombination Testing the Mode of Inheritance of Shaven Bristles in Drosophila Using FlyLab Objectives The motivation of this lab report is to use Mendel’s Laws of Inheritance to analyze and predict the genotypes and phenotypes of an offspring generation (F2) after knowing the genotypes and phenotypes of the parent generation (F1). The hypothesis for this experiment is that the mode of inheritance for the shaven bristle allele in flies is autosomal recessive in both male and female flies.

Introduction: • This lab had 2 exercises. Exercise 9.1 involved observing pictures of 60 F2 offspring and recording the phenotypes for 6 different traits. Exercise 9.2 required us to perform the “chi-square test” to determine whether the data we collected matches the standard Mendelian ratio.

10. You are given the following genetic map: L------15 mu-------M---------28mu---------N You are also provided with a heterozygous female, and a homozygous recessive male for a genetic cross. In this particular female, all the dominant alleles are on one chromosome, and the recessive counterparts are on the other homologous chromosome. Due to a chromosomal condition, in the female no recombination occurs between the M and N loci. Normal recombination occurs between the L and M loci. Diagram this cross, and show the genotypes and frequencies of all offspring expected from this cross.

2.) Out of the genotypes PP, Pp, pp, the resultant flower colors are (as described above in exercise 1) are purple (for PP,) purple or purplish-white (for Pp- likely purple as it is dominant, or a mixture of the colors,) or white (for the case of pp.) PP and pp, the purple and white flowers, are referred to as homozygous. In the case of PP this is homozygous dominant, and in the case of pp this is homozygous recessive. The case of Pp must be considered different, and is

Brown = B (dominate gene) Blue = b (recessive gene) Using the laws of Mendelian of inheritance using complete blending, If Brown is dominate but Betty has Blue eyes then it is assumed that her phenotype is bb added that her parents eye color was not mentioned. Note: that Alex’s mother was. If Alex has Brown eyes and his mother had Blue eyes then it is indicated that Alex will be Heterozygous Allele paired or Bb. The chart (Punnett square), or figure 1 below shows the possible offspring Phenotypes of their