Female CC Part1.zip

In the 1990s, women were more visible in the legislature and the executive branch. Dozens ran for parliament in April 1992; they won 13 out of 270 seats. In 1996, a woman was selected as mayor of a Tehran district. By 1994, 30 percent of government employees were female. Rafsanjani appointed Shahla Habibi as an advisor on women's affairs. During his second term, he appointed a female deputy minister. Still, women could not serve as judges, divorce their husbands at will, or freely marry without the consent of a male guardian.

Female CC Part1.zip

In Homeric poetry, apart from family members there are other members of the household [oikos] who are described by many different Greek words, and carry out differing roles. We were interested in understanding what those words would have meant and how servitude was portrayed, in the context of ancient Greek song culture of the Iliad and Odyssey. We start our exploration with female slaves/servants.

What evidence can you find about the roles and tasks of females who are captives, slaves or servants, or non-family members of the household? How were they treated? How do their masters or mistresses speak to them or about them? Please share passages in the forum and join the discussion!

Here are a number of terms for female slaves or servants, taken both from the passages we quoted above, and based on other searches using online corpora. The glosses are summarized from definitions in LSJ and/or Autenrieth, on Perseus, or from Montanari, Franco The Brill Dictionary of Ancient Greek. Notes are our own.

The elite female athlete is not immune to these problems with up to 37% reporting HMB(5). Indeed the physical and psychosocial effects of excessive menstrual blood loss are potentially greater due to the demands of training, travel and competition, as well as the high profile status of many female athletes. Indeed, female athletes with HMB are nearly four times more likely to cite negative impacts on training and performance as a result of heavy menstruation(6). Physical effects of HMB include fatigue, iron deficiency and anaemia(6,7) with up to 69% of female marathon runners reporting that HMB adversely effects their training and performance(5). There are other less quantifiable effects on player wellbeing and performance such as reduced confidence and concern about clothing(4). For a professional athlete the colour of the kit in which they play is often outside of their control. Anxiety about publicly visible blood loss is therefore not insignificant.

The impact of HMB on athlete wellbeing is frequently underreported. In a study by Bruinvels et al (2016) only 40% of marathon runners with HMB sought medical help for this problem(8). Clinicians must take a proactive approach to identify female athletes suffering with HMB and encourage a culture that ensures they feel comfortable raising this subject with medical and coaching staff. Numerous validated screening tools to identify and quantify HMB exist and their use should be considered(9).

The clinician must enquire about IMB, which raises the concern about a structural or histological abnormality. In women under 30 years of age IMB is much less likely to be pathological. Endometrial sampling should be considered if IMB persists for greater than 3 months in women aged 30 years or older, or with risk factors for endometrial pathology. Where appropriate, enquiry should be made regarding PCB, which is more indicative of cervical pathology. The status of cervical smears and a history of abnormal screening should be determined. In a female athlete this is especially important given their often nomadic lifestyle, which places them at greater risk to loss of follow-up and failure to receive cervical screening reminders, due to frequent changes of address. Lastly, it is important to ask about the current method of contraception and fertility wishes, as this will impact on the options for treatment.

Classically, oligomenorrhoea and amenorrhoea are menstrual disorders that are perhaps more commonly-associated with the female athlete. HMB is prevalent in this particular population and it is conceivable that its effects have an even greater physical and psychosocial impact on sportswomen than in the general female population. It is likely that HMB is under-reported in this group and the situation is further compounded by a lack of clear guidance on the initial assessment and management of this common condition. The aim of Part one of this article is to increase clinician awareness of HMB in female athletes. Part two will provide detail on appropriate investigation and treatment of these athletes in primary care.

If, however, we have a mating pair consisting of a male heterozygote black rabbit (Bb) and a female heterozygote black rabbit (Bb), what will the expected offspring proportions be? In this cross, there is a 25% chance that the offspring will be homozygous black (BB), 50% chance that the offspring will be heterozygous black (Bb), and 25% chance that the offspring will be homozygous chocolate (bb). See Table 3.

Both the Union and Confederate armies forbade the enlistment of women. Women soldiers of the Civil War therefore assumed masculine names, disguised themselves as men, and hid the fact they were female. Because they passed as men, it is impossible to know with any certainty how many women soldiers served in the Civil War. Estimates place as many as 250 women in the ranks of the Confederate army.(1) Writing in 1888, Mary Livermore of the U.S. Sanitary Commission remembered that:

Most of the articles provided few specific details about the individual woman's army career. For example, the obituary of Satronia Smith Hunt merely stated she enlisted in an Iowa regiment with her first husband. He died of battle wounds, but she apparently emerged from the war unscathed.(5) An 1896 story about Mary Stevens Jenkins, who died in 1881, tells an equally brief tale. She enlisted in a Pennsylvania regiment when still a schoolgirl, remained in the army two years, received several wounds, and was discharged without anyone ever realizing she was female.(6) The press seemed unconcerned about the women's actual military exploits. Rather, the fascination lay in the simple fact that they had been in the army.

• "Beginnings, Part 1"InformationSeriesThe Legend of KorraBookSpiritsEpisode19/52Original air dateOctober 12, 2013 (NYCC attendees)[1]

• October 18, 2013

Written byMichael Dante DiMartinoDirected byColin HeckAnimationStudio MirGuest starsD. B. Sweeney (Aang), James Garrett (Roku), Jim Meskimen (Kuruk - hunter #1), Jennifer Hale (Kyoshi), Justin Prentice (Jaya), Marcus Toji (Little Chou), Dee Bradley Baker (Mula), Bailey Gambertoglio (kind spirit), Nicholas Braico (kind spirit #2), Serena Williams (female sage)Production number119Episode guidePrevious"The Sting"Next"Beginnings, Part 2"TranscriptImage gallery (42)"Beginnings, Part 1" is the seventh episode of Book Two: Spirits of The Legend of Korra and the 19th episode of the overall series. It was aired on October 12, 2013, for attendees of the The Legend of Korra panel at the New York Comic-Con and premiered on Nickelodeon alongside "Beginnings, Part 2" on October 18, 2013.

Updated: 21 August 2000MENDELIAN GENETICS PROBLEMS AND ANSWERSPROBLEM 1.Hypothetically, brown color (B) in naked mole rats is dominant to white color (b). Suppose youran across a brown, male, naked mole rat in class and decided to find out if he was BB or Bb byusing a testcross. You'd mate him to a white (totally recessive) female, and examine the offspringproduced. Now, if only 2-3 offspring were born and they were all brown, you'd still be uncertainwhether he was BB or Bb (for instance, even though the odds are 50:50 that you will produce aboy or girl, there are plenty of people that produce 4-5 girls and never a boy and vice versa). But,if the mole rats produce 50 offspring and all are brown, then it is likely that no hidden alleles arepresent and that the male is BB. But, what if white offspring are produced? You'd know that thebrown parent had a hidden little "b" allele. So, what you need to do is perform a testcross on thisbrown, male, heterozygous, naked mole rat. What are the expected genotypic and phenotypicratios of such a cross?Bb (heterozygous male) x bb (testcross female) bbBBbBb bbbbbIf the brown male had been BB, then all offspring would have been Bb and all brown. However,if the male is Bb as above and you perform a testcross, 50% of all offspring should have the bbgenotype and a white phenotype. A testcross to a heterozygous individual should always yieldabout a 1:1 ratio of the dominant to recessive phenotype. So, boththe genotypic andphenotypic ratios here are 50:50. PROBLEM 2.What if you bred some snap dragons and crossed a homozygous red plant (RR) with ahomozygous white plant (rr)? In botony, "true breeding" means homozygous. In this case,100% of the F1 individuals would be pink! This is an example of "incomplete dominance," whereboth alleles contribute to the outcome. In some cases of incomplete dominance, both allelesmight contribute equally so one allele would produce red pigment and the other white; thus, apink plant appears. In another case, one allele may be non-functional. Although in many casesonly a single allele is needed, perhaps in this case only one-half the amount of needed pigment isproduced and so pink is due the low amount of red pigment in the petals. Who knows. Anyway,use a Punnett's square and set up a cross between a homozygous red plantand a homozygouswhite plant. Then, take the resulting offspring and cross these among themselves as well (i.e. F1x F1). Then, determine the phenotypic and genotypic ratios.Cross #1RR x rr rrRRrRr RRrRrYou'll note that 100% of all offspring are Rr, which is the genotype. Since the genotype for alloffspring are the same, and the Rr genotype encodes a pink color, then100% of the phenotypes will be pink (NOT red - remember that this is incomplete dominance). Nowthen, you need to perform a secondcross between the offspring. Since all offspring are Rr, then the cross will be Rr x Rr.Cross #2Rr x Rr RrRRRRr rRrrrYou'll note here that all offspring are not pink. Your genotypic ratio is25% (RR), 50% (Rr), and25% (rr). The phenotypic ratio is also the same in this case, with 25%red (RR), 50% pink (Rr),and 25% white (rr).PROBLEM 3.You know that the possession of claws (WW or Ww) is dominant to lack of claws (ww). Youalso know that the presence of smelly feet (FF or Ff) is dominant to non-smelly feet (ff). Youcross a male who is clawed and has smelly feet with a female who is clawed and has non-smellyfeet. All 18 offspring produced have smelly feet, and 14 have claws and 4 are un-clawed. Whatare the genotypes of the parents?Answer: Start with what you know early in the story: Dad isclawed, so he has at least one big W. You don't know whether his second allele is big W or littlew at this point. He also has smelly feet, so again you know he has onebig F but you cannot decipherthe second allele at this time. Mom is clawed so she has at least one big W, but the other alleleremains unknown. She has non-smelly feet, so she has the recessive characters and can only be"ff." So, based on the above, we know this much: Dad is (W ? F ?) and Mom is (W ? f f). OK, lets look at the offspring. All children had smelly feet. If Dad had a hidden little f, then itwould match up with Mom's little f's and about about one-half of the children would have endedup with non-smelly feet (ff). That didn't happen, so Dad must be FF (homozygous dominant). Now then, look at any recessive individuals that may be un-clawed. There are four, and all mustbe ww. Each child got a little w from Dad and the other little w from Mom. So, both parentsmust be heterozygous (Ww). Note that just like the monohybrid crosses, how important therecessive offspring are in these types of problems. You automatically know that each parenthad that hidden recessive allele based solely on the phenotype of the offspring. So, you figuredout the problem without any Punnett squares and the parents are as follows:Dad is "WwFF" and Mom is "Wwff"PROBLEM 4.You have an individual who is totally heterozygous for 2 genes that are not linked (i.e., not on thesame chromosome). One gene is for ear size (AA or Aa being big ears whereas aa is for smallears) and the other gene is for buggy eyes (BB and Bb for buggy eyes whereas bb representsnormal eyes). If you testcross this individual, what are the resulting genotypes andphenotypes?Answer: Remember that a testcross represents a cross with atotally recessive individual. These types of crosses are useful in weeding out hidden recessivealleles from your unknown. Remember the information on recessives if you don't rememberanything else. By knowing the recessive, you automatically know both the phenotype andgenotype. In the monohybrid cross, a testcross of a heterozygous individual resulted in a 1:1ratio. With the dihybrid cross, you should expect a 1:1:1:1 ratio!AaBb x aabbababababABAaBbAaBbAaBbAaBbAbAabbAabbAabbAabbaBaaBbaaBbaaBbaaBbabaabbaabbaabbaabbThus, you get the following...PERCENTAGESGENOTYPEPHENOTYPE25%AaBbBig ears, buggy eyes25%AabbBig ears, normal eyes25%aaBbSmall ears, buggy eyes25%aabbSmall ears, normal eyesPROBLEM 5. Now then, after you've completed the problem above, lets ignore the Punnett's square and simplylook at the 4 types of offspring from the above cross. What if the actual ratios in your testcrosswere not 1:1:1:1, but were as follows. What would this represent?PERCENTAGESGENOTYPEPHENOTYPE48%AaBbBig ears, buggy eyes2%AabbBig ears, normal eyes2%aaBbSmall ears, buggy eyes48%aabbSmall ears, normal eyesAnswer: Whenever you know that you have a totallyheterozygous individual, and you get this type of lopsided percentage during the testcross, youhave discovered that the A and B genes are linked (i.e. they occur on the same chromosome). Thus, they are NOT assorting independently as Mendel states in his second law. If they were, youwould get the 1:1:1:1 ratios. The genotypes and phenotypes with the small percentages (Aabband aaBb) represent outcomes that were produced due to "crossing over" (during Meiosis I,some homologous chromosomes broke between the 2 genes and DNA was exchanged). Becausethe percentage of these oddball recombinants was low, then it is likely that the genes are fairlynear one another. If the percentages of these middle two combinations were 10-12% each, thenthe distance between the genes would be greater. In this case, "A" and"B" are on the samechromosome whereas "a" and "b" occur on the other chromosome (except forthe ones that justcrossed over).PROBLEM 6.The following is a genetic linkage problem involving 4 genes. You want to determine which ofthe genes are linked, and which occur on separate chromosomes. You cross two true breeding(i.e., remember that this means that they are homozygous) plants that have the followingcharacteristics:PLANT 1PLANT 2Red flowersWhite flowersSpiny seedsSmooth seedsLong pollen grainsShort pollen grainsLate bloomingEarly bloomingFollowing the above cross, all of the offspring have red flowers, spiny seeds, long pollen grains,and early blooming (meaning, that these traits are dominant). You then testcross the F1generation, which you should realize by now are totally heterozygous individuals, and obtain theratios below. What's going on?49% red-spiny25% red-long25% red-early25% long-early1% red-smooth25% red-short25% red-late25% long-late1% white-spiny25% white-long25% white-early25% short-early49% white-smooth25% white-short25% white-late25% short-lateAnswer: A little more difficult, but still something you should beable to figure out. Obviously from the above, the red/white flowers and the spiny/smooth seedtraits are not assorting independently. If they were, we would see the1:1:1:1 ratios (25%:25%:25%:25%) representedfor the other sets of genes. Therefore, the flower color gene and seed texture are linked. Becauseof the high percentage of red-spiny and white-smooth, the allele for red flowers and the allele forspiny seeds are on the same homologue (except for 2% of the offspring, which are a result of thecrossover). Conversely, the allele for white petal color and the allele for smooth seeds are on thesame chromosome (again, except for the 2% of the offspring that are a result of crossing over). Since all of the other crosses are 1:1:1:1, then all other genes are on chromosomes separate fromthe first 2. Therefore, 3 separate chromosomes are involved.PROBLEM 7. The following is a genetic linkage problem also involving 4 genes. You want to determine whichof the genes are linked, which occur on separate chromosomes, and the distances between thelinked genes. You cross 2 true breeding (i.e. homozygous) plants that have the following"unusual" characteristics:PLANT 1PLANT 2Red flowersWhite flowersLong pollen grainsShort pollen grainsDumb backtalkSmart backtalkMean dispositionNice dispositionAll of the offspring have red flowers, long pollen grains, give smart backtalk, and have a nicedisposition (meaning, that these traits are dominant). You then testcross the F1 generation, andobtain the ratios below. How many chromosomes are involved in the linkages, and what are thepositions of the linked genes relative to one another?45% red-long25% red-dumb25% long-dumb48% red-mean43% long-mean5% red-short25% red-smart25% long smart2% red-nice7% long-nice5% white-long25% white-dumb25% short-dumb2% white-mean7% short-mean45% white-short25% white-smart25% short-smart48% white-nice43% short-niceAnswer: As you can see from the above, some characteristicsbetween genes do not assort in the 1:1:1:1 fashion. Therefore, they are linked. In the firstcolumn, one can see that red/white and long/short are on the same chromosome and are 10 (5 +5) units apart (see below). Also, red/white and mean/nice in the third column are linked and are 4(2 + 2) units apart (see below). Since mean/nice and short/long are on the same chromosome asred/white, they too are linked as can be seen in column five and are 14 (7 + 7) units apart (seebelow). The gene for smart/dumb must exist on a second, separate chromosome by itself.The arrangement below is the only one possibleCHROMOSOME: ________ mean/nice ________ red/white ___________________long/short ________(mean/nice is separated from red/white by 4 linkage units)(red/white is separated from long/short by 10 linkage units)(mean/nice is separated from long/short by 14 linkage units)PROBLEM 8. In the ABO blood system in human beings, alleles A and B are codominant and both are dominantto the O allele. In a paternity dispute, a type AB woman claimed that one of four men was thefather of her type A child (the child would be type A with a genotype of either be AA or AO). Which of the following men could be the father of the child on the basis of the evidence given?The Type A father? Answer: In this case, a type A personwould have one of the following genotypes: AA or AO. A man with either of these genotypescould be the father as the mother would donate the A allele to the child and either an A allele fromthe father or an O allele from the father would produce a child with Type A blood.The Type B father? Answer: In this case a type B fatherwould have either the genotype BB or BO. A man with the genotype BO could be the father asthe mother would donate the A allele to the child and an O allele from the father would produce achild with Type A blood.The Type O father? Answer: In this case a type O personwould have the genotype OO. A man with this genotype could be the father as the mother woulddonate the A allele to the child and an O